Print JSON from Variable in Javascript

Question

I am attempting to print a JSON that I get like this:

jsonfields = $.ajax({
url: "ajax.php?getsensors="+raw.deviceId, 
async: false
}).responseText;

into an ExtJS datastore that looks like this:

Ext.grid.dummyData = [
    //jsonfields,
    //["ping"],["location"],["death"],["birth"],["DeviceInfo"],
    ['3m Co',71.72,0.02,0.03,'4/2 12:00am', 'Manufacturing'],
    ['Alcoa Inc',29.01,0.42,1.47,'4/1 12:00am', 'Manufacturing'], ...

So when I alert the variable 'jsonfields' it alerts

["ping"],["location"],["death"],["birth"],["DeviceInfo"]

But it doesn't render to the grid at all, but when I hard-code the line above into the json, it does render. I set my jsonfields var like:

var jsonfields;

This is the first thing I do in the script tag. I know that the value of it is updated due to the alert. So how is it different from displaying the JSON from a pre-set variable compared to when I hard-code it in?

Thanks!

UPDATES: I can also see the response of the ajax request in the chrome developer tools XHR, it seems to be structured the same: ["ping"],["location"],["death"],["birth"],["DeviceInfo"]

alert(typeof jsonfields);

Returns "string"

["ping"],["location"],["death"],["birth"],["DeviceInfo"] does not look like valid JSON. — Felix Kling
– Felix Kling, Commented Dec 9, 2010 at 22:38
I think they are multiple JSONs comma-delimited, thats how the sample data is structured, it renders properly when the data is manually entered into the store. — Doug Molineux
– Doug Molineux, Commented Dec 9, 2010 at 22:40
what happens if you add datatype:"json" to your $.ajax call? — nico
– nico, Commented Dec 9, 2010 at 23:09

Cameron Jordan · Accepted Answer · 2010-12-09 23:05:24Z

1

It would be better to correct the server so it produces valid JSON and then use an Ext.data.JsonStore

If you cannot correct the server here's a very manual solution:

jsonfields = Ext.decode('[' + $.ajax({
    url: "ajax.php?getsensors="+raw.deviceId, 
    async: false
}).responseText + ']');

Ext.grid.dummyData = jsonfields.concat([
    ['3m Co',71.72,0.02,0.03,'4/2 12:00am', 'Manufacturing'],
    ['Alcoa Inc',29.01,0.42,1.47,'4/1 12:00am', 'Manufacturing'], ...
]);

answered Dec 9, 2010 at 23:05

Cameron Jordan

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