Question

在PHP代码和Ajax中添加了标题，但仍然没有任何乐趣。

在使用Ajax处理表单提交时，我无法理解处理MySQL错误。我有以下代码：

var url = "save.php"; // the script where you handle the form input.

    $.ajax({
        type: "POST",
        url: url,
        data: $("#frmSurvey").serialize(), // serializes the form's elements.


        success: function(jqXHR, textStatus, errorThrown){

                $('.sequence-container div').hide().delay( 2000 );
                $next.show().delay( 2000 );



        },
        error: function(jqXHR, textStatus, errorThrown){
            if (jqXHR.status === 0) {
                            alert('Not connect.\n Verify Network.');
                        } else if (jqXHR.status == 503) {
                alert('Error: ' + jqHXR.responseText);
            }    else if (jqXHR.status == 404) {
                            alert('Requested page not found. [404] - Click \'OK\' and try to re-submit your responses - if this error box re-appears, call 01255 850051 and we will assist.');
                        } else if (jqXHR.status == 500) {
                            alert('Internal Server Error. [500] - Click \'OK\' and try to re-submit your responses - if this error box re-appears, call 01255 850051 and we will assist');
                        } else if (exception === 'parsererror') {
                            alert('Requested JSON parse failed - Click \'OK\' and try to re-submit your responses - if this error box re-appears, call 01255 850051 and we will assist');
                        } else if (exception === 'timeout') {
                            alert('Time out error - Click \'OK\' and try to re-submit your responses - if this error box re-appears, call 01255 850051 and we will assist');
                        } else if (exception === 'abort') {
                            alert('Ajax request aborted - Click \'OK\' and try to re-submit your responses - if this error box re-appears, call 01255 850051 and we will assist');
                        } else {
                            alert('Uncaught Error.\n' + jqXHR.responseText + ' - Click \'OK\' and try to re-submit your responses - if this error box re-appears, call 01255 850051 and we will assist');
                        }
          }

    });

这样可以正常工作并处理任何错误，例如页面丢失等。

但是，我完全不知道如何处理/向用户显示实际提交到数据库时可能出现的任何错误。

目前我已尝试过这个：

if ($mysql_error!="") {
        header('HTTP/1.1 503 Service Unavailable');
                    printf("Unexpected database error: %s\n", $mysql_error);
        mysqli_stmt_close($proc);
        mysqli_clean_connection($link);
        exit();
    }

但由于页面未显示（因为它是由Ajax提交的），因此不会显示错误消息。我想我必须将该错误消息带回Ajax脚本以显示给用户（是吗？） - 我不知道该怎么做。

任何指针/建议？

Answer 1

发送503-Header

<?php
header('HTTP/1.1 503 Service Unavailable');

并且你的ajax代码应该可以工作。

修改它将使用您的失败代码，您可以在函数中检查状态代码503.

jqXHR.status

将是503和

jqXHR.responseText

将包含您的消息。

<强> EDIT2： js.file

var url = "save.php"; // the script where you handle the form input. $.ajax({ type: "POST", url: url, data: $("#frmSurvey").serialize(), // serializes the form's elements. success: function(jqXHR, textStatus, errorThrown){ if (jqXHR.status == 404) { alert('Error: ' + jqHXR.responseText); }else{ $('.sequence-container div').hide().delay( 2000 ); $next.show().delay( 2000 ); } }, error: function(jqXHR, textStatus, errorThrown){ if (jqXHR.status === 0) { alert('Not connect.\n Verify Network.'); } else if (jqXHR.status == 404) { alert('Requested page not found. [404] - Click \'OK\' and try to re-submit your responses - if this error box re-appears, call 01255 850051 and we will assist.'); } else if (jqXHR.status == 500) { alert('Internal Server Error. [500] - Click \'OK\' and try to re-submit your responses - if this error box re-appears, call 01255 850051 and we will assist'); } else if (errorThrown === 'parsererror') { alert('Requested JSON parse failed - Click \'OK\' and try to re-submit your responses - if this error box re-appears, call 01255 850051 and we will assist'); } else if (errorThrown === 'timeout') { alert('Time out error - Click \'OK\' and try to re-submit your responses - if this error box re-appears, call 01255 850051 and we will assist'); } else if (errorThrown === 'abort') { alert('Ajax request aborted - Click \'OK\' and try to re-submit your responses - if this error box re-appears, call 01255 850051 and we will assist'); } else { alert('Uncaught Error.\n' + jqXHR.responseText + ' - Click \'OK\' and try to re-submit your responses - if this error box re-appears, call 01255 850051 and we will assist'); } }

}）;

save.php：

<?php header('HTTP/1.1 503 Service Unavailable'); echo('hallo welt'); exit();

结果：

Uncaught Error. hallo welt - Click 'OK' and try to re-submit your responses - if this error box re-appears, call 01255 850051 and we will assist

Answer 2

我会把你的所有错误逻辑从jQuery移到PHP。您可以使用一个简单的JSON对象进行响应，该对象可以保存status（成功或错误），code（如果需要），message，甚至data，如果您需要提供具体结果。

例如，您提出这样的请求：

$.ajax({
  type: 'POST',
  url: url,
  data: $("#frmSurvey").serialize(),
  success: function(result){
    var json = $.parseJSON(result);
    if(json.response.status == 'success') {
      // do something
    } else {
      // look at message or code to perform specific actions
    }
  }
});

然后在处理此请求的PHP文件中，构建一个包含所需上述所有元素的数组（状态，代码，消息等）。最终，你会echo这样的事情：

$result = array(
  'response' => array(
    'status' => 'error',
    'code' => '1', // whatever you want
    'message' => 'Could not connect to the database.'
  )
);    

echo json_encode($result);

$result数组将包含基于您在PHP中进行的检查的相关数据。

希望这有帮助！

使用Ajax时处理MySQL错误

2 个答案: