Jquery and Json

JavaScript
1

Hi everyone, i got a problem with a json response from an ajax request.

I have the following code :


$.get(  "ajax_sous_famille.php",
        {"id": $(this).attr("value")},
        function(data)
        {
            // whatever !!!
        },
        "json"
);

And the called page (ajax_sous_famille.php) is giving me this result :


[{"id":"25","nom":"Pouf slim"},{"id":"26","nom":"Pouf g\\u00e9ant"},{"id":"27","nom":"Pouf XXL"},{"id":"28","nom":"Pouf Int\\u00e9rieur"}]

The info in the resulting string is correct, but here you can see the at the start and end of the line.
It seems not to be a correct Json string and then it crashes my script !

I have another request like this which gives me the good format of resulting data (without the ) and which works perfectly when i decode the Json string to display it.

The difference between the 2 requests is that the one which works returns only 1 array of values like


array(
   "key1" => "value1",
   "key2" => "value2",
   "key3" => "value3"
);

and the second one which doesn’t work returns something like


array(
   record1(array(
      "key1" => "value1",
      "key2" => "value2",
      "key3" => "value3"
      );
   ),
   record2(array(
      "key1" => "value1",
      "key2" => "value2",
      "key3" => "value3"
      );
   ),
   record3(array(
      "key1" => "value1",
      "key2" => "value2",
      "key3" => "value3"
      );
   )
);

Actually, all the records are coming from a database and in the case it doesn’t work, i wanna get several results from this database.

Does anyone have an idea of how i can make my Json answer correctly formated, or how i can achieve this if Json format is not the answer ?
Thanx

2

You can trying using jQuery.parseJSON(), the API seems to be down but see the example below for a quick reference.

var obj = jQuery.parseJSON('{"name":"John"}');
alert( obj.name === "John" );
3

I agree but it works well for 1 record.
Actually, i have this type of string format using json_encode() php function on an array with info from mysql, like this :


$id = $_GET['id'];
$retour = array();

if($id)
{
    $db = new DB();
    $db->query("SELECT * FROM familles WHERE idParent=".$id." ORDER BY ordre");
    $db->close();
    while($r = $db->next_record())
    {
        $temp['id'] = $r['id'];
        $temp['nom'] = $r['nom'];
        $retour[] = $temp;
    }
}
else
{
    $retour[0]['id'] = 'void';
}

echo(json_encode($retour));

, but i have those that seem to screw it all :

[{“id”:“25”,“nom”:“Pouf slim”},{“id”:“26”,“nom”:“Pouf g\u00e9ant”},{“id”:“27”,“nom”:“Pouf XXL”},{“id”:“28”,“nom”:“Pouf Int\u00e9rieur”}]

With only {“id”:“25”,“nom”:“Pouf slim”} (for instance), obj.id gives 25, but i wanna have several lines of DB records and not only 1.
How can i achieve this plz ?

4

Because your appending array values without setting an array key i believe that’s what is causing it. If you use the following does it work?

$id = $_GET['id'];
$retour = array('records' => array());

if($id)
{
    $db = new DB();
    $db->query("SELECT * FROM familles WHERE idParent=".$id." ORDER BY ordre");
    $db->close();
    while($r = $db->next_record())
    {
        $temp['id'] = $r['id'];
        $temp['nom'] = $r['nom'];
        $retour['records'] = $temp;
    }
}
else
{
    $retour[0]['id'] = 'void';
}

echo json_encode($retour);
5

So doing what you say, i will have (assuming there are 4 records in my table) :
1st loop ==>
$retour[‘records’][‘id’] = “25”;
$retour[‘records’][‘nom’] = “Pouf slim”;
2nd loop ==>
$retour[‘records’][‘id’] = “26”;
$retour[‘records’][‘nom’] = “Pouf g\u00e9ant”;
3rd loop ==>
$retour[‘records’][‘id’] = “27”;
$retour[‘records’][‘nom’] = “Pouf XXL”;
4th loop ==>
$retour[‘records’][‘id’] = “28”;
$retour[‘records’][‘nom’] = “Pouf Int\u00e9rieur”;
And in the end, i’ll only have 1 record in my variable with id=28 and nom=Pouf Int\u00e9rieur inside.

So i tried something else doing this :


$id = $_GET['id'];
$retour = array();

if($id)
{
    $db = new DB();
    $db->query("SELECT * FROM familles WHERE idParent=".$id." ORDER BY ordre");
    $db->close();
    $i = 0;
    while($r = $db->next_record())
    {
        $temp = array();
        $temp['id'] = $r['id'];
        $temp['nom'] = $r['nom'];
        $retour['record'.$i] = $temp;
        $i++;
    }
}
else
{
    $retour['record0']['id'] = 'void';
}

echo(json_encode($retour));

Then i obtain :

{“record0”:{“id”:“25”,“nom”:“Pouf slim”},“record1”:{“id”:“26”,“nom”:“Pouf g\u00e9ant”},“record2”:{“id”:“27”,“nom”:“Pouf XXL”},“record3”:{“id”:“28”,“nom”:“Pouf Int\u00e9rieur”}}

So i’ve managed to achieve what i wanted but there’s one remaining problem : when i wanna browse my record table, doing for(i = 0 ; i < data.length ; i++), where data is the return of “$retour”, data.length is undefined !
Doing it by hand setting for(i = 0 ; i < 4 ; i++), it works perfectly.
I’ve searched for info on JS objects but it seems it’s the right syntax.
object.length !!!
Have you got an idea ?

PS : BTW thanx a lot for your help :wink:

6

Just to test, i tried this (this is the format of my data) :


var test = new Object();
test['record0'] = new Object();
test['record0']['id'] = "1";
test['record0']['nom'] = "test1";
test['record1'] = new Object();
test['record1']['id'] = "2";
test['record1']['nom'] = "test2";
test['record2'] = new Object();
test['record2']['id'] = "3";
test['record2']['nom'] = "test3";

alert(test.length);

and it’s undefined !

7

OK here’s the trick (for the ones who are interested) :
The “length” propriety doesn’t exist for objects (associative arrays) !
You have to “simulate” it by doing a counting loop like this :


var n=0;
for(x in data)
{
    n++;
}

‘n’ is your data.length !
I hope it works on all platforms and all browsers (i only tested on WinXP/FF3.6) !!!

Thx for your support SgtLegend.

8

No problem, glad you got it sorted.