การส่งค่ากลับเป็น json [header('Content-Type: application/javascript');]

 <?php
           $link = mysql_pconnect("localhost","root","1234") or die("Unable to connect server");
           mysql_selectdb("test") or die("Unable to connect database");
           mysql_query("SET NAMES 'UTF8'");
           mysql_query("SET character_set_results='UTF8'");
           
           $arr = array();
           
           $rs = mysql_query("SELECT id,name,address,username,password FROM tbl_test");
           
           $i = 0;
           while($obj = mysql_fetch_object($rs))
           {
               $arr[] = $obj;
               //echo $i." ".$obj->address."<br/>";
               //$i+=1;
           }
           // add the header line to specify that the content type is JSON
           header('Content-Type: application/javascript'); // <--- เพิ่มนี้เข้าไปแล้ว error เลย
           echo "{\"NetBeansTest1\":" .json_encode($arr). "}";
?>