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change just the date portion of a sql record

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1

Using SQL (Azure Sql)
I have a number of records that contain datetime values such as:
3/16/19 9:25 AM
3/16/19 10:15 AM

I need these to all be todays date but retain their time portion.

I am a C# programmer so my SQL is a bit rusty.

So pseudo code would be something like (asuming requested_delivery_date is the field I need updated.

update set requested_delivery_date = requested_delivery_date.dateadd(7 days)

so anything after this that had a date of 3/20/19 9:30 am would now be 3/27/19 9:30 am

sql sql-server azure-sql-database

share|improve this question

edited Mar 28 at 7:42

Joe Ruder

asked Mar 28 at 7:16

Joe RuderJoe Ruder

1,00111 gold badge1111 silver badges3434 bronze badges

add a comment | 

1

Using SQL (Azure Sql)
I have a number of records that contain datetime values such as:
3/16/19 9:25 AM
3/16/19 10:15 AM

I need these to all be todays date but retain their time portion.

I am a C# programmer so my SQL is a bit rusty.

So pseudo code would be something like (asuming requested_delivery_date is the field I need updated.

update set requested_delivery_date = requested_delivery_date.dateadd(7 days)

so anything after this that had a date of 3/20/19 9:30 am would now be 3/27/19 9:30 am

sql sql-server azure-sql-database

share|improve this question

edited Mar 28 at 7:42

Joe Ruder

asked Mar 28 at 7:16

Joe RuderJoe Ruder

1,00111 gold badge1111 silver badges3434 bronze badges

add a comment | 

Using SQL (Azure Sql)
I have a number of records that contain datetime values such as:
3/16/19 9:25 AM
3/16/19 10:15 AM

I need these to all be todays date but retain their time portion.

I am a C# programmer so my SQL is a bit rusty.

So pseudo code would be something like (asuming requested_delivery_date is the field I need updated.

update set requested_delivery_date = requested_delivery_date.dateadd(7 days)

so anything after this that had a date of 3/20/19 9:30 am would now be 3/27/19 9:30 am

sql sql-server azure-sql-database

share|improve this question

edited Mar 28 at 7:42

Joe Ruder

asked Mar 28 at 7:16

Joe RuderJoe Ruder

1,00111 gold badge1111 silver badges3434 bronze badges

sql sql-server azure-sql-database

sql sql-server azure-sql-database

share|improve this question

edited Mar 28 at 7:42

Joe Ruder

asked Mar 28 at 7:16

Joe RuderJoe Ruder

1,00111 gold badge1111 silver badges3434 bronze badges

share|improve this question

edited Mar 28 at 7:42

Joe Ruder

asked Mar 28 at 7:16

Joe RuderJoe Ruder

1,00111 gold badge1111 silver badges3434 bronze badges

asked Mar 28 at 7:16

Joe RuderJoe Ruder

1,00111 gold badge1111 silver badges3434 bronze badges

asked Mar 28 at 7:16

Joe RuderJoe Ruder

1,00111 gold badge1111 silver badges3434 bronze badges

3 Answers
3

active

oldest

votes

1

It looks like a simple UPDATE statement that updates all rows of YourTable.

UPDATE YourTable
SET requested_delivery_date = 
 DATEADD(DAY, DATEDIFF(DAY, requested_delivery_date, GETDATE()), requested_delivery_date)
;

If you want to add a constant number of days, then the query is even simpler:

UPDATE YourTable
SET requested_delivery_date = 
 DATEADD(DAY, 7, requested_delivery_date)
;

share|improve this answer

answered Mar 28 at 8:05

Vladimir BaranovVladimir Baranov

24.1k55 gold badges3131 silver badges6767 bronze badges

add a comment | 

1

Use a combination of DateAdd and DateDiff:

DECLARE @Date As datetime = '2019-03-01T15:32:44'

SELECT @Date As Source,
 GETDATE() As Today,
 DATEADD(DAY, DATEDIFF(DAY, @Date, GETDATE()), @Date) As Result

Result:

Source Today Result
2019-03-01 15:32:44 2019-03-28 08:27:29 2019-03-28 15:32:44

To update the column in the table you simply need to write an update statement:

update TableName
set requested_delivery_date = DATEADD(DAY, DATEDIFF(DAY, requested_delivery_date , GETDATE()), requested_delivery_date)
-- add a where clause to only update some of the records

share|improve this answer

edited Mar 28 at 8:20

answered Mar 28 at 7:28

Zohar PeledZohar Peled

62.1k77 gold badges3939 silver badges8080 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  thank you....my question is how do I update the current records? I can hard code the date even, they all have the same date (3/16) just different times.
  
  – Joe Ruder
  Mar 28 at 7:32
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Something like: update set requested_delivery_date = requested_delivery_date.adddays(7) But with SQL code.
  
  – Joe Ruder
  Mar 28 at 7:35
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  The manipulation on the date stays the same, you only change the select to an update - I've edited my answer to reflect that.
  
  – Zohar Peled
  Mar 28 at 8:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thank you Zohar -- I had already accepted a answer, but I do appreciate your time.
  
  – Joe Ruder
  Mar 28 at 9:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Glad to help :-)
  
  – Zohar Peled
  Mar 28 at 10:23
  

  
  
  
  

add a comment | 

0

I tried this and successfully. This is my table.

startTime endTime
2019-03-28 15:55:10.690 2019-01-01 09:00:00.000

My SQL statement:

UPDATE [MyDatabase].[dbo].[deom1] SET [endTime]=DATEADD(DAY, DATEDIFF(DAY, [endTime], GETDATE()), [endTime])

And this the result:

startTime endTime
2019-03-28 15:55:10.690 2019-03-28 09:00:00.000

For more details, please see DATEADD (Transact-SQL).

Hope this helps.

share|improve this answer

edited Mar 28 at 8:17

answered Mar 28 at 8:05

Leon YueLeon Yue

2,31611 gold badge11 silver badge77 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  thank you Leon -- Vladimir's answer happened to be exactly what I was looking for. I appreciate your time and answer.
  
  – Joe Ruder
  Mar 28 at 8:22
  

  
  
  
  

add a comment | 

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1

It looks like a simple UPDATE statement that updates all rows of YourTable.

UPDATE YourTable
SET requested_delivery_date = 
 DATEADD(DAY, DATEDIFF(DAY, requested_delivery_date, GETDATE()), requested_delivery_date)
;

If you want to add a constant number of days, then the query is even simpler:

UPDATE YourTable
SET requested_delivery_date = 
 DATEADD(DAY, 7, requested_delivery_date)
;

share|improve this answer

answered Mar 28 at 8:05

Vladimir BaranovVladimir Baranov

24.1k55 gold badges3131 silver badges6767 bronze badges

add a comment | 

1

It looks like a simple UPDATE statement that updates all rows of YourTable.

UPDATE YourTable
SET requested_delivery_date = 
 DATEADD(DAY, DATEDIFF(DAY, requested_delivery_date, GETDATE()), requested_delivery_date)
;

If you want to add a constant number of days, then the query is even simpler:

UPDATE YourTable
SET requested_delivery_date = 
 DATEADD(DAY, 7, requested_delivery_date)
;

share|improve this answer

answered Mar 28 at 8:05

Vladimir BaranovVladimir Baranov

24.1k55 gold badges3131 silver badges6767 bronze badges

add a comment | 

It looks like a simple UPDATE statement that updates all rows of YourTable.

UPDATE YourTable
SET requested_delivery_date = 
 DATEADD(DAY, DATEDIFF(DAY, requested_delivery_date, GETDATE()), requested_delivery_date)
;

If you want to add a constant number of days, then the query is even simpler:

UPDATE YourTable
SET requested_delivery_date = 
 DATEADD(DAY, 7, requested_delivery_date)
;

share|improve this answer

answered Mar 28 at 8:05

Vladimir BaranovVladimir Baranov

24.1k55 gold badges3131 silver badges6767 bronze badges

UPDATE YourTable
SET requested_delivery_date = 
 DATEADD(DAY, DATEDIFF(DAY, requested_delivery_date, GETDATE()), requested_delivery_date)
;

UPDATE YourTable
SET requested_delivery_date = 
 DATEADD(DAY, 7, requested_delivery_date)
;

share|improve this answer

answered Mar 28 at 8:05

Vladimir BaranovVladimir Baranov

24.1k55 gold badges3131 silver badges6767 bronze badges

answered Mar 28 at 8:05

Vladimir BaranovVladimir Baranov

24.1k55 gold badges3131 silver badges6767 bronze badges

answered Mar 28 at 8:05

Vladimir BaranovVladimir Baranov

24.1k55 gold badges3131 silver badges6767 bronze badges

1

Use a combination of DateAdd and DateDiff:

DECLARE @Date As datetime = '2019-03-01T15:32:44'

SELECT @Date As Source,
 GETDATE() As Today,
 DATEADD(DAY, DATEDIFF(DAY, @Date, GETDATE()), @Date) As Result

Result:

Source Today Result
2019-03-01 15:32:44 2019-03-28 08:27:29 2019-03-28 15:32:44

To update the column in the table you simply need to write an update statement:

update TableName
set requested_delivery_date = DATEADD(DAY, DATEDIFF(DAY, requested_delivery_date , GETDATE()), requested_delivery_date)
-- add a where clause to only update some of the records

share|improve this answer

edited Mar 28 at 8:20

answered Mar 28 at 7:28

Zohar PeledZohar Peled

62.1k77 gold badges3939 silver badges8080 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  thank you....my question is how do I update the current records? I can hard code the date even, they all have the same date (3/16) just different times.
  
  – Joe Ruder
  Mar 28 at 7:32
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Something like: update set requested_delivery_date = requested_delivery_date.adddays(7) But with SQL code.
  
  – Joe Ruder
  Mar 28 at 7:35
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  The manipulation on the date stays the same, you only change the select to an update - I've edited my answer to reflect that.
  
  – Zohar Peled
  Mar 28 at 8:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thank you Zohar -- I had already accepted a answer, but I do appreciate your time.
  
  – Joe Ruder
  Mar 28 at 9:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Glad to help :-)
  
  – Zohar Peled
  Mar 28 at 10:23
  

  
  
  
  

add a comment | 

1

Use a combination of DateAdd and DateDiff:

DECLARE @Date As datetime = '2019-03-01T15:32:44'

SELECT @Date As Source,
 GETDATE() As Today,
 DATEADD(DAY, DATEDIFF(DAY, @Date, GETDATE()), @Date) As Result

Result:

Source Today Result
2019-03-01 15:32:44 2019-03-28 08:27:29 2019-03-28 15:32:44

To update the column in the table you simply need to write an update statement:

update TableName
set requested_delivery_date = DATEADD(DAY, DATEDIFF(DAY, requested_delivery_date , GETDATE()), requested_delivery_date)
-- add a where clause to only update some of the records

share|improve this answer

edited Mar 28 at 8:20

answered Mar 28 at 7:28

Zohar PeledZohar Peled

62.1k77 gold badges3939 silver badges8080 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  thank you....my question is how do I update the current records? I can hard code the date even, they all have the same date (3/16) just different times.
  
  – Joe Ruder
  Mar 28 at 7:32
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Something like: update set requested_delivery_date = requested_delivery_date.adddays(7) But with SQL code.
  
  – Joe Ruder
  Mar 28 at 7:35
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  The manipulation on the date stays the same, you only change the select to an update - I've edited my answer to reflect that.
  
  – Zohar Peled
  Mar 28 at 8:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thank you Zohar -- I had already accepted a answer, but I do appreciate your time.
  
  – Joe Ruder
  Mar 28 at 9:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Glad to help :-)
  
  – Zohar Peled
  Mar 28 at 10:23
  

  
  
  
  

add a comment | 

Use a combination of DateAdd and DateDiff:

DECLARE @Date As datetime = '2019-03-01T15:32:44'

SELECT @Date As Source,
 GETDATE() As Today,
 DATEADD(DAY, DATEDIFF(DAY, @Date, GETDATE()), @Date) As Result

Result:

Source Today Result
2019-03-01 15:32:44 2019-03-28 08:27:29 2019-03-28 15:32:44

To update the column in the table you simply need to write an update statement:

update TableName
set requested_delivery_date = DATEADD(DAY, DATEDIFF(DAY, requested_delivery_date , GETDATE()), requested_delivery_date)
-- add a where clause to only update some of the records

share|improve this answer

edited Mar 28 at 8:20

answered Mar 28 at 7:28

Zohar PeledZohar Peled

62.1k77 gold badges3939 silver badges8080 bronze badges

DECLARE @Date As datetime = '2019-03-01T15:32:44'

SELECT @Date As Source,
 GETDATE() As Today,
 DATEADD(DAY, DATEDIFF(DAY, @Date, GETDATE()), @Date) As Result

Source Today Result
2019-03-01 15:32:44 2019-03-28 08:27:29 2019-03-28 15:32:44

update TableName
set requested_delivery_date = DATEADD(DAY, DATEDIFF(DAY, requested_delivery_date , GETDATE()), requested_delivery_date)
-- add a where clause to only update some of the records

share|improve this answer

edited Mar 28 at 8:20

answered Mar 28 at 7:28

Zohar PeledZohar Peled

62.1k77 gold badges3939 silver badges8080 bronze badges

answered Mar 28 at 7:28

Zohar PeledZohar Peled

62.1k77 gold badges3939 silver badges8080 bronze badges

answered Mar 28 at 7:28

Zohar PeledZohar Peled

62.1k77 gold badges3939 silver badges8080 bronze badges

0

I tried this and successfully. This is my table.

startTime endTime
2019-03-28 15:55:10.690 2019-01-01 09:00:00.000

My SQL statement:

UPDATE [MyDatabase].[dbo].[deom1] SET [endTime]=DATEADD(DAY, DATEDIFF(DAY, [endTime], GETDATE()), [endTime])

And this the result:

startTime endTime
2019-03-28 15:55:10.690 2019-03-28 09:00:00.000

For more details, please see DATEADD (Transact-SQL).

Hope this helps.

share|improve this answer

edited Mar 28 at 8:17

answered Mar 28 at 8:05

Leon YueLeon Yue

2,31611 gold badge11 silver badge77 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  thank you Leon -- Vladimir's answer happened to be exactly what I was looking for. I appreciate your time and answer.
  
  – Joe Ruder
  Mar 28 at 8:22
  

  
  
  
  

add a comment | 

0

I tried this and successfully. This is my table.

startTime endTime
2019-03-28 15:55:10.690 2019-01-01 09:00:00.000

My SQL statement:

UPDATE [MyDatabase].[dbo].[deom1] SET [endTime]=DATEADD(DAY, DATEDIFF(DAY, [endTime], GETDATE()), [endTime])

And this the result:

startTime endTime
2019-03-28 15:55:10.690 2019-03-28 09:00:00.000

For more details, please see DATEADD (Transact-SQL).

Hope this helps.

share|improve this answer

edited Mar 28 at 8:17

answered Mar 28 at 8:05

Leon YueLeon Yue

2,31611 gold badge11 silver badge77 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  thank you Leon -- Vladimir's answer happened to be exactly what I was looking for. I appreciate your time and answer.
  
  – Joe Ruder
  Mar 28 at 8:22
  

  
  
  
  

add a comment | 

I tried this and successfully. This is my table.

startTime endTime
2019-03-28 15:55:10.690 2019-01-01 09:00:00.000

My SQL statement:

UPDATE [MyDatabase].[dbo].[deom1] SET [endTime]=DATEADD(DAY, DATEDIFF(DAY, [endTime], GETDATE()), [endTime])

And this the result:

startTime endTime
2019-03-28 15:55:10.690 2019-03-28 09:00:00.000

For more details, please see DATEADD (Transact-SQL).

Hope this helps.

share|improve this answer

edited Mar 28 at 8:17

answered Mar 28 at 8:05

Leon YueLeon Yue

2,31611 gold badge11 silver badge77 bronze badges

startTime endTime
2019-03-28 15:55:10.690 2019-01-01 09:00:00.000

UPDATE [MyDatabase].[dbo].[deom1] SET [endTime]=DATEADD(DAY, DATEDIFF(DAY, [endTime], GETDATE()), [endTime])

startTime endTime
2019-03-28 15:55:10.690 2019-03-28 09:00:00.000

share|improve this answer

edited Mar 28 at 8:17

answered Mar 28 at 8:05

Leon YueLeon Yue

2,31611 gold badge11 silver badge77 bronze badges

answered Mar 28 at 8:05

Leon YueLeon Yue

2,31611 gold badge11 silver badge77 bronze badges

answered Mar 28 at 8:05

Leon YueLeon Yue

2,31611 gold badge11 silver badge77 bronze badges