Variable data in ajax call JQUERY

Question

I'm trying to use variable in AJAX call in jquery but it is not working. move variable contains different values. Please check the following code:

var $move = 'next';

$.ajax({
   type: "POST",
   url: "somephp.php",
   data: {$move:1},
});

Suggest any way to use $move variable in data.

data accepts key value pairs. For now you your $move is acting as key and that is why it is not working. Make it a value and it should work.. Like data: { "data" : $move} — AdityaParab, Jul 27 '12 at 11:55
I think what the user is asking is how to pass something different for the key part of the value pair. — Kris.Mitchell, Jul 27 '12 at 11:58
If your problem is that the variable should be used as property name, and not its value, please see [my answer](http://stackoverflow.com/a/11687303/95033). — Wolfram, Jul 27 '12 at 11:59
Why would anyone want to have "dynamic" keys? I am not getting any idea why is that necessary. — AdityaParab, Jul 27 '12 at 12:01
It allows to use variables and [characters that are not allowed in dot notation](http://stackoverflow.com/a/4968448/95033). There are definitely examples where you want this. For example, I experienced this once with a jQuery UI dialog, for which I needed dynamic multilanguage button labels. — Wolfram, Jul 27 '12 at 12:12
Thanks Wolf. I never came across such scenario so was just wondering... :) — AdityaParab, Jul 27 '12 at 12:36

Wolfram · Answer 1 · 2017-10-09T10:10:04.787

14

If you want to use the variable as the name of the property, use array notation.

There are two ways to access object members: dot notation and bracket notation (a.k.a. subscript operator).

Your code with array notation:

var $move = 'next';
var data = {};
data[$move] = 1;

$.ajax({
   type: "POST",
   url: "somephp.php",
   data: data,
});

The example on jsfiddle (the post obviously doesn't work, so check the console to see what gets posted.)

edited Oct 09 '17 at 10:10

answered Jul 27 '12 at 11:56

Wolfram

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1

I had a similar problem (needed to allow a variable within the name) and this is exactly what was needed. – Voodoo Apr 05 '13 at 21:45
Perfect answer ! – carry0987 May 26 '19 at 05:39

Sem · Answer 2 · 2012-07-27T12:06:55.940

8

If you want to have a variable-variable in your POST request, you will need to make a seperate JSON object:

var name = 'next';
var dataObject = {};
dataObject[name] = 1;

$.ajax({
   type: "POST",
   url: "somephp.php",
   data: dataObject,
   complete : function() {
     // success!
   }
});

edited Jul 27 '12 at 12:06

answered Jul 27 '12 at 11:54

Sem

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I had to figure this out with a lot of research, so for reference, when the `data` is a PlainObject and not a String, using `complete:function(response)` returns a PlainObject. For example, `alert(response)` returns `[object Object]` and `console.log(response)` returns `Object {readyState: 4, responseText: "example", status: 200, statusText: "OK"}`. So to retrieve `responseText`, use `success:function(response)`. – Jan 05 '16 at 19:36

score 3 · Accepted Answer · answered Jul 27 '12 at 11:54

3

It should be

$.ajax({
   type: "POST",
   url: "somephp.php",
   data: {"data":$move},
});

answered Jul 27 '12 at 11:54

AdityaParab

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3
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$move is the key not the value. – Subir Kumar Sao Jul 27 '12 at 11:56
I guess he wants to post value of $move. Lets wait for him to clarify. :) – AdityaParab Jul 27 '12 at 11:59
1

Actually, looks to me like `$move` is a variable, and the OP whishes to use it as the key, and the value should be `1`. So far only Wolfram got it right! – adeneo Jul 27 '12 at 12:01

score 2 · Answer 4 · answered Jul 27 '12 at 11:56

2

You can do something like data: $move+"=1"

answered Jul 27 '12 at 11:56

Subir Kumar Sao

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Responded with the assumption user wants the variable as a key. – Subir Kumar Sao Jul 27 '12 at 12:00
+1, this works and in contrast to most other answers (not mine of course :)), it solves the right problem. – Wolfram Jul 27 '12 at 12:02
I think its just a JS variable. – Subir Kumar Sao Jul 27 '12 at 12:04
Yep, the `$` prefix is sometimes used to identify jQuery objects. However, in this case it is a simple string but might not represent the real code, but a simple example for the sake of this question. So this is a JavaScript variable. Out of context it is confusing. – Wolfram Jul 27 '12 at 12:09
@Kris $ is a valid character to be used as variable name in javascript.. $,$$,$$$ $$$$$$$ are all valid javascript variable names.. Just like aa bbb ccc etc – AdityaParab Jul 27 '12 at 12:11
This is what I was looking. Thank you. :) @SubirKumarSao – Fahad Ahammed Jul 28 '17 at 13:26

score 1 · Answer 5 · answered Jun 18 '19 at 22:59

1

I was looking for something like this. I wanted to have a variable for the key AND a variable for the value.

let dataPair = {};
dataPair[dataKey] = dataValue;
$.ajax({
  url: TheAPIPath,
  data: dataPair,
  method: "GET",
(etc)

answered Jun 18 '19 at 22:59

John Grabauskas

310
2
5

In light of the answer by Subir, I tried `data:dataKey+"="+dataValue,` and it worked. – John Grabauskas Jun 18 '19 at 23:05

dexter.ba · Answer 6 · 2012-07-27T11:55:36.880

0

To post the value of $move variable, do this:

$.ajax({
   type: "POST",
   url: "somephp.php",
   data: {move: $move}
});

edited Jul 27 '12 at 11:55

answered Jul 27 '12 at 11:54

dexter.ba

292
2
5

score -1 · Answer 7 · answered Jul 27 '12 at 11:58

It looks like you are trying to use a PHP variable in javascript. You can use .serialize to gather data and pass it to the ajax call. But as for making calls with different named variables for the name of the value pair, you will need to gather that information from the php page you are passing.

Example:

$.get("jQueryFunctions.php?action=checkCaptula",$('#resetPasswordDiv :input').serialize()).done(function(data){ ....

While this is a .get instead of a .ajax, it's just shorthand for the .ajax call. The passwordDiv contains HTML inputs with names and id. It gathers the information and passes it to the php page for processing.

Variable data in ajax call JQUERY

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