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Is there any way to get source code inside context manager as string?

Tags:

python

Source code of function can be received with inspect.getsourcelines(func) function. Is there any way to do same for context manager?

with test():
    print('123')

# How to get "print('123')" as line here?
like image 894
Mikhail Gerasimov Avatar asked Oct 31 '22 05:10

Mikhail Gerasimov


1 Answers

What do you think about this solution?

import traceback


class ContextManagerContent(object):

    def __enter__(self):
        return

    def __exit__(self, _type, value, _traceback):
        stack = traceback.extract_stack()
        f, last_line = self._get_origin_info(stack)

        with open(f) as fin:
            lines = list(fin)

        search = 'with {cls_name}'.format(cls_name=self.__class__.__name__)
        for i, x in enumerate(lines[:last_line + 1][::-1]):
            if search in x:
                first_line = len(lines) - i
                break

        selected_lines = lines[first_line:last_line + 1]
        print ''.join(selected_lines)

    def _get_origin_info(self, stack):
        origin = None
        for i, x in enumerate(stack[::-1]):
            if x[2] == '__exit__':
                origin = stack[::-1][i + 1]
                break

        return origin[0], origin[1] - 1


with ContextManagerContent():
    print '123'
    print '456'
    print '789'

If you save this in a .py file and run it you can see being printed the numbers 123, 456 and 789, after that you can see the block of the context manager.

Notice that I did not handle possibile exceptions or formatting of the output, and some parts can be improved, but I think that this is a good starting point.

like image 109
se7entyse7en Avatar answered Nov 09 '22 23:11

se7entyse7en