我对Python和多线程编程非常陌生。基本上,我有一个脚本,将文件复制到另一个位置。我想把这个放在另一个线程,这样我就可以输出....表示脚本仍在运行。
我遇到的问题是,如果文件不能复制,它将抛出异常。如果在主线程中运行,这是可以的;但是,使用以下代码是无效的:
try:
threadClass = TheThread(param1, param2, etc.)
threadClass.start() ##### **Exception takes place here**
except:
print "Caught an exception"
在线程类本身中,我试图重新抛出异常,但它不起作用。我在这里看到有人问类似的问题,但他们似乎都在做一些比我试图做的更具体的事情(我不太理解所提供的解决方案)。我看到有人提到sys.exc_info()的用法,但我不知道在哪里或如何使用它。
编辑:线程类的代码如下:
class TheThread(threading.Thread):
def __init__(self, sourceFolder, destFolder):
threading.Thread.__init__(self)
self.sourceFolder = sourceFolder
self.destFolder = destFolder
def run(self):
try:
shul.copytree(self.sourceFolder, self.destFolder)
except:
raise
我做的是,简单的覆盖连接和运行线程的方法:
class RaisingThread(threading.Thread):
def run(self):
self._exc = None
try:
super().run()
except Exception as e:
self._exc = e
def join(self, timeout=None):
super().join(timeout=timeout)
if self._exc:
raise self._exc
用途如下:
def foo():
time.sleep(2)
print('hi, from foo!')
raise Exception('exception from foo')
t = RaisingThread(target=foo)
t.start()
try:
t.join()
except Exception as e:
print(e)
结果:
hi, from foo!
exception from foo!
我喜欢这门课:
https://gist.github.com/earonesty/b88d60cb256b71443e42c4f1d949163e
import threading
from typing import Any
class PropagatingThread(threading.Thread):
"""A Threading Class that raises errors it caught, and returns the return value of the target on join."""
def __init__(self, *args, **kwargs):
self._target = None
self._args = ()
self._kwargs = {}
super().__init__(*args, **kwargs)
self.exception = None
self.return_value = None
assert self._target
def run(self):
"""Don't override this if you want the behavior of this class, use target instead."""
try:
if self._target:
self.return_value = self._target(*self._args, **self._kwargs)
except Exception as e:
self.exception = e
finally:
# see super().run() for why this is necessary
del self._target, self._args, self._kwargs
def join(self, timeout=None) -> Any:
super().join(timeout)
if self.exception:
raise self.exception
return self.return_value
我知道我在这里有点晚了,但我有一个非常类似的问题,但它包括使用tkinter作为GUI,并且主循环使它不可能使用依赖于.join()的任何解决方案。因此,我调整了原问题EDIT中给出的解决方案,但使其更一般,以便于其他人更容易理解。
下面是运行中的新线程类:
import threading
import traceback
import logging
class ExceptionThread(threading.Thread):
def __init__(self, *args, **kwargs):
threading.Thread.__init__(self, *args, **kwargs)
def run(self):
try:
if self._target:
self._target(*self._args, **self._kwargs)
except Exception:
logging.error(traceback.format_exc())
def test_function_1(input):
raise IndexError(input)
if __name__ == "__main__":
input = 'useful'
t1 = ExceptionThread(target=test_function_1, args=[input])
t1.start()
当然,您总是可以让它以日志以外的其他方式处理异常,例如将其打印出来,或将其输出到控制台。
这允许您像使用Thread类一样使用ExceptionThread类,无需任何特殊修改。
虽然不可能直接捕获在不同线程中抛出的异常,但下面的代码可以相当透明地获取与此功能非常接近的内容。子线程必须继承ExThread类而不是线程。线程和父线程在等待线程完成任务时必须调用child_thread.join_with_exception()方法,而不是child_thread.join()方法。
此实现的技术细节:当子线程抛出异常时,它将通过Queue传递给父线程,并在父线程中再次抛出。注意,在这种方法中没有忙碌等待。
#!/usr/bin/env python
import sys
import threading
import Queue
class ExThread(threading.Thread):
def __init__(self):
threading.Thread.__init__(self)
self.__status_queue = Queue.Queue()
def run_with_exception(self):
"""This method should be overriden."""
raise NotImplementedError
def run(self):
"""This method should NOT be overriden."""
try:
self.run_with_exception()
except BaseException:
self.__status_queue.put(sys.exc_info())
self.__status_queue.put(None)
def wait_for_exc_info(self):
return self.__status_queue.get()
def join_with_exception(self):
ex_info = self.wait_for_exc_info()
if ex_info is None:
return
else:
raise ex_info[1]
class MyException(Exception):
pass
class MyThread(ExThread):
def __init__(self):
ExThread.__init__(self)
def run_with_exception(self):
thread_name = threading.current_thread().name
raise MyException("An error in thread '{}'.".format(thread_name))
def main():
t = MyThread()
t.start()
try:
t.join_with_exception()
except MyException as ex:
thread_name = threading.current_thread().name
print "Caught a MyException in thread '{}': {}".format(thread_name, ex)
if __name__ == '__main__':
main()
如果在线程中发生异常,最好的方法是在连接期间在调用线程中重新引发它。您可以使用sys.exc_info()函数获取当前正在处理的异常的信息。此信息可以简单地存储为线程对象的属性,直到调用join,此时可以重新引发它。
注意,队列。队列(在其他回答中建议)在这个简单的情况下是不必要的,因为线程最多抛出1个异常,并且在抛出一个异常后立即完成。我们通过简单地等待线程完成来避免竞争条件。
例如,扩展ExcThread(如下),覆盖excRun(而不是run)。
Python 2. x:
import threading
class ExcThread(threading.Thread):
def excRun(self):
pass
def run(self):
self.exc = None
try:
# Possibly throws an exception
self.excRun()
except:
import sys
self.exc = sys.exc_info()
# Save details of the exception thrown but don't rethrow,
# just complete the function
def join(self):
threading.Thread.join(self)
if self.exc:
msg = "Thread '%s' threw an exception: %s" % (self.getName(), self.exc[1])
new_exc = Exception(msg)
raise new_exc.__class__, new_exc, self.exc[2]
Python 3. x:
在Python 3中,raise的参数形式为3,因此将最后一行更改为:
raise new_exc.with_traceback(self.exc[2])
并发。Futures模块使得在单独的线程(或进程)中工作并处理任何由此产生的异常变得简单:
import concurrent.futures
import shutil
def copytree_with_dots(src_path, dst_path):
with concurrent.futures.ThreadPoolExecutor(max_workers=1) as executor:
# Execute the copy on a separate thread,
# creating a future object to track progress.
future = executor.submit(shutil.copytree, src_path, dst_path)
while future.running():
# Print pretty dots here.
pass
# Return the value returned by shutil.copytree(), None.
# Raise any exceptions raised during the copy process.
return future.result()
并发。futures包含在Python 3.2中,并可作为早期版本的反向移植futures模块使用。