jQuery not receiving JSON encoded data from PHP script

Question

I'm trying to get jQuery to retrieve PHP variables and insert them into input values. I was looking everywhere trying to find a solution.

Here's jQuery code.

$(document).ready() {
$(function() {
    $.ajax({
      type: 'POST',
      url: 'loadstuff.php',
      data: {
            'something': something, 
            'different': different, 
            'another': another
            },
      dataType: 'json',
      success: function(data) {
        $('input[name=get_seomething_here]').val( 'something' );
        $('input[name=get_different_here]').val( 'different' );
        $('input[name=get_another_here]').val( 'another' );
}
    });

});
});

And here's PHP side.

//connecting to db etc.

$query = "SELECT something, different, another FROM stuff WHERE id='1'";  
$result = mysql_query($query) or die(mysql_error());
$row = mysql_fetch_array( $result );

echo(json_encode(array('something' => $row['something'], 
'different' => $row['different'], 
'another' => $row['another']
)));

Have you tried setting the header to text/json? Before your echo in the PHP, put: header('Content-type: text/json'); — kingcoyote
– kingcoyote, Commented Mar 14, 2012 at 17:48
@kingcoyote The proper Content-type for json is application/json; so it should look like header('Content-type: application/json'); — Edwin
– Edwin, Commented Mar 14, 2012 at 18:13
Another thing to keep in mind is force JSON_NUMBERIC_CHECK, or all PHP numbers will be converted to JavaScript string types. If the cols you're querying contain no numbers, you won't have to worry, but you should do it since it's a good habit. — Edwin
– Edwin, Commented Mar 14, 2012 at 18:15
Thanks for your response. In fact I wanted the script to pass some numbers as well. Right now, as I modified the script using aletzo advise, it turned out in strange way. The purpose of this script (as presented one is just an example) was to fill 9 input text fields with numbers and at the same time fill drop down select input values for a sort-of calculator - however in the end it gave all selected inputs the same value (adding the values together and giving each the same) and still didnt fill the input text — user1269386
– user1269386, Commented Mar 14, 2012 at 18:27
Could you possibly create a jsfiddle with the actual code, and tell us the exact error that you get? Probably it's gonna very easy to solve the problem that way. — aletzo
– aletzo, Commented Mar 15, 2012 at 1:11

aletzo · Accepted Answer · 2012-03-14 17:48:32Z

2

I think it should be:

$('input[name=get_seomething_here]').val( data.something );
$('input[name=get_different_here]').val( data.different );
$('input[name=get_another_here]').val( data.another );

answered Mar 14, 2012 at 17:48

aletzo

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1 Comment

user1269386 Over a year ago

Thank you for your response. I changed the code following your suggestion. However, the data still cant get through for some reason...

Smit · Accepted Answer · 2012-03-14 18:05:56Z

-1

aletzo is right.. but before doing that you have to do like:

data = eval("(" + data + ")");

Then you can use data.something, data.different and data.another

answered Mar 14, 2012 at 18:05

Smit

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3 Comments

Edwin Over a year ago

I've been doing this myself, and eval() is not only unnecessary, but can leave a web app vulnerable to XSS attacks.

Smit Over a year ago

you can use JSON.parse rather then eval

Edwin Over a year ago

Nonetheless, calling an extra method when it's unnecessary is inefficient.

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