Question

Been reading up on php and json and am trying this piece of code which is not working and I can’t figure out why. Any help is appreciated.

I call a JavaScript function from my html file onLoad(). That function is

    function getSched()
    {
        $.post("schedlz.php", {dat: ""+inputString+"", action: "searchSched"}, function(data)
    {
            var y="";
    if(data.length >0) 
    {
        var obj = JSON.parse(data);
        y = getRes (data);
        }
        else //some error handler
}
function getRes(data)
{
    var str="";
    var obj = JSON.parse (data);
    alert (data.length + " | " +obj + " | ");
    return str;
}

In schedlz.php file

$conn = getDbConn();
mysql_select_db("myschedulez", $conn);
$result = mysql_query($sql,$conn) or die('Error: ' . mysql_error());
    $rows = array();
    while ($r = mysql_fetch_assoc($result))
    {
        // $rows['schedulez'][] = $r;
        $rows[] = $r;
    }
    echo json_encode($rows);
 mysql_close($conn);

The problem is that I get the data back in correct json format but when I parse it, there is no Object. What could be going wrong here?

Answer 1

You need to add the dataType parameter to your $.post, which in your case is json.
I’m also not sure if you closed $.post correctly in your sample so I’m adding the closing ); as well.

function getSched()
{
    $.post("schedlz.php", 
    {dat: ""+inputString+"", action: "searchSched"}, 
    function(data) {
      var y="";
      if(data.length >0) 
      {
        var obj = JSON.parse(data);
        y = getRes (data);
      }
      else {} //some error handler
    },
    'json'
    ); 
}