JSON.parse意外令牌多维数组

Question

Hello I have simple question. I have some data fetched from database and json encoded and sent to html file over AJAX from PHP script. I want to JSON parse that data. Data looks like this:

And in PHP file like this:

$profesors = array
                (
                'id' => ($id),
                'name' => ($profesor)
                );

When I try to JSON parse it with JSON.parse() command I get "unexpected token {". Does anyone have some suggestion what to do? Also if it would be easier for achieving my goal I can use some other data type than array I am using, if someone knows better way...

EDIT

my PHP code that generates 2d array

<?php
$var=$_POST["oblast"];
$connection = mysql_connect('localhost','root','pass');
if (!$connection) {die("not successfull" . mysql_error());}

$result = mysql_query('set character set utf8', $connection); 
$result = mysql_query('set names utf8', $connection);
$db_select = mysql_select_db("fakultet",$connection);
$brojac=0;
$profesor="";
//$data = array(array());
$queryData = mysql_query("SELECT * FROM predmet WHERE idpred = '$var'");
while($result1 = mysql_fetch_array($queryData))
{
$prof=$result1['idprof'];
    $queryData1 = mysql_query("SELECT * FROM profesori WHERE idprof = '$prof'");
        while($result2 = mysql_fetch_array($queryData1))
        {

            $id=$result2['idprof'];
            $profesor=$result2['ime']." ".$result2['prezime'];
            $profesors = array
                (
                'id' => ($id),
                'name' => ($profesor)
                );

            echo json_encode($profesors);

        }
    //echo($result1['idprof']);
    //$data[]=$result['idprof'];

}

//echo "$data";
?>

Answer 1

The unexpected token is the second opening curly brace. It's expecting one object because there are no array brackets. Either have your response look like this:

[{
    "id"   : "1",
    "name" : "Ivan Nikolaj"
 },
 {
    "id"   : "2",
    "name" : "Zdravko Topic"
 }]

or this :

{
   "id"   : "1",
   "name" : "Ivan Nikolaj"
}

or even better, this:

 {
     "status"   : 200,
     "response" : [
         {
             "id"   : 1,
             "name" : "Ivan Nikolaj"
         },
         {
             "id"   : 2,
             "name" : "Zdravko Topic"
         }
     ]
 }

Answer 2

You are doing echo json_encode($profesors); inside your while loop. Don't do that. You should only ever call json_encode once .

Try something like this:

$profesors = array();

while($result2 = mysql_fetch_array($queryData1))
{
    $id=$result2['idprof'];
    $profesor=$result2['ime']." ".$result2['prezime'];
    $profesors[] = array(
        'id' => $id,
        'name' => $profesor
    );
}

echo json_encode($profesors);