擅长:python、mysql、java
<p><strong>此解决方案影响所有视图:</strong></p>
<p>当然,您可以提供自定义异常处理程序:<a href="http://www.django-rest-framework.org/api-guide/exceptions/#custom-exception-handling" rel="noreferrer">Custom exception handling</a></p>
<pre><code>from rest_framework.views import exception_handler
from rest_framework import status
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.status_code == status.HTTP_404_NOT_FOUND:
response.data['custom_field'] = 'some_custom_value'
return response
</code></pre>
<p>确定您可以跳过默认值<code>rest_framework.views.exception_handler</code>并使其完全原始。在</p>
<p>注意:请记住在<code>django.conf.settings.REST_FRAMEWORK['EXCEPTION_HANDLER']</code>中提到您的处理程序</p>
<p><strong>针对特定视图的解决方案:</strong></p>
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