CODEWITHSUNDEEP
phpajaxarrays
Kenneth P.
Kenneth P.

Reputation: 1816

Error when returning array object

I have a problem when returning an array object and then display it to the user, please look at the demo code. A basic snippet but it has the same idea has been, I just can't post the very long code here.

Class foobar{
   public function foo()
   {
     return array( 'bar' => 'value' );
   }
}

This php code was used by another class

Class foobar_fetcher{
   public function getFoo()
   {
     $fb = new foobar();
     $result = $fb->foo();
     return $result;
   }
}

foobar_fetcher is again called by a main executioner file( ajaxdispatcher.php ) - with a json header.

if( isset( $_POST['fetch'] ) ){
   $httpresponse = new stdClass();
   $fb_fetch = new foobar_fetcher();
   $httpresponse->data = $fb_fetch->getFoo();
}

echo json_encode( $httpresponse );

Finally this ajaxdispatcher was called by a jquery ajax.

$.ajax({
  url: 'ajaxdispatcher.php',
  type: 'post',
  data: {fetch:'fetch'},
  success: function( data ){
      if( data ) console.log( data );
  }
});

Now, the when I try to print out the data , it has no response from the server. But when I change the return value of the foo() under foobar Class to an integer or string. Things will work fine.

Upvotes: 1

Views: 87

Answers (2)

Phil
Phil

Reputation: 164901

Some things I would do that may improve your chances of success

  1. Set appropriate HTTP headers and exit immediately after sending your JSON code

    header('Content-type: application/json');
echo json_encode($httpresponse);
exit;

    Also make sure you haven't sent any data to the output buffer prior to this.

  2. Tell jQuery the data-type to expect

    $.ajax({
    dataType: 'json',
    // and the rest

  3. Add an error callback

    $.ajax({
    // snip
    error: function(jqXHR, textStatus, errorThrown) {
        console.log(jqXHR, textStatus, errorThrown);
    }
});

Upvotes: 0

mcottingham
mcottingham

Reputation: 2056

You should try to change you ajaxdispatcher to accept a GET request and navigate there from a browser to see what is returned.

if( isset( $_GET['fetch'] ) ){
   $httpresponse = new stdClass();
   $fb_fetch = new foobar_fetcher();
   $httpresponse->data = $fb_fetch->getFoo();
}

echo json_encode( $httpresponse );

Navigate to /ajaxdispatcher.php?fetch=fetch

Upvotes: 2

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