我对python还比较陌生,我一直在处理以下问题。
我有一个脚本,它逐个处理文件,并根据输入文件名将输出写入单独的文件。有时我需要破坏脚本,但我想让它完成当前文件的处理,然后终止(以避免结果文件的信息不完整)。如何在python中编码这种行为?
这是我试过的。
a)试除块

x = 1
print "Script started."
while True:
 try:
  print "Processing file #",x,"started...",
  # do something time-cosnuming
  time.sleep(1)
  x += 1
  print " finished."
 except KeyboardInterrupt:
  print "Bye"
  print "x=",x
  sys.exit()

sys.exit()

输出:
Script started.
Processing file # 1 started...  finished.
Processing file # 2 started...  finished.
Processing file # 3 started... Bye
x= 3

迭代3没有优雅地完成。
b)系统例外挂钩
OriginalExceptHook = sys.excepthook
def NewExceptHook(type, value, traceback):
global Terminator
    Terminator = True
    if type == KeyboardInterrupt:
        #exit("\nExiting by CTRL+C.")   # this line was here originally
        print("\n\nExiting by CTRL+C.\n\n")
    else:
        OriginalExceptHook(type, value, traceback)
sys.excepthook = NewExceptHook

global Terminator
Terminator = False

x = 1
while True:
  print "Processing file #",x,"started...",
  # do something time-cosnuming
  time.sleep(1)
  x += 1
  print " finished."
  if Terminator:
   print "I'll be back!"
   break

print "Bye"
print "x=",x
sys.exit()

输出:
Script started.
Processing file # 1 started...  finished.
Processing file # 2 started...  finished.
Processing file # 3 started...

Exiting by CTRL+C.

迭代3没有优雅地完成。
UPD 1
@mguijar,我稍微修改了一下代码,如下所示:
import time, sys

x = 1
print "Script started."
stored_exception=None

while True:
    try:
        print "Processing file #",x,"started...",
        # do something time-cosnuming
        time.sleep(1)
        print "Processing file #",x,"part two...",
        time.sleep(1)
        print " finished."
        if stored_exception:
            break
        x += 1
    except KeyboardInterrupt:
        print "[CTRL+C detected]",
        stored_exception=sys.exc_info()

print "Bye"
print "x=",x

if stored_exception:
    raise stored_exception[0], stored_exception[1], stored_exception[2]

sys.exit()

输出(在win7-64位上使用“python 2.7.6::anaconda 2.0.0(64位)”进行测试):
Script started.
Processing file # 1 started... Processing file # 1 part two...  finished.
Processing file # 2 started... Processing file # 2 part two...  finished.
Processing file # 3 started... [CTRL+C detected] Processing file # 3 started... Processing file # 3 part two...  finished.
Bye
x= 3
Traceback (most recent call last):
  File "test2.py", line 12, in <module>
    time.sleep(1)
KeyboardInterrupt

在这种情况下,有效地重新启动了迭代3,这看起来很奇怪,并且不是所需的行为。有没有可能避免这种情况?
我删除了“print”语句中的逗号,并添加了更多内容,以查看迭代是否真正重新启动:
import time, sys

x = 1
y = 0
print "Script started."
stored_exception=None

while True:
    try:
        y=x*1000
        y+=1
        print "Processing file #",x,y,"started..."
        y+=1
        # do something time-cosnuming
        y+=1
        time.sleep(1)
        y+=1
        print "Processing file #",x,y,"part two..."
        y+=1
        time.sleep(1)
        y+=1
        print " finished.",x,y
        y+=1
        if stored_exception:
            break
        y+=1
        x += 1
        y+=1
    except KeyboardInterrupt:
        print "[CTRL+C detected]",
        stored_exception=sys.exc_info()

print "Bye"
print "x=",x
print "y=",y

if stored_exception:
    raise stored_exception[0], stored_exception[1], stored_exception[2]

sys.exit()

输出为:
Script started.
Processing file # 1 1001 started...
Processing file # 1 1004 part two...
 finished. 1 1006
Processing file # 2 2001 started...
Processing file # 2 2004 part two...
[CTRL+C detected] Processing file # 2 2001 started...
Processing file # 2 2004 part two...
 finished. 2 2006
Bye
x= 2
y= 2007
Traceback (most recent call last):
  File "test2.py", line 20, in <module>
    time.sleep(1)
KeyboardInterrupt

最佳答案

我只需要使用一个异常处理程序,它将捕获
存储异常。然后,在迭代完成时,如果出现异常
挂起时,我将中断循环并重新引发异常(以允许正常异常
把握机会)。
这项工作(用python 2.7测试):

x = 1
print "Script started."
stored_exception=None

while True:
    try:
        print "Processing file #",x,"started...",
        # do something time-cosnuming
        time.sleep(1)
        print " finished."
        if stored_exception:
            break
        x += 1
    except KeyboardInterrupt:
        stored_exception=sys.exc_info()

print "Bye"
print "x=",x

if stored_exception:
    raise stored_exception[0], stored_exception[1], stored_exception[2]

sys.exit()

编辑:正如在评论中所看到的,这个答案对于最初的海报是不满意的,这里有一个基于线程的解决方案:
import time
import sys
import threading

print "Script started."

class MyProcessingThread(threading.Thread):
    def __init__(self):
        threading.Thread.__init__(self)

    def run(self):
        print "Processing file #",x,"started...",
        # do something time-cosnuming
        time.sleep(1)
        print " finished."

for x in range(1,4):
    task = MyProcessingThread()
    task.start()
    try:
        task.join()
    except KeyboardInterrupt:
        break

print "Bye"
print "x=",x

sys.exit()

09-10 08:03
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