在flask应用程序中破解伪造者

2024-03-29 08:15:57 发布

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我有一个带有websockets的flask应用程序,当一个人点击一个套接字开始线程时,我希望它运行一个线程,比如:

@socketio.on('start', namespace='/ws')
def patrol():
    asset = {'x': 0, 'y': 1}
    while True:
        thread_patrol(asset, [[0, 0], [400, 400]])


def patrol(asset, coordinates):
    count = 0
    import itertools
    for coordinate in itertools.cycle(coordinates):
        val = True
        while val:
            asset, val = take_step(asset, coordinate[0], coordinate[1])
            emit('asset',
                 {'data': asset, 'count': count},
                 broadcast=True)
            count += 1
            time.sleep(1)


import threading
def thread_patrol(asset, coordinates):
    print('threading!')
    patrolling_thread = threading.Thread(target=patrol, args=(asset, coordinates))
    patrolling_thread.start()

def take_step(asset, x, y):
    asset[x] = x
    asset[y] = y

但是我得到一个错误,因为它在请求上下文之外。我需要做什么才能让我的应用程序线程化?公司名称:

threading!
Exception in thread Thread-2005:
Traceback (most recent call last):
  File "/usr/local/Cellar/python/2.7.10_2/Frameworks/Python.framework/Versions/2.7/lib/python2.7/threading.py", line 810, in __bootstrap_inner
    self.run()
  File "/usr/local/Cellar/python/2.7.10_2/Frameworks/Python.framework/Versions/2.7/lib/python2.7/threading.py", line 763, in run
    self.__target(*self.__args, **self.__kwargs)
  File "app2.py", line 270, in patrol
    broadcast=True)
  File "/usr/local/lib/python2.7/site-packages/flask_socketio/__init__.py", line 520, in emit
    namespace = flask.request.namespace
  File "/usr/local/lib/python2.7/site-packages/werkzeug/local.py", line 338, in __getattr__
    return getattr(self._get_current_object(), name)
  File "/usr/local/lib/python2.7/site-packages/werkzeug/local.py", line 297, in _get_current_object
    return self.__local()
  File "/usr/local/lib/python2.7/site-packages/flask/globals.py", line 20, in _lookup_req_object
    raise RuntimeError('working outside of request context')
RuntimeError: working outside of request context

Tags: inpyselftrueflasklibusrlocal
1条回答
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1楼 · 发布于 2024-03-29 08:15:57

您(我)必须包含thread.daemon = True以通知应用程序将其作为后台进程运行,我删除了broadcast=True,因为这无论如何都不是必需的。你知道吗

def thread_patrol(asset, coordinates):
    patrolling_thread = Thread(target=patrol, args=(asset, coordinates))
    thread.daemon = True
    thread.start()

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