Question

我有这个尝试/除了代码：

document = raw_input ('Your document name is ')

try:
    with open(document, 'r') as a:
        for element in a:
           print element

except:
    print document, 'does not exist'

打印“[filename]不存在”后如何退出程序？ break和pass显然不起作用，我不希望出现任何崩溃错误，因此sys.exit不是一个选项。

请忽略try部分 - 它只是一个假人。

Answer 1

使用sys.exit：

import sys

try:
    # do something
except Exception, e:
    print >> sys.stderr, "does not exist"
    print >> sys.stderr, "Exception: %s" % str(e)
    sys.exit(1)

一个好的做法是打印出现的异常，以便以后进行调试。

您还可以使用traceback模块打印堆栈跟踪。

请注意，您在sys.exit中返回的int将是程序的返回码。要查看程序返回的退出代码（这将为您提供有关已发生的事件和可自动化的信息），您可以执行以下操作：

echo $?

Answer 2

使用

sys.exit(1)

不是崩溃错误，退出程序是一种非常正常的方法。退出代码1是一个约定意味着出错（在成功运行的情况下你会返回0）。

Answer 3

您还可以将代码放入函数中并发出返回。您可以将其称为main，您可以从脚本中调用它。

def main():
    document = raw_input ('Your document name is ')

    try:
        with open(document, 'r') as a:
            for element in a:
               print element

    except:
        print document, 'does not exist'
        return

if __name__ == "__main__":
    main()

Answer 4

重新加注吧。它对开发人员来说更友好了

document = raw_input ('Your document name is ')

try:
    with open(document, 'r') as a:
        for element in a:
           print element

except:
    print document, 'does not exist'
    raise

在“提升例外”部分中查看except中有关重新引发错误的python document。

Answer 5

如果您在if内使用try语句，则需要多个sys.exit()才能真正退出该程序。

例如，您在调用某个文件的执行时正在解析一个参数，例如$./do_instructions.py 821，例如：

import sys

# index number 1 is used to pass a set of instructions to parse
# allowed values are integer numbers from 1 to 4, maximum number of instructions is 3
arg_vector = "821" # <- pretending to be an example of sys.argv[1]

if len(arg_vector) > 3:
    sys.exit(2) # <- this will take you out, but the following needs an extra step.

# for an invalid input (8). 
for i in arg_vector:

    # to validate that only numbers are passed as args.
    try:
        int(i) # <- 8 is valid so far
        
        # value (8) is not  valid, since is greater than 4
        if (int(i) == 0) or (int(i) > 4): 
            print("Values must be 1-4")
            # the following call does not takes you out from the program,
            # but rise the SystemExit exception.
            sys.exit(2) 
            
    except SystemExit: # <- needed to catch the previous as the first evaluation
        # The following parameter "2" is just for this example
        sys.exit(2) # <- needed to actually interrupt the execution of the program/script. 

    # if there is no "except SystemExit:", the following will be executed when the 
    # previous "if" statement evaluates to True and the sys.exit(2) is called.
    #
    # and the "print("Only num...") function will be called, even when the intention 
    # of it is to advice the *user* to use only numbers, since 8 is a number this
    # shouldn't be executed.
    except:
        print("Only numbers are allowed.")
        sys.exit(2)

否则，您要对每个评估使用一个 try-except 块。

Answer 6

可能不是最佳做法，但对我有用：

import sys    

close = False
try:
    if SomethingBadHappend:
        close = True
except:
    pass
if close:
    sys.exit(1)

接近剂量似乎在“尝试”中不起作用。

如何在try / except中退出程序？

6 个答案: