Question

我已经在python中编写了一个脚本，除了处理所有运行时异常的处理（catch块）之外。如果我把try块放在与脚本相同的文件中，那么它打印异常但我需要的是try块是在另一个文件中，然后是它将使用脚本中编写的catch块的过程。

import traceback
import sys
import linecache


try:
    # execfile(rahul2.py)

    def first():
        second()

    def second():
        i=1/0;


    def main():
        first()

    if __name__ == "__main__":
        main()    

except SyntaxError as e:
    exc_type, exc_value, exc_traceback = sys.exc_info()
    filename = exc_traceback.tb_frame.f_code.co_filename
    lineno = exc_traceback.tb_lineno
    line = linecache.getline(filename, lineno)
    print("exception occurred at %s:%d: %s" % (filename, lineno, line))
    print("**************************************************** ERROR ************************************************************************")
    print("You have encountered an error !! no worries ,lets try figuring it out together")
    print(" It looks like there is a syntax error in the statement:" , formatted_lines[2], " at line number " , exc_traceback.tb_lineno)
    print("Make sure you look up the syntax , this may happen because ")
    print(" Remember this is the error message always thrown " "'" ,e , "'")

同样，我也为其他例外而写...

现在我的问题是假设我想对所有程序使用这个脚本，或者比如假设try块在另一个文件中...然后我如何链接我的脚本和尝试阻止的程序..

或者如果我用不同的词语，那么我想要的是每当有一个try catch块时，catch块应该按照我的脚本而不是内置库来执行..

Answer 1

如果要在调用此脚本的脚本中处理此异常，则需要引发异常。例如：

except SyntaxError as e:
       exc_type, exc_value, exc_traceback = sys.exc_info()
       filename = exc_traceback.tb_frame.f_code.co_filename
       lineno = exc_traceback.tb_lineno
       line = linecache.getline(filename, lineno)
       print("exception occurred at %s:%d: %s" % (filename, lineno, line))
       print("**************************************************** ERROR ************************************************************************")
       print("You have encountered an error !! no worries ,lets try figuring it out together")
       print(" It looks like there is a syntax error in the statement:" , formatted_lines[2], " at line number " , exc_traceback.tb_lineno)
       print("Make sure you look up the syntax , this may happen because ")
       print(" Remember this is the error message always thrown " "'" ,e , "'")
       #### Raise up the exception for handling in a calling script ####
       raise e

然后在你的调用脚本中你只需要另外一个try-except块（假设你编写的'库文件'被称为mymodule.py并且两个文件都位于同一个工作目录中），就像这样

try:
   import mymodule
   mymodule.main()
except SyntaxError as e:
   print("Exception found") # Optional: Add code to handle the exception

请记住，如果您无法处理此重新引发的异常，则会导致脚本失败并退出（打印异常消息和堆栈跟踪）。 这是一件好事。如果恢复方法未知或无法处理，遇到致命错误的脚本会以引起用户注意的方式失败编程。

用于python中异常处理的库文件

1 个答案: