Getting JSON encoded variables from php

Question

i have this php code:

$query = mysql_query("SELECT * FROM table WHERE id=$id");
while($item = mysql_fetch_assoc($query)){
    $dataArray['var1'] = $item['var1'];
    $dataArray['var2'] = $item['var2'];
    $dataArray['var3'] = $item['var3'];
    array_push($return_arr,$dataArray);
}

echo json_encode($return_arr);

it's in my ajax.php file

Then I'm using jquery to send there id via ajax request:

$.ajax({
    type: "POST",
    url: 'ajax.php',
    data: 'id='+id,
    success: function(data) { 
        alert(data);
    }
 });

and it alerts me this: [{"var1":"somevarhere","var2":"somevarhere2","var3":"somevarhere3"}]

can someone tell me please how I can access for ex. var3 to get somevarhere3?

thanks

So your question is how to transform that json encoded string to an object? — Cronco
– Cronco, Commented Mar 27, 2011 at 19:47

Community · Accepted Answer · 2017-05-23 12:26:51Z

4

According to the documentation of jQuery.ajax, the data it receives as first parameter is interpreted depending on the value of the dataType option passed to jQuery.Ajax.

What if you pass 'json' for that parameter ?
Like this :

$.ajax({
    type: "POST",
    url: 'ajax.php',
    data: 'id='+id,
    dataType: 'json', 
    success: function(data) { 
        alert(data);
    }
 });

Doing this, jQuery.ajax should interpret the data returned from the server as JSON.

Else, the documentation says that, if no dataType is specified (quoting) :

jQuery will try to infer it based on the MIME type of the response

So, the other solution would be to return some JSON content-type from the PHP code (see The right JSON content type?), with something like this :

header('Content-type: application/json');

Before doing the echoing.

And, again, jQuery.ajax should interpret the data returned from the server as JSON.

edited May 23, 2017 at 12:26

CommunityBot

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answered Mar 27, 2011 at 19:47

Pascal MARTIN

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2 Comments

simPod Over a year ago

I have added dataType: 'json', - this looks almost great, now it says that data is Object but when I alert(data.toSource()) it still returns [{"var1":"somevarhere","var2":"somevarhere2","var3":"somevarhere3"}] and therefore it doesnt work. It think that it's because of those brackets ([])

Pascal MARTIN Over a year ago

well, now that your data is interpreted as JSON, you can access it like any other Javascript array that would contain one object ; for example : data[0].var3

simPod · Accepted Answer · 2011-03-27 20:33:14Z

1

Ok, I found the solution. Thanks to Pascal MARTIN I got the idea to parse the result so I used this:

data = data.substring(1,data.length-1);
data = $.parseJSON(data);
alert(data.var3);

first line removes [], second line parses JSON to Object, and third line accesses the property

answered Mar 27, 2011 at 20:33

simPod

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1 Comment

Pascal MARTIN Over a year ago

Well, if your SQL query always return one, and only one, result, you'd better return a single object from PHP, and not an array that contains one object -- which is why those [] are there : your PHP code returns a JSON-encoded array.

Richard Dalton · Accepted Answer · 2011-03-27 19:47:00Z

0

Should be able to access it like this:

data[0].var3

Where 0 is the index of the item in the array you want to access.

answered Mar 27, 2011 at 19:47

Richard Dalton

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