jQuery ajax returning undefined

Question

I am trying to return a row from my database where the id sent via jquery to the php matches a field value. I am getting back undefined and can't seem to work my way out of it.

My jquery:

function renderPreview( event ) {
    target = event.target.id;

    console.log(target) // should say the id name

            $('#results').removeClass();
            $('#results').addClass(target).html( $('#textInput').val() );
            $('html, body').animate({ scrollTop: 600}, "slow");

    console.log('start ajax')

    $.ajax({
        url: '../test.php',
        type: 'POST',
        data: [{'class' : target}],
        dataType: 'json',
        success: function(data) {
            var id = data[0];
            var name = data[1];
            var style = data[2];
            $('#codeTest').html("<b>id: </b><br />"+id+"<br /><b> name: </b><br />"+name+"<br /><b> style: </b><br />"+style);
        }
    });
};

PHP:

$dbstylename = $_POST['class'];
$result = mysql_query("SELECT * FROM style where stylename like '$dbstylename'");
$array = mysql_fetch_row($result);

echo json_encode($array);

mysql_close($con);
?>

Also is there a line of code I can put in my jquery or php to see what query is going through in my chrome developer console?...like the console.logs I already have.

have you tried putting debugger in your success section, or tried alerting data[0], if yes what is the result — rahul
– rahul, Commented Nov 19, 2012 at 7:05
Is $dbstylename the right value? What is data? Note that your code is prone to SQL injection! — Felix Kling
– Felix Kling, Commented Nov 19, 2012 at 7:13
@rahul data.d[0] returns an for cannot read property of undefined. — tehaaron
– tehaaron, Commented Nov 19, 2012 at 7:15
@tehaaron is your test.php returning the correct data. problem is not in jquery-ajax check your test.php — rahul
– rahul, Commented Nov 19, 2012 at 7:17
@FelixKling what do you mean by the right value? =\ I am very new to php and read through a few tutorials to get to this point. I had a basic ajax function working and then when attempting to merge it into this existing function encountered problems I havent been able to work out (obviously). Maybe data should be array since that is what holds the data in the php? Also, I figured it was prone to SQL Injection just based on my inexperience. Can you point me somewhere that can teach me about preventing that? — tehaaron
– tehaaron, Commented Nov 19, 2012 at 7:19

Felix Kling · Accepted Answer · 2012-11-19 07:44:33Z

2

The problem is that you are not sending the data in the correct way. jQuery is passing the value you assign to the data: property to jQuery.param. This function converts the data to a proper query string.

But if you pass an array instead of an object, .param expects it to be an array of objects, where each of the objects has a name and value property.

You can see what jQuery generates in your case by calling .param directly:

> jQuery.param([{'class' : target}])
  "undefined=undefined"

You get the correct query string, if you pass either

[{name: 'class', value: target}]

or

{'class': target}

Both generate:

"class=<whatever the value of target is>"

answered Nov 19, 2012 at 7:44

Felix Kling

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