Question

我有一个简单的ajax json请求。

这是客户端javascript：

    function ajax(){
    //alert('hello');
    $.ajax({
         url: '../ajax/goal_ajax.php',
         dataType: 'json',
         success: function( data ){
          // success! :D
          alert('hello');
         }, error: function( data ){
          // data.responseText is what you want to display, that's your error.
          alert('die');
         }
    })
    //progressBar.set('value',data.total);
    //document.getElementById('txtCDInfo').innerHTML=txt; 
}

这是php处理程序：

    <?php  
include ("../includes/db_con.php");

    $itemResults = mysql_query("SELECT `sold_for` FROM `items` WHERE `item_did_sell`='1'") or die();
    $mIResults = mysql_query("SELECT `mi_price`, `mi_total_sold` FROM `misc_items` WHERE `mi_total_sold`>'1'") or die();
    $donationResults = mysql_query("SELECT `amount` FROM `donations`") or die(mysql_error());

$total = 0;
$itemTotal = 0;
$mITotal = 0;
$donationTotal = 0;

while($row = mysql_fetch_assoc($itemResults)){  
    $itemTotal += $row['sold_for'];
    $total += $itemTotal;
}

while($row = mysql_fetch_assoc($mIResults)){  
    $mITotal += ($row['mi_price'] * $row['mi_total_sold']);
    $total += $mITotal;
}

while($row = mysql_fetch_assoc($donationResults)){  
    $donationTotal += $row['amount'];
    $total += $donationTotal;
}

header("Content-Type: application/json");
echo json_encode(array("items" => $itemTotal, "mitems" => $mITotal, "donations" => $donationTotal, "total" => $total));

include ("../includes/db_discon.php"); 
?>

当我在客户端页面上调用函数ajax()时，会出现一个警告框，显示die，显然意味着请求失败。但是，我没有在页面上收到404错误，所以它找到了php处理程序。

php脚本echo：

{"items":1000,"mitems":0,"donations":0,"total":1000}

编辑：

data.responseText返回{"items":1000,"mitems":0,"donations":0,"total":1000}

这个JSON请求有什么问题？

0 个答案: