JSON冗余数据检索

Question

$('.editBtn').click(function(){//EDIT BUTTON EVENT 
     $.getJSON('edit.php?url='+encodeURI($(this).siblings('a').attr('id'))+'&action=edit',function(data){ 
      $.each(data, function(key, val) { 
       alert(key+': '+val); 
      }); 
     }); 
    });//EDIT BUTTON END 

，这里是有问题的PHP部分：

elseif($_GET['action']=='edit'){ 
$output=$mysql->getDb()->query("select * from video 
    where url='{$_GET['url']}'")->fetchAll(); 
header("content-type: application/json"); 
echo json_encode($output[0]); 
} 

什么情况是，当点击.editBtn按钮，警报弹出0:value0, 1:value1,...，然后再进入，但在某种程度上，我只希望它要name0:value0, name1:value1,...

发生了什么事？

p.s. PHP的独立运行：这里返回

{"url":"www.vimeo.com\/20721308","0":"www.vimeo.com\/20721308","title":"Dis-patch Festival R.I.P.","1":"Dis-patch Festival R.I.P.","description":"Sadly, the last goodbyes to the Dis-patch Festival Belgrade edition in this tribute \"R.I.P.\" video collage. The end is always the beginning...","2":"Sadly, the last goodbyes to the Dis-patch Festival Belgrade edition in this tribute \"R.I.P.\" video collage. The end is always the beginning...","country":"serbia","3":"serbia","postDate":"2011-05-07 05:56:04","4":"2011-05-07 05:56:04","views":null,"5":null}

Answer 1

不完全确定的格式。通常框架SQL对象返回的数据在它自己的对象格式设置。你可能会希望在SQL数据传递到一个数组对象，然后将数组json_encode返回给客户端。

$returnData = array(
    'name' => $output[0]->name, 
    'videopath' => $output[1]->videopath 
); 

echo json_encode(array('success'=>1, 'data'=>$returnData));