如果视图代码抛出异常,所有django 事务的东西都会回滚事务。HttpResponse
如果我返回带有错误状态代码的错误状态代码,即 4xx 或 5xx,如何让它回滚?
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1 回答
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好吧,我根本没有测试过,我不知道它是否正确,但这样的事情可能会奏效。我将不胜感激任何反馈!
from functools import wraps
from django.utils.decorators import available_attrs
from django.db import transaction
from django.http import HttpResponse
def commit_on_http_success(function=None, using=None, successStatuses=[200]):
"""Like commit_on_success, but rolls back the commit if we return an HttpResponse
with a status that isn't 200 (by default).
Args:
using: The database to use. Uses the default if None.
successStatuses: The HTTP statuses for which we will commit the transaction.
Raises:
It can raise an exception that overrides the view's exception if commit() or
rollback() fails.
"""
def decorator(view_func):
@wraps(view_func, assigned=available_attrs(view_func))
def _wrapped_view(request, *args, **kwargs):
try:
transaction.enter_transaction_management(using=using)
transaction.managed(True, using=using)
exc = None
commit = False
try:
response = view_func(request, *args, **kwargs)
if isinstance(response, HttpResponse) and response.status in successStatuses:
commit = True
except Exception, exc:
pass
if transaction.is_dirty(using=using):
if commit:
try:
transaction.commit(using=using)
except:
transaction.rollback(using=using)
raise
else:
transaction.rollback(using=using)
finally:
transaction.leave_transaction_management(using=using)
if exc is not None:
raise exc
return response
return _wrapped_view
if function:
return decorator(function)
return decorator
于 2012-11-28T18:20:56.497 回答