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Question

JavaScript: Variable Value gets lost between functions

Answer 1

A:

Are you sure json is an object. Maybe its a json-string? Than you should eval it before use.
If you get it via $.ajax() call don't forget to add dataType:'json' as an option...

NilColor 2009-12-02 08:59:13

Try this: function updateSliderContent(json) { alert(typeof json); //<-- here you get an alert with json type. if its string then you found your error jQuery.each(screen_order, function(i, item) { ...

NilColor 2009-12-02 09:02:53

I used dataType and set it to json. I actually have it defined in jQuery.ajaxSetup() since 90% of my calls all expect json as result.

Mike 2009-12-02 09:19:16

typeof returns "object"

Mike 2009-12-02 09:21:01

Answer 2

A:

tray in the if() clause to use item.action, because if the json data is an array of objects, item will contain each object, but this is only my assumption

bozhidarc 2009-12-02 09:19:14

Answer 3

+1 A:

There seems to be nothing wrong with the jQuery.each as I can't reproduce your problem.

        function alertName (json) {
            var a = new Array();
            a.push(1);
            a.push(2);

            jQuery.each(a, function(i, item) {
                alert(json[0].name);
                alert(json[1].name);
            });
        }

        var andre = { name: "André" }
        var joana = { name: "Joana" }

        var arrayOfJson = new Array();
        arrayOfJson.push(andre);
        arrayOfJson.push(joana);

        alertName(arrayOfJson);

alertName() function works exactly as it should. The json parameter isn't lost within jQuery.each function

It seems to be a problem on your implementation, something you're not telling us about. Please try to "compress" your issue to a working sample as I did and show us so that we can try in on our own :)

Ciwee 2009-12-02 10:41:02

That's odd, because this is basically what I am trying to do with the object. The function I provided is actually almost the entire thing. What I removed was the actual call, which you simulated in your example, except thatjson is an Object and not an Array.

Mike 2009-12-02 11:45:42

Answer 4

A:

I think you should make sure the function updateSliderContent is called after the json response from the server side.

The code below shouldn't work:

    var json = {};

    $.getJSON("/url/", {}, function(data){
        json = data;
    });

    //we call our function before the server response
    //and the json will be just blank object
    updateSliderContent ( json );

So, please have a look at your code because sometimes we did something like that without intention.

The code below should work and the json object must not be blank if the server really response the correct json.

    $.getJSON("/url/", {}, function(json){
        //we call only after the response from server
        updateSliderContent ( json );
    });

max 2009-12-02 14:26:16

I have done that. See Edit 1.

Mike 2009-12-02 14:56:07

Answer 5

+2 A:

I've attempted, and failed, to reproduce your problem on JS Bin:
http://jsbin.com/ereha (editable via http://jsbin.com/ereha/edit)

The code you've shown us so far seems perfectly fine, so the problem must be caused by some other part of your code or system. We're all just shooting in the dark if we don't know what the issue is.

Please try and reproduce the problem on http://jsbin.com and we can help you from there.

Complete Source Code

index.html

<!doctype html>
<html lang="en">
  <head>
    <meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
    <title>http://stackoverflow.com/questions/1831384/javascript-variable-value-gets-lost-between-functions&lt;/title&gt;
    <script type="text/javascript" src="http://getfirebug.com/releases/lite/1.2/firebug-lite-compressed.js"&gt;&lt;/script&gt;
    <script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.3.2/jquery.min.js"&gt;&lt;/script&gt;
    <script type="text/javascript">
      $.ajaxSetup({url: 'test.json'});

      function updateSliderContent(json) {  //<- json defined here is correct
        var screen_order = json.screen_order.split('_');
        jQuery.each(screen_order, function(i, item) {
          var screen_id = item;
          //at this point it is not, thus the function does not execute whatever is in the if blocks
          if (json[screen_id].action == 'add') {
            console.log(screen_id, 'action add');
          } else if (json[screen_id].action == 'remove') {
            console.log(screen_id, 'action remove');
          };
        });
      }

      function updateSlider() {
        var ajaxData = {};
        jQuery.ajax({
          data: ajaxData,
          dataType: 'json',
          success: function (json) {
            updateSliderContent(json);
          }
        });
        // theDate = Math.round(new Date().getTime()/1000.0); //UNIX Timestamp
        sliderTimer = setTimeout('updateSlider();',15000);
      };

      $(updateSlider);
    </script>
  </head>
  <body>
  </body>
</html>

test.json

{
  'screen_order': 'foo_bar_baz',
  'foo': {
    'action': 'add'
  },
  'bar': {
    'action': 'add'
  },
  'baz': {
    'action': 'remove'
  }
}

brianpeiris 2009-12-03 04:36:43

I have tried to also reproduce the problem on my local server but couldn't... Even if I revert to the version I had my problems with I cannot seem to reproduce it. I'll give you a correct answer for your trouble... still, I cannot understand why it works now...

Mike 2009-12-03 08:21:50

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answers:

JavaScript: Variable Value gets lost between functions

Edit 1

Complete Source Code

index.html

test.json

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