检索到json冗余数据

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

下面是正在讨论的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]);
}

在单击.editBtn按钮时，会弹出警告0:value0, 1:value1,...，然后再次发出警告，但在某种程度上，我只希望它是name0:value0, name1:value1,...

怎么一回事？

附注：php独立运行：

{"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

不完全确定此处返回的格式..通常，框架SQL对象以其自己的对象格式返回数据集。您可能希望将SQL数据传递到数组对象中，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));