Redirect to file

[vndywarhol]

I would like to redirect all the function outputs and errors to a separate file using the context manager. I wrote the code for a function that executes without errors. It works perfectly.

import sys def decorator(func): def wrapper(): with open('output', 'w') as f: # sys.stderr = f sys.stdout = f print(func()) return wrapper # @decorator # def function_with_exc(): # return 5 / 0 @decorator def function_without_exc(): return 5 * 2 # function_with_exc() function_without_exc()

But when I try to do the same for outputting errors, nothing is stored in the file, why?

import sys def decorator(func): def wrapper(): with open('output', 'w') as f: sys.stderr = f # sys.stdout = f print(func()) return wrapper @decorator def function_with_exc(): return 5 / 0 # @decorator # def function_without_exc(): # return 5 * 2 function_with_exc() # function_without_exc()

[DeaD_EyE]

This happens, because you don't catch any exception while calling func().
You can catch all exceptions and log them to a file.
Instead of overwriting sys.stdout, use the print function.
You can define with file the outputfile which is print using.

def decorator(func): def wrapper(): with open('output', 'w') as fd: try: result = func() except Exception as error: print(error, file=fd) else: print(result, file=fd) return wrapper

To return a value, you have to add after the print in the else block return result