CODEWITHSUNDEEP
phpajaxjson
Patrick J
Patrick J

Reputation: 23

Issue with a JSON Query

I Use a json onclick script for triggering a php function. I call the php function with:

$('#dtable').on('click', '[name=option3]', function () {
    var select = $(this);
    var id = select.attr('id');
    $.post('index.php/stop/'+id+'', function(json) {
            if (json && json.status) {
                $("#failure").show().delay(2500).fadeOut(1500);
            } else {
                $("#success").show().delay(2500).fadeOut(1500);
            }
        }
    );
} );

If the item is trigger he calls a slim property in my index.php

$admin->slim->post('/stop/:action', function($action) use ($admin) {
    $admin->slim->contentType('application/json');
    echo json_encode($admin->cont->Stop($action));

After this he triggert the function

function Stop($action) {
    $port = $action;
    $connection = @fsockopen($this->cfg->base_host, $port, $errno, $errstr, 1);
    if (!$connection) {
        return false;
    } else {
        $pid = $this->db->query("SELECT pid FROM testdb WHERE port='" . $port . "'", SQL_ALL, SQL_ASSOC);
        if ($pid == "") {
            return false;
        } else {
                  ......
                }
            }
        }
    }
}

the only thing is that the answer from json is always success but there is no return from true or false from the function

Does someone knows what i doing wrong or can someone give me some advice.

Upvotes: 0

Views: 85

Answers (1)

furas
furas

Reputation: 142631

Use alert(json); before if (json && json.status) to see what you get. I'm sure json is not what you expect. It seems json == false (because function Stop() return false) and there is no json.status - or maybe rest of your Stop function return something more.

Upvotes: 1

Related Questions