up vote 2 down vote favorite

I'm using the $.post() function from jQuery to make a Ajax call with a JSON string. The call looks like this:

$.post(
    urlVar,
    jsonVar,
    function(data){
        //do stuff
    },
    'json'
)
.complete(function(){
    //do other stuff
});

To create jsonVar I'm using this code

var1 = {};
var1.id = fooId;
var1.amount = fooAmount;
var1.zoom = fooZoom;
jsonVar = JSON.stringify(var1);

To make the call work, jsonVar should look like this

{id:fooId, amount:fooAmount, zoom:fooZoom}

but it looks like this

{"id":fooId, "amount":fooAmount, "zoom":fooZoom}

Now my code will not work, because of the double quotes. I couldn't figure out how to get rid of those. Can anyone help me out?

IMPORTANT:

the code does work if I put the $.post() function like this:

$.post(
    urlVar,
    {id: fooId, amount: fooAmount, zoom: fooZoom},
    function(data){
        //do stuff
    },
    'json'
)
.complete(function(){
    //do other stuff
});
share|improve this question
4  
I'd interject that there might be something fundamentally wrong with the receiving end of the AJAX call, because that second JSON string is legal JSON syntax. In fact, I think that's actually standard (over the no-quotation-mark version). – Richard Neil Ilagan Jun 27 at 12:11
You example appears to be flawed in that "zoom" is never a part of var1 but is actually a part of var2. – Mark Schultheiss Jun 27 at 12:15
i had the same problem in different browsers. it worked after qouting the values in the json-string. like: {"id":"fooId", "amount":"fooAmount", "zoom":"fooZoom"} – some_coder Jun 27 at 12:15
@MarkSchultheiss My bad, that was a typo. – LuudJacobs Jun 27 at 12:17
1  
@LuudJacobs ~ if that works, then passing var1 directly into the AJAX call should work as well. Have you given that a try? – Richard Neil Ilagan Jun 27 at 12:35
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3 Answers

up vote 0 down vote

Given this code:

var fooId = 'foodyfood';
var fooAmount = 10.20;
var fooZoom = 'zoomer';
var var2 = {};

var var1 = {}; 
var1.id = fooId; 
var1.amount = fooAmount; 
var2.zoom = fooZoom; 
jsonVar = JSON.stringify(var1); 
$('#showme').text(jsonVar);

the value shown in the showme would be:

{"id":"foodyfood","amount":10.2}

so your example appears to be flawed and actually is not JSON standard which specifies the double quotes.

EDIT: now your use of the post is the equivelent of:

$.ajax({
  type: 'POST',
  url: url,
  data: data,
  success: success,
  dataType: dataType
});

except that it handles the object as objects in your second example after your edit.

see the example in the documentation here: http://api.jquery.com/jQuery.post/

with this form:

$.post("test.php", { name: "John", time: "2pm" },   function(data) {
     alert("Data Loaded: " + data);   
});

so if you have:

var mytest =  { name: "John", time: "2pm" };

it becomes: on the stringify

{"name":"John","time":"2pm"}

see working example here:http://jsfiddle.net/v4NHv/

share|improve this answer
one note, if you need a parameter to pass you might use $.param(var1); form which creates: id=foodyfood&amount=10.2 – Mark Schultheiss Jun 27 at 12:47
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up vote 0 down vote

Change the post call like this

$.post(
    urlVar,
    {var1:jsonVar},
    function(data){
      //do stuff
   },
   'json'
)
.complete(function(){
   //do other stuff
});

And then of course you will need to make a little tweek to your receiving app

share|improve this answer
feedback
up vote 1 down vote

The JSON specification states that the keys must have double-quotes.

What do you mean your code won't work because of double quotes? Parse the JSON back into an object using JSON.parse; which is built-in to many modern browsers or you can shim it using the json2 library.

share|improve this answer
Regarding the parsing: $.parseJOSN() works as well – Johan Jun 27 at 12:26
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