jquery ajax array response handling

Question

How to handle a array response from .getjson. Following code handles ajax request.

function getinfo()
 {
     $query="select field_uname_value,field_uaddress_value,field_uphone_value from {content_type_udetails}";
     $result= db_query($query);
     $i=0;     
     while($item = db_fetch_object($result))
     {
      $info[$i++]=array($item->field_uname_value,$item->field_uaddress_value,$item->field_uphone_value);
     }
     print_r($info);
 }

and returns array as follows

Array
(
    [0] => Array
        (
            [0] => bharath
            [1] => 12th street ,ram nagar
            [2] => 213124442
        )

    [1] => Array
        (
            [0] => Christina
            [1] => 77,five corner
            [2] => 76874323
        )

    [2] => Array
        (
            [0] => Ranjan
            [1] => queen towers, 4th layout
            [2] => 45745747
        )

)

but following ajax handling not works. how to get the array response

 $.getJSON('/knpgetuserinfo.json&timestamp'+new Date().getTime(), function(data) {
   alert(data[0][0]); 
}

RobertPitt · Accepted Answer · 2010-10-02 11:26:55Z

You cant do that, print_r just prints teh array to readable text, you still some key/value separation form in there, and as your using getJSON i would say that you use json_encode()

function getinfo()
    {
        $query="select field_uname_value,field_uaddress_value,field_uphone_value from {content_type_udetails}";
         $result= db_query($query);
         $info = array();
         while($item = db_fetch_object($result))
         {
             $info[] = array(
                 'name' => $item->field_uname_value,
                 'address' => $item->field_uaddress_value,
                 'phone' => $item->field_uphone_value
             );
         }
         echo json_encode($info);
     }

and then use javascript like so:

$.getJSON('/knpgetuserinfo.json&timestamp'+new Date().getTime(), function(data) {
    $.each(data,function(item){
       //use item.name or item.address here
    });
}

Nick Craver · Accepted Answer · 2010-10-02 11:25:09Z

1

Instead of printing, you need it in JSON format via json_encode, by replacing:

print_r($info);

With:

echo json_encode($info);

The problem is you're not returning the data in a JSON format currently, so when jQuery goes to parse, it fails.

answered Oct 2, 2010 at 11:25

Nick Craver

631k138 gold badges1.3k silver badges1.2k bronze badges

2 Comments

RobertPitt Over a year ago

the json_encode returns a string, not an object of any type in PHP, should just be echo'd !

Nick Craver Over a year ago

@RobertPitt - Yup, when I slapped his code in a quick test it reminded me, already updated :)

Collectives™ on Stack Overflow

jquery ajax array response handling

2 Answers 2

Comments

2 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Comments

2 Comments

Your Answer

Sign up or log in

Post as a guest

Related