jQuery Ajax失败并返回异常? [英] JQuery Ajax fail and return exception?

问题描述

我已经阅读了几篇有关JQuery Ajax调用失败参数的文章，但没有一篇直接回答我的问题.如果您想在这里阅读我的出发点，那么这篇文章将是一个不错的开始:

I have read a few posts on fail parameters for a JQuery Ajax call, but none that have directly answered my question. If you want to read up on my jumping off point here, this post would be a good start:

jquery:是否有$的失败处理程序.post在Jquery中?

我的问题是，有一些事情可能导致我的应用程序失败-如果我的脚本返回false，则上述方法对我来说就可以工作(如果我正确理解的话)，但是大多数情况下，我的脚本会因为踢出我使用Zend Framework处理的异常而失败.我希望返回该异常，以便为用户提供更详细的消息.可以让我的PHP脚本返回一个值，同时仍然让Ajax调用知道这是一个失败吗?

My problem is that there are a handful of things that may cause my application to fail - if my script returns false the above method would work just fine for me (if I understand it correctly), but most of the time my script will fail by kicking out an exception that I handle using the Zend Framework. I would prefer to return the exception so that I can provide the user with a more detailed message. Is it possible to have my PHP script return a value while still letting the Ajax call know that it was a failure?

推荐答案

可以.首先，您需要对错误进行分类，例如:

Sure you can. First of all you need to categorize you errors, for example:

• 致命
• 例外
• 错误/错误状态

我建议您将返回值用于正确且无错误的处理-0 .在所有其他情况中，这都是错误.

I would advise you to take as a return value for correct and with no errors processing - 0. In all other case that would be an error.

另一个有用的建议是将JSON用作客户端-服务器对话.

Another useful advise would be to use JSON as a client-server conversation.

在PHP中，它将是:

function prepareJSONResponse($code, $message, array $extra = array())
{
    return json_encode(array_merge(
        $extra, array(
            'code'    => (int) $code,
            'message' => $message)));
}

在这种情况下，您可以为此调用传递错误代码和消息以及$ extra中的其他参数:

In this case you could pass error code and message, and additional params in $extra, for example, for this call:

prepareJSONResponse(1, 'Not enough data passed', array('debug' => true));

来自服务器端的响应将是:

Response from server side would be:

{code:1,message:'Not enough data passed','debug': true}

对于客户端，您需要$ .ajax的包装函数:

For the client side you need a wrapper function for $.ajax:

// calback(result, error);
function call(url, params, callback)
{
    if (typeof params == 'undefined') {
        params = {};    
    }

    $.ajax({
        'type'      : "POST",
        'url'       : url,
        'async'     : true,
        'data'      : params,
        'complete'  : function(xhr) {
            if (xhr.status != 200) {
                if (typeof callback == 'function') {
                    callback(xhr.responseText, true);
                }
            } else {
                if (typeof callback == 'function') {
                    callback(xhr.responseText, false);
                }
            }
        }
    });
}

并具有验证JSON的功能，以便确定格式是否损坏.

and function to validate JSON, in order if the corrupted format comes.

function toJSON(data){
    try {
        data = JSON.parse(data);
    } catch (err) {
        data = { 'code' : -999, 'message' : 'Error while processing response' };
    }

    if (typeof data.debug != 'undefined') {
        console.log(data.debug);
    }

    return data;
}

在try-catch中包装您的代码，然后在catch语句中执行以下操作:

Wrap in try-catch you code, and in catch statement do something like:

try {
    ...
} catch (Exception $e) {
    exit(prepareJSONResponse(1, $e->getMessage(), array(
        'debug' => 'There was an error while I were processing your request')));
}

结果将是您在浏览器控制台中收到调试信息，并且可以处理错误/异常(prepareJSONResponse())和致命错误(通过读取HTTP状态标头，如果不是200，则表示错误).

The result would be that you receive debug information in browser console, and could handle errors / exception (prepareJSONResponse()) and fatals (by reading HTTP-status header, if it's not 200, then there was error).

希望这就是您的要求.

这篇关于jQuery Ajax失败并返回异常?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！