Question

PHP

$table = mysql_real_escape_string($_REQUEST['table']);

$query  = "SELECT COLUMN_NAME FROM INFORMATION_SCHEMA.Columns where TABLE_NAME = '$table'";
$result = mysql_query($query);
$arr1 = array();
while($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
    $arr1[] = $row['COLUMN_NAME'];
}
echo json_encode($arr1);

}

Javascript/Jquery

$("#verticalSelect").change(function() {
    var table = $("#verticalSelect option:selected").attr('value');

    $.post(PROCESSORFILE, {"task": "getTableDupeFields", "table": table}, function(data) {
        $.each(data, function(key, value){
            $("#test-area").append(key+' is: '+value+'<br/>');
        });
        alert(data);
    });
});

Now with the alert of the returned JSON variable I get the result I expect.

["lead_id", "callcenter", "monkey" ...]

But with the Jquery.each() function I get this:

0:is: [
1:is: “
2:is: l
3:is: e
4:is: a
5:is: d
6:is: _
7:is: i
8:is: d
9:is: “
10:is: ,
11:is: “
12:is: c
13:is: a
14:is: l
15:is: l
16:is: c
17:is: e
18:is: n
19:is: t
20:is: e
21:is: r

Every character of the returned array is looped instead of every discreet value. What is the proper way of handling JSON returned arrays?

Answer 1

You either need to set your response header to application/json or specify the dataType in $.post to be ‘json’