u013074424
Discuz3
2017-06-01 09:20

PHP ajax不执行success回调而是执行了error回调

  • php
  • ajax
<script>
 $.ajax({
        url: 'ajax.php?act=personal',
        type: 'post',
        dataType: 'json',
        data: { name:name, phone:$('#phone').val(), alipay:$('#alipay').val(), weixin:$('#weixin').val(), qq:$('#qq').val(), email:$('#email').val() },
        success: function(data){ 
            if(data.success){
                toast_tip(保存成功);
            } else {
                toast_tip(保存失败);
            }
        },
        error: function(XMLHttpRequest, textStatus, errorThrown) {
            alert(XMLHttpRequest.status);
            alert(XMLHttpRequest.readyState);
            alert(textStatus);
        },
</script>
    });

php代码

if(isset($_GET['act'])){

    switch ($_GET['act']) {

//保存资料////////////////////////////////////////////////
        case 'personal':
            try {
                //修改数据
                $stmt = $db->prepare('UPDATE members SET Name=:Name, Phone=:Phone, AliPay=:AliPay, WeiXin=:WeiXin, QQ=:QQ, EMail=:EMail WHERE UserName=:UserName') ;
                $stmt->execute(array(':Name' => $name, ':Phone' => $phone, ':AliPay' => $alipay, ':WeiXin' => $weixin, ':QQ' => $qq, ':EMail' => $email, ':UserName' => $_SESSION['UserName'] ));
            } catch(PDOException $e) {
                $error[] = $e->getMessage();
            }

            echo '{"message":"保存成功","success":true}';
        break;
        }
}

修改数据能成功, 但是老是回调error
返回代码 200, 4, parsererror

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