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Set parameter in FormData for dynamic fields (Javascript)

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0

I am getting an issue on setting up the form data for dynamic fields included input[type="file"].

In my word, Dynamic fields mean generating the fields with the loop of an array.

I tried the for loop than data is not going to the server.

I am using Vue and Vuex.

Only the value from the last loop is going on the database.

let formdata = new FormData();
for(var i = 0; i< this.assignmentForm.length; i++) 
 formdata.append('file', this.$refs.assignmentFile[i].files[0]);
 formdata.append('name', this.$refs.assignmentName[i].value);
 formdata.append('comment', this.$refs.assignmentComment[i].value);
 formdata.append('assignment_solution', this.respondId);

javascript vue.js vuex

share|improve this question

edited Mar 23 at 7:17

Styx

5,74762738

asked Mar 23 at 4:46

Vick SainVick Sain

6711

add a comment | 

0

I am getting an issue on setting up the form data for dynamic fields included input[type="file"].

In my word, Dynamic fields mean generating the fields with the loop of an array.

I tried the for loop than data is not going to the server.

I am using Vue and Vuex.

Only the value from the last loop is going on the database.

let formdata = new FormData();
for(var i = 0; i< this.assignmentForm.length; i++) 
 formdata.append('file', this.$refs.assignmentFile[i].files[0]);
 formdata.append('name', this.$refs.assignmentName[i].value);
 formdata.append('comment', this.$refs.assignmentComment[i].value);
 formdata.append('assignment_solution', this.respondId);

javascript vue.js vuex

share|improve this question

edited Mar 23 at 7:17

Styx

5,74762738

asked Mar 23 at 4:46

Vick SainVick Sain

6711

add a comment | 

I am getting an issue on setting up the form data for dynamic fields included input[type="file"].

In my word, Dynamic fields mean generating the fields with the loop of an array.

I tried the for loop than data is not going to the server.

I am using Vue and Vuex.

Only the value from the last loop is going on the database.

let formdata = new FormData();
for(var i = 0; i< this.assignmentForm.length; i++) 
 formdata.append('file', this.$refs.assignmentFile[i].files[0]);
 formdata.append('name', this.$refs.assignmentName[i].value);
 formdata.append('comment', this.$refs.assignmentComment[i].value);
 formdata.append('assignment_solution', this.respondId);

javascript vue.js vuex

share|improve this question

edited Mar 23 at 7:17

Styx

5,74762738

asked Mar 23 at 4:46

Vick SainVick Sain

6711

let formdata = new FormData();
for(var i = 0; i< this.assignmentForm.length; i++) 
 formdata.append('file', this.$refs.assignmentFile[i].files[0]);
 formdata.append('name', this.$refs.assignmentName[i].value);
 formdata.append('comment', this.$refs.assignmentComment[i].value);
 formdata.append('assignment_solution', this.respondId);

share|improve this question

edited Mar 23 at 7:17

Styx

5,74762738

asked Mar 23 at 4:46

Vick SainVick Sain

6711

share|improve this question

edited Mar 23 at 7:17

Styx

5,74762738

asked Mar 23 at 4:46

Vick SainVick Sain

6711

edited Mar 23 at 7:17

Styx

5,74762738

edited Mar 23 at 7:17

Styx

5,74762738

asked Mar 23 at 4:46

Vick SainVick Sain

6711

asked Mar 23 at 4:46

Vick SainVick Sain

6711

2 Answers
2

active

oldest

votes

0

You're correctly using .append instead of .set, but you forgot that variables should be arrays, thus their keys should be file[], name[], comment[] and assignment_solution[] respectively.

This way your backend will correctly recognize that they are indeed arrays.

share|improve this answer

answered Mar 23 at 7:20

StyxStyx

5,74762738

add a comment | 

0

You are appending multiple files using the same name. Only the last addition will reach your server.

You have at least 2 options:

1. 
   

   Give unique names to each field. Something similar to this

   
   
   

   formdata.append('file' + i, this.$refs.assignmentFile[i].files[0]);

   
   

Notice the concatenation of file with the index variable.

2. 
   

   Use the array notation for the fields’ names. This is compatible with how PHP handled POST variables.

   
   
   

   formdata.append('file[]', this.$refs.assignmentFile[i].files[0]);

   
   

Notice the square brackets in the name file[].

You need to do this for all the fields: file, name, comments.

share|improve this answer

answered Mar 23 at 7:22

Radu DițăRadu Diță

4,36311321

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I did same. But not working.
  
  – Vick Sain
  Mar 23 at 14:46
  

  
  
  
  

add a comment | 

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0

You're correctly using .append instead of .set, but you forgot that variables should be arrays, thus their keys should be file[], name[], comment[] and assignment_solution[] respectively.

This way your backend will correctly recognize that they are indeed arrays.

share|improve this answer

answered Mar 23 at 7:20

StyxStyx

5,74762738

add a comment | 

0

You're correctly using .append instead of .set, but you forgot that variables should be arrays, thus their keys should be file[], name[], comment[] and assignment_solution[] respectively.

This way your backend will correctly recognize that they are indeed arrays.

share|improve this answer

answered Mar 23 at 7:20

StyxStyx

5,74762738

add a comment | 

You're correctly using .append instead of .set, but you forgot that variables should be arrays, thus their keys should be file[], name[], comment[] and assignment_solution[] respectively.

This way your backend will correctly recognize that they are indeed arrays.

share|improve this answer

answered Mar 23 at 7:20

StyxStyx

5,74762738

share|improve this answer

answered Mar 23 at 7:20

StyxStyx

5,74762738

answered Mar 23 at 7:20

StyxStyx

5,74762738

answered Mar 23 at 7:20

StyxStyx

5,74762738

0

You are appending multiple files using the same name. Only the last addition will reach your server.

You have at least 2 options:

1. 
   

   Give unique names to each field. Something similar to this

   
   
   

   formdata.append('file' + i, this.$refs.assignmentFile[i].files[0]);

   
   

Notice the concatenation of file with the index variable.

2. 
   

   Use the array notation for the fields’ names. This is compatible with how PHP handled POST variables.

   
   
   

   formdata.append('file[]', this.$refs.assignmentFile[i].files[0]);

   
   

Notice the square brackets in the name file[].

You need to do this for all the fields: file, name, comments.

share|improve this answer

answered Mar 23 at 7:22

Radu DițăRadu Diță

4,36311321

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I did same. But not working.
  
  – Vick Sain
  Mar 23 at 14:46
  

  
  
  
  

add a comment | 

0

You are appending multiple files using the same name. Only the last addition will reach your server.

You have at least 2 options:

1. 
   

   Give unique names to each field. Something similar to this

   
   
   

   formdata.append('file' + i, this.$refs.assignmentFile[i].files[0]);

   
   

Notice the concatenation of file with the index variable.

2. 
   

   Use the array notation for the fields’ names. This is compatible with how PHP handled POST variables.

   
   
   

   formdata.append('file[]', this.$refs.assignmentFile[i].files[0]);

   
   

Notice the square brackets in the name file[].

You need to do this for all the fields: file, name, comments.

share|improve this answer

answered Mar 23 at 7:22

Radu DițăRadu Diță

4,36311321

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I did same. But not working.
  
  – Vick Sain
  Mar 23 at 14:46
  

  
  
  
  

add a comment | 

You are appending multiple files using the same name. Only the last addition will reach your server.

You have at least 2 options:

1. 
   

   Give unique names to each field. Something similar to this

   
   
   

   formdata.append('file' + i, this.$refs.assignmentFile[i].files[0]);

   
   

Notice the concatenation of file with the index variable.

2. 
   

   Use the array notation for the fields’ names. This is compatible with how PHP handled POST variables.

   
   
   

   formdata.append('file[]', this.$refs.assignmentFile[i].files[0]);

   
   

Notice the square brackets in the name file[].

You need to do this for all the fields: file, name, comments.

share|improve this answer

answered Mar 23 at 7:22

Radu DițăRadu Diță

4,36311321

share|improve this answer

answered Mar 23 at 7:22

Radu DițăRadu Diță

4,36311321

answered Mar 23 at 7:22

Radu DițăRadu Diță

4,36311321

answered Mar 23 at 7:22

Radu DițăRadu Diță

4,36311321