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

当前回答

我使用这个版本,它是最小的,它工作得很好。

class SafeThread(threading.Thread):
    def __init__(self, *args, **kwargs):
        super(SafeThread, self).__init__(*args, **kwargs)
        self.exception = None

    def run(self) -> None:
        try:
            super(SafeThread, self).run()
        except Exception as ex:
            self.exception = ex
            traceback.print_exc()

    def join(self, *args, **kwargs) -> None:
        super(SafeThread, self).join(*args, **kwargs)
        if self.exception:
            raise self.exception

要使用它,只需替换线程。带安全线程的线程

t = SafeThread(target = some_function, args = (some, args,))
t.start()
# do something else here if you want as the thread runs in the background
t.join()

其他回答

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])

使用裸例外并不是一个好的实践,因为您通常会获得比您讨价还价时更多的东西。

我建议修改except以只捕获您想要处理的异常。我不认为引发它有预期的效果,因为当你在外层try中实例化TheThread时,如果它引发一个异常,赋值永远不会发生。

相反,你可能只想提醒它,然后继续前进,比如:

def run(self):
    try:
       shul.copytree(self.sourceFolder, self.destFolder)
    except OSError, err:
       print err

然后,当异常被捕获时,您可以在那里处理它。然后,当外部try从TheThread捕获异常时,您知道它不是您已经处理过的异常,并将帮助您隔离流程流。

我认为其他的解决方案有点复杂,如果你唯一想要的是真正看到某个异常,而不是完全无视和盲目。

解决方案是创建一个自定义线程,从主线程获取记录器并记录任何异常。

class ThreadWithLoggedException(threading.Thread):
    """
    Similar to Thread but will log exceptions to passed logger.

    Args:
        logger: Logger instance used to log any exception in child thread

    Exception is also reachable via <thread>.exception from the main thread.
    """

    def __init__(self, *args, **kwargs):
        try:
            self.logger = kwargs.pop("logger")
        except KeyError:
            raise Exception("Missing 'logger' in kwargs")
        super().__init__(*args, **kwargs)
        self.exception = None

    def run(self):
        try:
            if self._target is not None:
                self._target(*self._args, **self._kwargs)
        except Exception as exception:
            thread = threading.current_thread()
            self.exception = exception
            self.logger.exception(f"Exception in child thread {thread}: {exception}")
        finally:
            del self._target, self._args, self._kwargs

例子:

logger = logging.getLogger(__name__)
logger.setLevel(logging.INFO)
logger.addHandler(logging.StreamHandler())

def serve():
    raise Exception("Earth exploded.")

th = ThreadWithLoggedException(target=serve, logger=logger)
th.start()

主线程输出:

Exception in child thread <ThreadWithLoggedException(Thread-1, started 139922384414464)>: Earth exploded.
Traceback (most recent call last):
  File "/core/utils.py", line 108, in run
    self._target(*self._args, **self._kwargs)
  File "/myapp.py", line 105, in serve
    raise Exception("Earth exploded.")
Exception: Earth exploded.

问题是thread_obj.start()立即返回。您所生成的子线程在它自己的上下文中使用自己的堆栈执行。在那里发生的任何异常都在子线程的上下文中,并且在它自己的堆栈中。我现在能想到的一种将此信息传递给父线程的方法是使用某种消息传递,因此您可以研究一下。

试试这个尺寸:

import sys
import threading
import queue


class ExcThread(threading.Thread):

    def __init__(self, bucket):
        threading.Thread.__init__(self)
        self.bucket = bucket

    def run(self):
        try:
            raise Exception('An error occured here.')
        except Exception:
            self.bucket.put(sys.exc_info())


def main():
    bucket = queue.Queue()
    thread_obj = ExcThread(bucket)
    thread_obj.start()

    while True:
        try:
            exc = bucket.get(block=False)
        except queue.Empty:
            pass
        else:
            exc_type, exc_obj, exc_trace = exc
            # deal with the exception
            print exc_type, exc_obj
            print exc_trace

        thread_obj.join(0.1)
        if thread_obj.isAlive():
            continue
        else:
            break


if __name__ == '__main__':
    main()

Pygolang提供同步功能。工作组,特别是将异常从派生的工作线程传播到主线程。例如:

#!/usr/bin/env python
"""This program demostrates how with sync.WorkGroup an exception raised in
spawned thread is propagated into main thread which spawned the worker."""

from __future__ import print_function
from golang import sync, context

def T1(ctx, *argv):
    print('T1: run ... %r' % (argv,))
    raise RuntimeError('T1: problem')

def T2(ctx):
    print('T2: ran ok')

def main():
    wg = sync.WorkGroup(context.background())
    wg.go(T1, [1,2,3])
    wg.go(T2)

    try:
        wg.wait()
    except Exception as e:
        print('Tmain: caught exception: %r\n' %e)
        # reraising to see full traceback
        raise

if __name__ == '__main__':
    main()

在运行时给出以下结果:

T1: run ... ([1, 2, 3],)
T2: ran ok
Tmain: caught exception: RuntimeError('T1: problem',)

Traceback (most recent call last):
  File "./x.py", line 28, in <module>
    main()
  File "./x.py", line 21, in main
    wg.wait()
  File "golang/_sync.pyx", line 198, in golang._sync.PyWorkGroup.wait
    pyerr_reraise(pyerr)
  File "golang/_sync.pyx", line 178, in golang._sync.PyWorkGroup.go.pyrunf
    f(pywg._pyctx, *argv, **kw)
  File "./x.py", line 10, in T1
    raise RuntimeError('T1: problem')
RuntimeError: T1: problem

问题的原始代码将是:

    wg = sync.WorkGroup(context.background())

    def _(ctx):
        shul.copytree(sourceFolder, destFolder)
    wg.go(_)

    # waits for spawned worker to complete and, on error, reraises
    # its exception on the main thread.
    wg.wait()