自定义异常处理程序无法按django-rest-framework中所述工作 [英] Custom exception handler not working as documented in django-rest-framework

问题描述

我正在尝试在django-rest-framework中编写一个自定义异常处理程序，并且代码与示例中给出的相同：

I'm trying to write a custom exception handler in django-rest-framework, and the code is the same as given in the example:

from rest_framework.views import exception_handler

def custom_exception_handler(exc, context):
    # Call REST framework's default exception handler first,
    # to get the standard error response.
    response = exception_handler(exc, context)

    # Now add the HTTP status code to the response.
    if response is not None:
        response.data['status_code'] = response.status_code

    return response

但是在视图中引发异常时，这不起作用，而是抛出以下消息：

But on raising an exception from the view, this does not work, it instead throws this message:

custom_exception_handler（）缺少1个必需的位置参数：'context'

我尝试将第一个参数设置为无，例如：

I've tried setting the first argument to None, like so:

def custom_exception_handler（exc，context = None）：

但这会发生：

exception_handler（）需要1位置参数，但给出了2个

因此，似乎 rest_framework.views.exception_handler 需要

确实是这种情况：

So it seems rest_framework.views.exception_handler takes only one argument.
Indeed this is the case:

def exception_handler(exc):
    """
    Returns the response that should be used for any given exception.

    By default we handle the REST framework `APIException`, and also
    Django's built-in `ValidationError`, `Http404` and `PermissionDenied`
    exceptions.

    Any unhandled exceptions may return `None`, which will cause a 500 error
    to be raised.
    """

所以我的问题是，这是一个错误吗？还是我错过了。

So my question is, is this a bug? Or am i missing something, and there is another way to do this?..

编辑：

此信息已由rest_framework团队正式确认，已在最新版本中添加，因此似乎使用v3.0.2不会反映新文档。
https://github.com/tomchristie/django-rest-framework/issues/2737

This has been confirmed officially by the rest_framework team. This has been added in the latest version so it seems using v3.0.2 will not reflect the new documentation.
https://github.com/tomchristie/django-rest-framework/issues/2737

推荐答案

我们可以使用API​​Exception从API视图引发异常，或创建扩展APIException的自定义Exception类。

We can raise exception from API View using APIException or create custom Exception class that extends APIException.

raise APIException("My Error Message")

现在自定义异常处理程序为

Now Custom Exception handler is

def custom_exception_handler(exc, context):
    # Call REST framework's default exception handler first,
    # to get the standard error response.
    response = exception_handler(exc, context)
    if isinstance(exc, APIException):
        response.data = {}
        response.data['error'] = str(exc)
    elif isinstance(exc, MyCustomException):
        response.data = {}
        response.data['error'] = str(exc)
    return response

这篇关于自定义异常处理程序无法按django-rest-framework中所述工作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！