Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
100 views
in Technique[技术] by (71.8m points)

javascript - jQuery AJAX Call to PHP Script with JSON Return

I've been smashing my head against a brick wall with this one, i've tried loads of the solutions on stackoverflow but can't find one that works!

Basically when I POST my AJAX the PHP returns JSON but the AJAX shows Undefined instead of the value:

JS:

  /* attach a submit handler to the form */
  $("#group").submit(function(event) {

  /* stop form from submitting normally */
  event.preventDefault();

  /*clear result div*/
  $("#result").html('');

  /* get some values from elements on the page: */
  var val = $(this).serialize();

  /* Send the data using post and put the results in a div */
  $.ajax({
      url: "inc/group.ajax.php",
      type: "post",
      data: val,
  datatype: 'json',
      success: function(data){
            $('#result').html(data.status +':' + data.message);   
            $("#result").addClass('msg_notice');
            $("#result").fadeIn(1500);           
      },
      error:function(){
          $("#result").html('There was an error updating the settings');
          $("#result").addClass('msg_error');
          $("#result").fadeIn(1500);
      }   
    }); 
});

PHP:

  $db = new DbConnector();
  $db->connect();
  $sql='SELECT grp.group_id, group_name, group_enabled, COUNT('.USER_TBL.'.id) AS users, grp.created, grp.updated '
        .'FROM '.GROUP_TBL.' grp '
        .'LEFT JOIN members USING(group_id) '
        .'WHERE grp.group_id ='.$group_id.' GROUP BY grp.group_id';

    $result = $db->query($sql);     
    $row = mysql_fetch_array($result);
    $users = $row['users'];
    if(!$users == '0'){
        $return["json"] = json_encode($return);
        echo json_encode(array('status' => 'error','message'=> 'There are users in this group'));
    }else{

        $sql2= 'DELETE FROM '.GROUP_TBL.' WHERE group_id='.$group_id.'';
        $result = $db->query($sql2);

        if(!$result){
            echo json_encode(array('status' => 'error','message'=> 'The group has not been removed'));
        }else{
            echo json_encode(array('status' => 'success','message'=> 'The group has been removed'));
        }
    }

JSON Result from firebug:

{"status":"success","message":"success message"}

AJAX Displays the JSON result as Undefined and I dont have a clue why. I have tried displaying adding dataType='json' and datatype='json'. I have also tried changing it to data.status and data['status']: still no joy though.

Any help would be really appreciated.

question from:https://stackoverflow.com/questions/19155192/jquery-ajax-call-to-php-script-with-json-return

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

Make it dataType instead of datatype.

And add below code in php as your ajax request is expecting json and will not accept anything, but json.

header('Content-Type: application/json');

Correct Content type for JSON and JSONP

The response visible in firebug is text data. Check Content-Type of the response header to verify, if the response is json. It should be application/json for dataType:'json' and text/html for dataType:'html'.


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...