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How to make (patient id) forgein key in table of bill?


How does database indexing work?How can I prevent SQL injection in PHP?Add a column with a default value to an existing table in SQL ServerGet list of all tables in Oracle?Use of null values in related tables with foreign key constraintsHow do I UPDATE from a SELECT in SQL Server?mySql errno: 150 Create table statement insideDatabase design for patients and diseasesDoctors, Patients and Contact information for bothHow to join sql tables correctly to avoid data repetition?






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2















I try to create three tables by using website suport compiler any code but I have a problem in the table of the bill.
When I run it I get to error show me it is near in foreign key



enter image description here



These are codes of three tables



CREATE TABLE patient (
Patient Id (5) Primary key,
Name Varchar (20) Not null ,
Age Int Not null ,
Weight Int Not null ,
Gender Varchar (10) Not null,
Address Varchar (50) Not null ,
Disease Varchar (20) Not null
);

CREATE TABLE doctors (
DoctorId Varchar (5) Primary key,
Doctorname Varchar (15) Not null,
Dept Varchar (15) Not null
);


CREATE TABLE bill (
Bill_no Varchar (50) Primary key,
Patient_Id Varchar (5) Foreign key,,
doctor_charge Int Not null,
patient_type Varchar (10) null,
no_of_days Int null,
lab_charge Int null,
bill Int Not null
);









share|improve this question
























  • Please accept my answer if it helped you. Thanks!

    – Josh Katz
    Mar 22 at 16:46

















2















I try to create three tables by using website suport compiler any code but I have a problem in the table of the bill.
When I run it I get to error show me it is near in foreign key



enter image description here



These are codes of three tables



CREATE TABLE patient (
Patient Id (5) Primary key,
Name Varchar (20) Not null ,
Age Int Not null ,
Weight Int Not null ,
Gender Varchar (10) Not null,
Address Varchar (50) Not null ,
Disease Varchar (20) Not null
);

CREATE TABLE doctors (
DoctorId Varchar (5) Primary key,
Doctorname Varchar (15) Not null,
Dept Varchar (15) Not null
);


CREATE TABLE bill (
Bill_no Varchar (50) Primary key,
Patient_Id Varchar (5) Foreign key,,
doctor_charge Int Not null,
patient_type Varchar (10) null,
no_of_days Int null,
lab_charge Int null,
bill Int Not null
);









share|improve this question
























  • Please accept my answer if it helped you. Thanks!

    – Josh Katz
    Mar 22 at 16:46













2












2








2


1






I try to create three tables by using website suport compiler any code but I have a problem in the table of the bill.
When I run it I get to error show me it is near in foreign key



enter image description here



These are codes of three tables



CREATE TABLE patient (
Patient Id (5) Primary key,
Name Varchar (20) Not null ,
Age Int Not null ,
Weight Int Not null ,
Gender Varchar (10) Not null,
Address Varchar (50) Not null ,
Disease Varchar (20) Not null
);

CREATE TABLE doctors (
DoctorId Varchar (5) Primary key,
Doctorname Varchar (15) Not null,
Dept Varchar (15) Not null
);


CREATE TABLE bill (
Bill_no Varchar (50) Primary key,
Patient_Id Varchar (5) Foreign key,,
doctor_charge Int Not null,
patient_type Varchar (10) null,
no_of_days Int null,
lab_charge Int null,
bill Int Not null
);









share|improve this question
















I try to create three tables by using website suport compiler any code but I have a problem in the table of the bill.
When I run it I get to error show me it is near in foreign key



enter image description here



These are codes of three tables



CREATE TABLE patient (
Patient Id (5) Primary key,
Name Varchar (20) Not null ,
Age Int Not null ,
Weight Int Not null ,
Gender Varchar (10) Not null,
Address Varchar (50) Not null ,
Disease Varchar (20) Not null
);

CREATE TABLE doctors (
DoctorId Varchar (5) Primary key,
Doctorname Varchar (15) Not null,
Dept Varchar (15) Not null
);


CREATE TABLE bill (
Bill_no Varchar (50) Primary key,
Patient_Id Varchar (5) Foreign key,,
doctor_charge Int Not null,
patient_type Varchar (10) null,
no_of_days Int null,
lab_charge Int null,
bill Int Not null
);






sql oracle






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Mar 22 at 18:23









DxTx

1,7921822




1,7921822










asked Mar 22 at 16:34









Mohamed AbdallaMohamed Abdalla

185




185












  • Please accept my answer if it helped you. Thanks!

    – Josh Katz
    Mar 22 at 16:46

















  • Please accept my answer if it helped you. Thanks!

    – Josh Katz
    Mar 22 at 16:46
















Please accept my answer if it helped you. Thanks!

– Josh Katz
Mar 22 at 16:46





Please accept my answer if it helped you. Thanks!

– Josh Katz
Mar 22 at 16:46












4 Answers
4






active

oldest

votes


















0














Patient Table



 CREATE TABLE patient
(
patient_id VARCHAR (5) PRIMARY KEY,
name VARCHAR (20) NOT NULL,
age INT NOT NULL,
weight INT NOT NULL,
gender VARCHAR (10) NOT NULL,
address VARCHAR (50) NOT NULL,
disease VARCHAR (20) NOT NULL
);


Errors



  • No data type has been assigned in Patient id column (Patient Id (5)
    Primary key
    )

  • Patient id column name contains spaces. You need to
    enclose the column name in double quotes or replace space with something else
    (ex: _). It's not recommended to use spaces.

A column name with space



CREATE TABLE tablename ("column name" datatype); 


Doctors Table



CREATE TABLE doctors
(
doctorid VARCHAR (5) PRIMARY KEY,
doctorname VARCHAR (15) NOT NULL,
dept VARCHAR (15) NOT NULL
);


Bill Table



CREATE TABLE bill
(
bill_no VARCHAR (50) PRIMARY KEY,
patient_id VARCHAR (5),
doctor_charge INT NOT NULL,
patient_type VARCHAR (10) NULL,
no_of_days INT NULL,
lab_charge INT NULL,
bill INT NOT NULL,
FOREIGN KEY (patient_id) REFERENCES patient(patient_id)
);


Errors



  • The way you have assigned foreign key is wrong. Please refer this and this article for more information. (Patient_Id Varchar (5) Foreign key,,)

  • There are two commans in the Patient_Id column (Patient_Id Varchar (5) Foreign key,,)





share|improve this answer

























  • How to write code as you . CREATE TABLE bill ( Bill_no Varchar (50) Primary key, Patient_Id Varchar (5), doctor_charge Int Not null, patient_type Varchar (10) null, no_of_days Int null, lab_charge Int null, bill Int Not null, CONSTRAINT fk_patient_id FOREIGN KEY (Patient_Id) REFERENCES patient(Patient_Id) );

    – Mohamed Abdalla
    Mar 22 at 17:16











  • What do you mean..? Are you asking about the formatting or something else..? BTW, you can check these code using this website.

    – DxTx
    Mar 22 at 17:38











  • You commit at my post but I want to write code as you

    – Mohamed Abdalla
    Mar 22 at 17:47











  • You're talking about formatting. :) That's easy actually. Check these articles: link1, link2, link3, link4. If you want to format your SQL query, you can use this site to do that.

    – DxTx
    Mar 22 at 17:58



















0














You have to provide reference of the table for which you want to use the reference key.
For example, you have table Persons which has Primary key PersonID, in that case if you want to use that as foreign key in another table, lets say Orders.
In Oracle



CREATE TABLE Orders (
OrderID numeric(10) not null,
OrderNumber numeric(10) not null,
PersonID numeric(10) not null,
CONSTRAINT fk_person_id
FOREIGN KEY (PersonID )
REFERENCES Persons(PersonID )

Your Case :
CREATE TABLE bill
( Bill_no Varchar (50) Primary key,
Patient_Id Varchar (5),
doctor_charge Int Not null,
patient_type Varchar (10) null,
no_of_days Int null,
lab_charge Int null,
bill Int Not null,
CONSTRAINT fk_patient_id
FOREIGN KEY (Patient_Id)
REFERENCES patient(Patient_Id)
);





share|improve this answer























  • How to write code as you

    – Mohamed Abdalla
    Mar 22 at 17:14


















0














  1. Remove the 'Foreign Key' from the table creation script.


  2. Add this to your SQL script:



ALTER TABLE [Bill] WITH CHECK ADD CONSTRAINT [FK_Bill_Patient] FOREIGN KEY([Patient_Id])
REFERENCES [Patient] ([Patient_Id])
GO
ALTER TABLE [Bill] CHECK CONSTRAINT [FK_Bill_Patient]
GO





share|improve this answer
































    0














    The words FOREIGN KEY are only needed for introducing the name of the FK constraint. Since your other constraints are not named, you might as well skip that part and go straight to REFERENCES.



    If you specify a foreign key constraint as part of the column definition, you can omit the datatype to allow it to inherit from its parent at the time of creation, which I think is good practice as the types will automatically match.



    We use VARCHAR2 in Oracle, not VARCHAR.



    You don't need to specify NULL for columns that are allowed to be null.



    I am not sure a 5-character string is a good datatype for a unique ID. How will you generate the values? Normally an auto-incrementing sequence number simplifies this.



    create table doctors
    ( doctorid varchar2(5) primary key
    , doctorname varchar2(15) not null
    , dept varchar2(15) not null );

    create table patients
    ( patient_id varchar2(5) primary key
    , name varchar2(20) not null
    , age integer not null
    , weight integer not null
    , gender varchar2(10) not null
    , address varchar2(50) not null
    , disease varchar2(20) not null );

    create table bills
    ( bill_no varchar2(50) primary key
    , patient_id references patients -- Allow datatype to inherit from parent
    , patient_type varchar2(10)
    , no_of_days integer
    , lab_charge integer
    , bill integer not null );





    share|improve this answer

























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      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      0














      Patient Table



       CREATE TABLE patient
      (
      patient_id VARCHAR (5) PRIMARY KEY,
      name VARCHAR (20) NOT NULL,
      age INT NOT NULL,
      weight INT NOT NULL,
      gender VARCHAR (10) NOT NULL,
      address VARCHAR (50) NOT NULL,
      disease VARCHAR (20) NOT NULL
      );


      Errors



      • No data type has been assigned in Patient id column (Patient Id (5)
        Primary key
        )

      • Patient id column name contains spaces. You need to
        enclose the column name in double quotes or replace space with something else
        (ex: _). It's not recommended to use spaces.

      A column name with space



      CREATE TABLE tablename ("column name" datatype); 


      Doctors Table



      CREATE TABLE doctors
      (
      doctorid VARCHAR (5) PRIMARY KEY,
      doctorname VARCHAR (15) NOT NULL,
      dept VARCHAR (15) NOT NULL
      );


      Bill Table



      CREATE TABLE bill
      (
      bill_no VARCHAR (50) PRIMARY KEY,
      patient_id VARCHAR (5),
      doctor_charge INT NOT NULL,
      patient_type VARCHAR (10) NULL,
      no_of_days INT NULL,
      lab_charge INT NULL,
      bill INT NOT NULL,
      FOREIGN KEY (patient_id) REFERENCES patient(patient_id)
      );


      Errors



      • The way you have assigned foreign key is wrong. Please refer this and this article for more information. (Patient_Id Varchar (5) Foreign key,,)

      • There are two commans in the Patient_Id column (Patient_Id Varchar (5) Foreign key,,)





      share|improve this answer

























      • How to write code as you . CREATE TABLE bill ( Bill_no Varchar (50) Primary key, Patient_Id Varchar (5), doctor_charge Int Not null, patient_type Varchar (10) null, no_of_days Int null, lab_charge Int null, bill Int Not null, CONSTRAINT fk_patient_id FOREIGN KEY (Patient_Id) REFERENCES patient(Patient_Id) );

        – Mohamed Abdalla
        Mar 22 at 17:16











      • What do you mean..? Are you asking about the formatting or something else..? BTW, you can check these code using this website.

        – DxTx
        Mar 22 at 17:38











      • You commit at my post but I want to write code as you

        – Mohamed Abdalla
        Mar 22 at 17:47











      • You're talking about formatting. :) That's easy actually. Check these articles: link1, link2, link3, link4. If you want to format your SQL query, you can use this site to do that.

        – DxTx
        Mar 22 at 17:58
















      0














      Patient Table



       CREATE TABLE patient
      (
      patient_id VARCHAR (5) PRIMARY KEY,
      name VARCHAR (20) NOT NULL,
      age INT NOT NULL,
      weight INT NOT NULL,
      gender VARCHAR (10) NOT NULL,
      address VARCHAR (50) NOT NULL,
      disease VARCHAR (20) NOT NULL
      );


      Errors



      • No data type has been assigned in Patient id column (Patient Id (5)
        Primary key
        )

      • Patient id column name contains spaces. You need to
        enclose the column name in double quotes or replace space with something else
        (ex: _). It's not recommended to use spaces.

      A column name with space



      CREATE TABLE tablename ("column name" datatype); 


      Doctors Table



      CREATE TABLE doctors
      (
      doctorid VARCHAR (5) PRIMARY KEY,
      doctorname VARCHAR (15) NOT NULL,
      dept VARCHAR (15) NOT NULL
      );


      Bill Table



      CREATE TABLE bill
      (
      bill_no VARCHAR (50) PRIMARY KEY,
      patient_id VARCHAR (5),
      doctor_charge INT NOT NULL,
      patient_type VARCHAR (10) NULL,
      no_of_days INT NULL,
      lab_charge INT NULL,
      bill INT NOT NULL,
      FOREIGN KEY (patient_id) REFERENCES patient(patient_id)
      );


      Errors



      • The way you have assigned foreign key is wrong. Please refer this and this article for more information. (Patient_Id Varchar (5) Foreign key,,)

      • There are two commans in the Patient_Id column (Patient_Id Varchar (5) Foreign key,,)





      share|improve this answer

























      • How to write code as you . CREATE TABLE bill ( Bill_no Varchar (50) Primary key, Patient_Id Varchar (5), doctor_charge Int Not null, patient_type Varchar (10) null, no_of_days Int null, lab_charge Int null, bill Int Not null, CONSTRAINT fk_patient_id FOREIGN KEY (Patient_Id) REFERENCES patient(Patient_Id) );

        – Mohamed Abdalla
        Mar 22 at 17:16











      • What do you mean..? Are you asking about the formatting or something else..? BTW, you can check these code using this website.

        – DxTx
        Mar 22 at 17:38











      • You commit at my post but I want to write code as you

        – Mohamed Abdalla
        Mar 22 at 17:47











      • You're talking about formatting. :) That's easy actually. Check these articles: link1, link2, link3, link4. If you want to format your SQL query, you can use this site to do that.

        – DxTx
        Mar 22 at 17:58














      0












      0








      0







      Patient Table



       CREATE TABLE patient
      (
      patient_id VARCHAR (5) PRIMARY KEY,
      name VARCHAR (20) NOT NULL,
      age INT NOT NULL,
      weight INT NOT NULL,
      gender VARCHAR (10) NOT NULL,
      address VARCHAR (50) NOT NULL,
      disease VARCHAR (20) NOT NULL
      );


      Errors



      • No data type has been assigned in Patient id column (Patient Id (5)
        Primary key
        )

      • Patient id column name contains spaces. You need to
        enclose the column name in double quotes or replace space with something else
        (ex: _). It's not recommended to use spaces.

      A column name with space



      CREATE TABLE tablename ("column name" datatype); 


      Doctors Table



      CREATE TABLE doctors
      (
      doctorid VARCHAR (5) PRIMARY KEY,
      doctorname VARCHAR (15) NOT NULL,
      dept VARCHAR (15) NOT NULL
      );


      Bill Table



      CREATE TABLE bill
      (
      bill_no VARCHAR (50) PRIMARY KEY,
      patient_id VARCHAR (5),
      doctor_charge INT NOT NULL,
      patient_type VARCHAR (10) NULL,
      no_of_days INT NULL,
      lab_charge INT NULL,
      bill INT NOT NULL,
      FOREIGN KEY (patient_id) REFERENCES patient(patient_id)
      );


      Errors



      • The way you have assigned foreign key is wrong. Please refer this and this article for more information. (Patient_Id Varchar (5) Foreign key,,)

      • There are two commans in the Patient_Id column (Patient_Id Varchar (5) Foreign key,,)





      share|improve this answer















      Patient Table



       CREATE TABLE patient
      (
      patient_id VARCHAR (5) PRIMARY KEY,
      name VARCHAR (20) NOT NULL,
      age INT NOT NULL,
      weight INT NOT NULL,
      gender VARCHAR (10) NOT NULL,
      address VARCHAR (50) NOT NULL,
      disease VARCHAR (20) NOT NULL
      );


      Errors



      • No data type has been assigned in Patient id column (Patient Id (5)
        Primary key
        )

      • Patient id column name contains spaces. You need to
        enclose the column name in double quotes or replace space with something else
        (ex: _). It's not recommended to use spaces.

      A column name with space



      CREATE TABLE tablename ("column name" datatype); 


      Doctors Table



      CREATE TABLE doctors
      (
      doctorid VARCHAR (5) PRIMARY KEY,
      doctorname VARCHAR (15) NOT NULL,
      dept VARCHAR (15) NOT NULL
      );


      Bill Table



      CREATE TABLE bill
      (
      bill_no VARCHAR (50) PRIMARY KEY,
      patient_id VARCHAR (5),
      doctor_charge INT NOT NULL,
      patient_type VARCHAR (10) NULL,
      no_of_days INT NULL,
      lab_charge INT NULL,
      bill INT NOT NULL,
      FOREIGN KEY (patient_id) REFERENCES patient(patient_id)
      );


      Errors



      • The way you have assigned foreign key is wrong. Please refer this and this article for more information. (Patient_Id Varchar (5) Foreign key,,)

      • There are two commans in the Patient_Id column (Patient_Id Varchar (5) Foreign key,,)






      share|improve this answer














      share|improve this answer



      share|improve this answer








      edited Mar 22 at 18:01

























      answered Mar 22 at 17:01









      DxTxDxTx

      1,7921822




      1,7921822












      • How to write code as you . CREATE TABLE bill ( Bill_no Varchar (50) Primary key, Patient_Id Varchar (5), doctor_charge Int Not null, patient_type Varchar (10) null, no_of_days Int null, lab_charge Int null, bill Int Not null, CONSTRAINT fk_patient_id FOREIGN KEY (Patient_Id) REFERENCES patient(Patient_Id) );

        – Mohamed Abdalla
        Mar 22 at 17:16











      • What do you mean..? Are you asking about the formatting or something else..? BTW, you can check these code using this website.

        – DxTx
        Mar 22 at 17:38











      • You commit at my post but I want to write code as you

        – Mohamed Abdalla
        Mar 22 at 17:47











      • You're talking about formatting. :) That's easy actually. Check these articles: link1, link2, link3, link4. If you want to format your SQL query, you can use this site to do that.

        – DxTx
        Mar 22 at 17:58


















      • How to write code as you . CREATE TABLE bill ( Bill_no Varchar (50) Primary key, Patient_Id Varchar (5), doctor_charge Int Not null, patient_type Varchar (10) null, no_of_days Int null, lab_charge Int null, bill Int Not null, CONSTRAINT fk_patient_id FOREIGN KEY (Patient_Id) REFERENCES patient(Patient_Id) );

        – Mohamed Abdalla
        Mar 22 at 17:16











      • What do you mean..? Are you asking about the formatting or something else..? BTW, you can check these code using this website.

        – DxTx
        Mar 22 at 17:38











      • You commit at my post but I want to write code as you

        – Mohamed Abdalla
        Mar 22 at 17:47











      • You're talking about formatting. :) That's easy actually. Check these articles: link1, link2, link3, link4. If you want to format your SQL query, you can use this site to do that.

        – DxTx
        Mar 22 at 17:58

















      How to write code as you . CREATE TABLE bill ( Bill_no Varchar (50) Primary key, Patient_Id Varchar (5), doctor_charge Int Not null, patient_type Varchar (10) null, no_of_days Int null, lab_charge Int null, bill Int Not null, CONSTRAINT fk_patient_id FOREIGN KEY (Patient_Id) REFERENCES patient(Patient_Id) );

      – Mohamed Abdalla
      Mar 22 at 17:16





      How to write code as you . CREATE TABLE bill ( Bill_no Varchar (50) Primary key, Patient_Id Varchar (5), doctor_charge Int Not null, patient_type Varchar (10) null, no_of_days Int null, lab_charge Int null, bill Int Not null, CONSTRAINT fk_patient_id FOREIGN KEY (Patient_Id) REFERENCES patient(Patient_Id) );

      – Mohamed Abdalla
      Mar 22 at 17:16













      What do you mean..? Are you asking about the formatting or something else..? BTW, you can check these code using this website.

      – DxTx
      Mar 22 at 17:38





      What do you mean..? Are you asking about the formatting or something else..? BTW, you can check these code using this website.

      – DxTx
      Mar 22 at 17:38













      You commit at my post but I want to write code as you

      – Mohamed Abdalla
      Mar 22 at 17:47





      You commit at my post but I want to write code as you

      – Mohamed Abdalla
      Mar 22 at 17:47













      You're talking about formatting. :) That's easy actually. Check these articles: link1, link2, link3, link4. If you want to format your SQL query, you can use this site to do that.

      – DxTx
      Mar 22 at 17:58






      You're talking about formatting. :) That's easy actually. Check these articles: link1, link2, link3, link4. If you want to format your SQL query, you can use this site to do that.

      – DxTx
      Mar 22 at 17:58














      0














      You have to provide reference of the table for which you want to use the reference key.
      For example, you have table Persons which has Primary key PersonID, in that case if you want to use that as foreign key in another table, lets say Orders.
      In Oracle



      CREATE TABLE Orders (
      OrderID numeric(10) not null,
      OrderNumber numeric(10) not null,
      PersonID numeric(10) not null,
      CONSTRAINT fk_person_id
      FOREIGN KEY (PersonID )
      REFERENCES Persons(PersonID )

      Your Case :
      CREATE TABLE bill
      ( Bill_no Varchar (50) Primary key,
      Patient_Id Varchar (5),
      doctor_charge Int Not null,
      patient_type Varchar (10) null,
      no_of_days Int null,
      lab_charge Int null,
      bill Int Not null,
      CONSTRAINT fk_patient_id
      FOREIGN KEY (Patient_Id)
      REFERENCES patient(Patient_Id)
      );





      share|improve this answer























      • How to write code as you

        – Mohamed Abdalla
        Mar 22 at 17:14















      0














      You have to provide reference of the table for which you want to use the reference key.
      For example, you have table Persons which has Primary key PersonID, in that case if you want to use that as foreign key in another table, lets say Orders.
      In Oracle



      CREATE TABLE Orders (
      OrderID numeric(10) not null,
      OrderNumber numeric(10) not null,
      PersonID numeric(10) not null,
      CONSTRAINT fk_person_id
      FOREIGN KEY (PersonID )
      REFERENCES Persons(PersonID )

      Your Case :
      CREATE TABLE bill
      ( Bill_no Varchar (50) Primary key,
      Patient_Id Varchar (5),
      doctor_charge Int Not null,
      patient_type Varchar (10) null,
      no_of_days Int null,
      lab_charge Int null,
      bill Int Not null,
      CONSTRAINT fk_patient_id
      FOREIGN KEY (Patient_Id)
      REFERENCES patient(Patient_Id)
      );





      share|improve this answer























      • How to write code as you

        – Mohamed Abdalla
        Mar 22 at 17:14













      0












      0








      0







      You have to provide reference of the table for which you want to use the reference key.
      For example, you have table Persons which has Primary key PersonID, in that case if you want to use that as foreign key in another table, lets say Orders.
      In Oracle



      CREATE TABLE Orders (
      OrderID numeric(10) not null,
      OrderNumber numeric(10) not null,
      PersonID numeric(10) not null,
      CONSTRAINT fk_person_id
      FOREIGN KEY (PersonID )
      REFERENCES Persons(PersonID )

      Your Case :
      CREATE TABLE bill
      ( Bill_no Varchar (50) Primary key,
      Patient_Id Varchar (5),
      doctor_charge Int Not null,
      patient_type Varchar (10) null,
      no_of_days Int null,
      lab_charge Int null,
      bill Int Not null,
      CONSTRAINT fk_patient_id
      FOREIGN KEY (Patient_Id)
      REFERENCES patient(Patient_Id)
      );





      share|improve this answer













      You have to provide reference of the table for which you want to use the reference key.
      For example, you have table Persons which has Primary key PersonID, in that case if you want to use that as foreign key in another table, lets say Orders.
      In Oracle



      CREATE TABLE Orders (
      OrderID numeric(10) not null,
      OrderNumber numeric(10) not null,
      PersonID numeric(10) not null,
      CONSTRAINT fk_person_id
      FOREIGN KEY (PersonID )
      REFERENCES Persons(PersonID )

      Your Case :
      CREATE TABLE bill
      ( Bill_no Varchar (50) Primary key,
      Patient_Id Varchar (5),
      doctor_charge Int Not null,
      patient_type Varchar (10) null,
      no_of_days Int null,
      lab_charge Int null,
      bill Int Not null,
      CONSTRAINT fk_patient_id
      FOREIGN KEY (Patient_Id)
      REFERENCES patient(Patient_Id)
      );






      share|improve this answer












      share|improve this answer



      share|improve this answer










      answered Mar 22 at 16:55









      SK -SK -

      898




      898












      • How to write code as you

        – Mohamed Abdalla
        Mar 22 at 17:14

















      • How to write code as you

        – Mohamed Abdalla
        Mar 22 at 17:14
















      How to write code as you

      – Mohamed Abdalla
      Mar 22 at 17:14





      How to write code as you

      – Mohamed Abdalla
      Mar 22 at 17:14











      0














      1. Remove the 'Foreign Key' from the table creation script.


      2. Add this to your SQL script:



      ALTER TABLE [Bill] WITH CHECK ADD CONSTRAINT [FK_Bill_Patient] FOREIGN KEY([Patient_Id])
      REFERENCES [Patient] ([Patient_Id])
      GO
      ALTER TABLE [Bill] CHECK CONSTRAINT [FK_Bill_Patient]
      GO





      share|improve this answer





























        0














        1. Remove the 'Foreign Key' from the table creation script.


        2. Add this to your SQL script:



        ALTER TABLE [Bill] WITH CHECK ADD CONSTRAINT [FK_Bill_Patient] FOREIGN KEY([Patient_Id])
        REFERENCES [Patient] ([Patient_Id])
        GO
        ALTER TABLE [Bill] CHECK CONSTRAINT [FK_Bill_Patient]
        GO





        share|improve this answer



























          0












          0








          0







          1. Remove the 'Foreign Key' from the table creation script.


          2. Add this to your SQL script:



          ALTER TABLE [Bill] WITH CHECK ADD CONSTRAINT [FK_Bill_Patient] FOREIGN KEY([Patient_Id])
          REFERENCES [Patient] ([Patient_Id])
          GO
          ALTER TABLE [Bill] CHECK CONSTRAINT [FK_Bill_Patient]
          GO





          share|improve this answer















          1. Remove the 'Foreign Key' from the table creation script.


          2. Add this to your SQL script:



          ALTER TABLE [Bill] WITH CHECK ADD CONSTRAINT [FK_Bill_Patient] FOREIGN KEY([Patient_Id])
          REFERENCES [Patient] ([Patient_Id])
          GO
          ALTER TABLE [Bill] CHECK CONSTRAINT [FK_Bill_Patient]
          GO






          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited Mar 22 at 17:43









          DxTx

          1,7921822




          1,7921822










          answered Mar 22 at 16:46









          Josh KatzJosh Katz

          83119




          83119





















              0














              The words FOREIGN KEY are only needed for introducing the name of the FK constraint. Since your other constraints are not named, you might as well skip that part and go straight to REFERENCES.



              If you specify a foreign key constraint as part of the column definition, you can omit the datatype to allow it to inherit from its parent at the time of creation, which I think is good practice as the types will automatically match.



              We use VARCHAR2 in Oracle, not VARCHAR.



              You don't need to specify NULL for columns that are allowed to be null.



              I am not sure a 5-character string is a good datatype for a unique ID. How will you generate the values? Normally an auto-incrementing sequence number simplifies this.



              create table doctors
              ( doctorid varchar2(5) primary key
              , doctorname varchar2(15) not null
              , dept varchar2(15) not null );

              create table patients
              ( patient_id varchar2(5) primary key
              , name varchar2(20) not null
              , age integer not null
              , weight integer not null
              , gender varchar2(10) not null
              , address varchar2(50) not null
              , disease varchar2(20) not null );

              create table bills
              ( bill_no varchar2(50) primary key
              , patient_id references patients -- Allow datatype to inherit from parent
              , patient_type varchar2(10)
              , no_of_days integer
              , lab_charge integer
              , bill integer not null );





              share|improve this answer





























                0














                The words FOREIGN KEY are only needed for introducing the name of the FK constraint. Since your other constraints are not named, you might as well skip that part and go straight to REFERENCES.



                If you specify a foreign key constraint as part of the column definition, you can omit the datatype to allow it to inherit from its parent at the time of creation, which I think is good practice as the types will automatically match.



                We use VARCHAR2 in Oracle, not VARCHAR.



                You don't need to specify NULL for columns that are allowed to be null.



                I am not sure a 5-character string is a good datatype for a unique ID. How will you generate the values? Normally an auto-incrementing sequence number simplifies this.



                create table doctors
                ( doctorid varchar2(5) primary key
                , doctorname varchar2(15) not null
                , dept varchar2(15) not null );

                create table patients
                ( patient_id varchar2(5) primary key
                , name varchar2(20) not null
                , age integer not null
                , weight integer not null
                , gender varchar2(10) not null
                , address varchar2(50) not null
                , disease varchar2(20) not null );

                create table bills
                ( bill_no varchar2(50) primary key
                , patient_id references patients -- Allow datatype to inherit from parent
                , patient_type varchar2(10)
                , no_of_days integer
                , lab_charge integer
                , bill integer not null );





                share|improve this answer



























                  0












                  0








                  0







                  The words FOREIGN KEY are only needed for introducing the name of the FK constraint. Since your other constraints are not named, you might as well skip that part and go straight to REFERENCES.



                  If you specify a foreign key constraint as part of the column definition, you can omit the datatype to allow it to inherit from its parent at the time of creation, which I think is good practice as the types will automatically match.



                  We use VARCHAR2 in Oracle, not VARCHAR.



                  You don't need to specify NULL for columns that are allowed to be null.



                  I am not sure a 5-character string is a good datatype for a unique ID. How will you generate the values? Normally an auto-incrementing sequence number simplifies this.



                  create table doctors
                  ( doctorid varchar2(5) primary key
                  , doctorname varchar2(15) not null
                  , dept varchar2(15) not null );

                  create table patients
                  ( patient_id varchar2(5) primary key
                  , name varchar2(20) not null
                  , age integer not null
                  , weight integer not null
                  , gender varchar2(10) not null
                  , address varchar2(50) not null
                  , disease varchar2(20) not null );

                  create table bills
                  ( bill_no varchar2(50) primary key
                  , patient_id references patients -- Allow datatype to inherit from parent
                  , patient_type varchar2(10)
                  , no_of_days integer
                  , lab_charge integer
                  , bill integer not null );





                  share|improve this answer















                  The words FOREIGN KEY are only needed for introducing the name of the FK constraint. Since your other constraints are not named, you might as well skip that part and go straight to REFERENCES.



                  If you specify a foreign key constraint as part of the column definition, you can omit the datatype to allow it to inherit from its parent at the time of creation, which I think is good practice as the types will automatically match.



                  We use VARCHAR2 in Oracle, not VARCHAR.



                  You don't need to specify NULL for columns that are allowed to be null.



                  I am not sure a 5-character string is a good datatype for a unique ID. How will you generate the values? Normally an auto-incrementing sequence number simplifies this.



                  create table doctors
                  ( doctorid varchar2(5) primary key
                  , doctorname varchar2(15) not null
                  , dept varchar2(15) not null );

                  create table patients
                  ( patient_id varchar2(5) primary key
                  , name varchar2(20) not null
                  , age integer not null
                  , weight integer not null
                  , gender varchar2(10) not null
                  , address varchar2(50) not null
                  , disease varchar2(20) not null );

                  create table bills
                  ( bill_no varchar2(50) primary key
                  , patient_id references patients -- Allow datatype to inherit from parent
                  , patient_type varchar2(10)
                  , no_of_days integer
                  , lab_charge integer
                  , bill integer not null );






                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited Mar 23 at 9:42

























                  answered Mar 23 at 9:31









                  William RobertsonWilliam Robertson

                  8,57632233




                  8,57632233



























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