python - Flask handle_exception 不适用于 test_request_context

Question

我注意到，当 app.test_request_context 发生异常时，app.handle_exception 似乎不会被调用:

from flask import *

app = Flask(__name__)
app.handle_exception = lambda e: 'exception!'

@app.route('/foo')
def foo():
    x = 1 / 0
    return 'ok'

if __name__ == '__main__':
    #app.run(port=81) # handle_exception works here
    with app.test_request_context('/foo'):
        print app.dispatch_request() # but not here

这是预期的行为吗？

Answer 1

您可以轻松地覆盖此行为并使用相同的处理程序强制处理异常。

def run_test(path=None,check_func=None,*args,**kwargs):
    with app.test_request_context(path,*args,**kwargs):
        try:
            data=app.dispatch_request() 
            if check_func is not None:
                 check_func()
            else:
                print data
        except Exception as e:
            print app.handle_exception(e)

run_test('/')
run_test('/other')

def current_test(data):
     assert 'has some content' in data
run_test('/should_be_checked',check_func=current_test)

还有一句话。

你的方法不起作用，因为你只是没有使用 Flask 的那部分，它实际上捕获了异常。您正在直接调用上下文。

引用文档:

If you look into how the Flask WSGI application internally works, you will find a piece of code that looks very much like this:

def wsgi_app(self, environ):
    with self.request_context(environ):
        try:
            response = self.full_dispatch_request()
        except Exception, e:
            response = self.make_response(self.handle_exception(e))
        return response(environ, start_response)

<小时/>

但是!以下是正确的方法，因为每个级别上的所有 Flask 方法都将以适当的方式调用:

with app.test_request_context():
    with app.test_client() as client:
        resp = client.get('/')
        #and if you need content of response: print resp.data

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