Question

I want to pass the result of a query through a $.post.

function GetAllTasks()
    {
        $sql = "select t.id as task_id,
                description,
                createdat,
                createdby,
                max_requests,
                max_duration,
                j.name as job_name 
        from darkfuture.tasks t, darkfuture.jobs j
        where t.job_id = j.id"; 



$sqlresult = mysql_query($sql)

or die("The list of works failed: ".mysql_error($this->con));

$result = array();

while($row = mysql_fetch_assoc($sqlresult))
{
    $task = new TasksResult();
    $task->id = $row["task_id"];
    $task->description = $row["description"];
    $task->createdat = $row["createdat"];
    $task->createdby = $row["createdby"];
    $task->max_requests = $row["max_requests"];
    $task->max_duration = $row["max_duration"];
    $task->job_id = $row["job_name"];
    array_push($result, $task);
}

mysql_free_result($sqlresult);
return $result;

}

Here is how i call it:

$tasksDB = new TasksDB();
$tasks = $tasksDB->GetAllTasks();

Now i want to pass $tasks through here:

$.post("views/insert_tasks.php",{'tasks[]': $tasks}, function(data)
    {

});

I know this {‘tasks[]’: $tasks} it’s wrong but i don’t know how to do it right.

Answer 1

<script type="text/javascript">
$.post("views/insert_tasks.php", {tasks:<?php echo json_encode($tasks); ?>}, function(data) {

});
</script>

On a side note you might want to look at mysql_fetch_object($result, 'TasksResult');