1

我正在开发一个由 ajax 请求调用的 php 脚本,它应该回显一些要解析的响应代码。

例如:

die('#1:200|Success|'.$data);

或者另一个:

die('#0:100|Access Denied|');

现在,我还想包括在脚本执行期间可能发生的任何错误或警告 - 但附加在消息的末尾。

那么,将所有错误捕获到某个变量中的优雅方法是什么?

编辑

嗯,理解它的用途并不容易,手册在很多方面都不清楚。

但是好的,我会尝试举例说明我是如何理解它的,如果我弄错了,请指出:-)

//So guess first I off error reporting that would naturally occur.
error_reporting(0);

//Then I will define array to stuff the errors in.
$errors=array();

//Then I make my handler function.
function handler($errno,$errstr){
    global $errors;
    $errors[]=$errno.': '.$errstr;    //Stuff it into array.
}

//Then I define handler.
set_error_handler('handler',E_ALL);

这是正确的用法吗?

它还说:

The following error types cannot be handled with a user defined function: E_ERROR, E_PARSE, E_CORE_ERROR, E_CORE_WARNING, E_COMPILE_ERROR, E_COMPILE_WARNING, and most of E_STRICT raised in the file where set_error_handler() is called.

还有一个问题,为什么它不能捕获严格的错误?

4

1 回答 1

1

我总是需要这个来在 ajaxing 时捕获错误:

header('content-type: application/json; charset=utf-8');
error_reporting(E_ALL);ob_start();

function error($msg,$do = false)
{
    //personal error message
    if(!isset($_SESSION))session_start();       
    trigger_error($msg."\n".(isset($_SESSION)?"[".$_SESSION['id']."|".$_SESSION['name']."]":"")."-----------");

    ob_clean(); 
    die(json_encode(array($msg,$do)));
}
function ob_error($msg = "Error!",$do = "ob_error")
{
    if($s = ob_get_clean())
    error("$msg\nDetails:\n$s",$do);
}

用法:

//require the php above

//do something

//call error("acces denied") if there is an fatal error

//do anything else

//call ob_error() at the end: if there was anything outputted like warning/notice it will shown

//call die(json_encode(array(1, ...anything you need*...))); - this will run only if there was nothing displayed

客户网站使用情况:

$.post('/_ajax/... .php',{
            'param1':...
            },function(a){
                if(a[0]=="1") //OK
                {                   
                    //* do something with a[n]                  
                }else{ //onError
                    alert(a[0]); 
                    //what to do client site after the error
                    if(a[1]=="refresh") 
                        location.href=location.href;
                }
            },"json").error(function() {alert("postError"));
于 2012-08-13T13:34:19.820 回答