jQuery $.get() Array Returns [object Object]

Question

test.php includes this:

echo json_encode( array(
  array("name"=>"John","time"=>"2pm"),
  array("name"=>"2","time"=>"1242pm"),
  array("name"=>"J231ohn","time"=>"2p213m"),
));

jQuery:

$.get("test.php", function(data) {
  $.each(data, function(n, val) {
    alert(n + ': ' + val)
  });
}, "json");

This is the result:

0: [object Object]
1: [object Object]
2: [object Object]

What am I doing wrong?

related: [Javascript - \[object Object\] means?](http://stackoverflow.com/q/8892465/1048572) — Bergi, Nov 04 '14 at 16:14

score 5 · Accepted Answer · answered Feb 23 '11 at 19:05

5

Try:

alert(n + ': name = ' + val.name + ' time = ' + val.time);

answered Feb 23 '11 at 19:05

Darin Dimitrov

1,023,142
271
3,287
2,928

score 2 · Answer 2 · answered Feb 23 '11 at 19:05

2

I dont know php but my guess is you need to do this instead, as each val is a json object.

$.get("test.php", function(data) {
  $.each(data, function(n, val) {
    alert(n + ': ' + val.name + ' ' + val.time)
  });
}, "json");

jsfiddle example

answered Feb 23 '11 at 19:05

Mark Coleman

40,542
9
81
101

score 2 · Answer 3 · answered Feb 23 '11 at 19:13

2

Use console.log(data) to get a decent look at what's inside your JSON from the console in Firebug or Webkit.

Great tutorial here: http://jqueryfordesigners.com/debugging-tools/

answered Feb 23 '11 at 19:13

Mark Holland

846
4
8

score 1 · Answer 4 · answered Sep 20 '17 at 03:26

1

Use .getJSON. Try:

$.getJSON("test.php", function(data) {
     $.each(data, function(n, val) {
         alert(n + ': ' + val)
     });
});

answered Sep 20 '17 at 03:26

TNS MUMET

11
1

score 1 · Answer 5 · answered Feb 23 '11 at 19:07

Technically, nothing. 'val' references the object, which I suspect isn't what you want. You probably want the values stored in each array. So, instead of alerting out 'val', you probably want to access the array key values:

alert(n + ': ' + val.name + ' ' + val.time)

score 0 · Answer 6 · answered Feb 23 '11 at 19:14

0

depending on your browser, you could call val.toSource() which will spill the objects contents instead of its type (which is the default behaviour for .toString() )

a nice, shorthand way to write this is alert ([n, val.toSource() ]);

answered Feb 23 '11 at 19:14

Ken Egozi

1,825
11
14

jQuery $.get() Array Returns [object Object]

6 Answers6