CODEWITHSUNDEEP
phpjqueryarraysjson
Norbert
Norbert

Reputation: 2771

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

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?

Upvotes: 2

Views: 8948

Answers (6)

TNS MUMET
TNS MUMET

Reputation: 11

Use .getJSON. Try:

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

Upvotes: 1

Ken Egozi
Ken Egozi

Reputation: 1835

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() ]);

Upvotes: 0

Mark Holland
Mark Holland

Reputation: 846

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/

Upvotes: 2

Brian Flanagan
Brian Flanagan

Reputation: 4642

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)

Upvotes: 1

Mark Coleman
Mark Coleman

Reputation: 40863

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

Upvotes: 2

Darin Dimitrov
Darin Dimitrov

Reputation: 1039598

Try:

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

Upvotes: 5

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