检查在 Python 中打开了哪些文件

问题描述

我在一个应该运行很长时间的程序中遇到错误，因为打开了太多文件.有什么方法可以让我跟踪哪些文件是打开的，这样我就可以偶尔打印该列表并查看问题出在哪里?

I'm getting an error in a program that is supposed to run for a long time that too many files are open. Is there any way I can keep track of which files are open so I can print that list out occasionally and see where the problem is?

推荐答案

我最终在我的程序入口点包装了内置文件对象.我发现我没有关闭我的记录器.

I ended up wrapping the built-in file object at the entry point of my program. I found out that I wasn't closing my loggers.

import io
import sys
import builtins
import traceback
from functools import wraps


def opener(old_open):
    @wraps(old_open)
    def tracking_open(*args, **kw):
        file = old_open(*args, **kw)

        old_close = file.close
        @wraps(old_close)
        def close():
            old_close()
            open_files.remove(file)
        file.close = close
        file.stack = traceback.extract_stack()

        open_files.add(file)
        return file
    return tracking_open


def print_open_files():
    print(f'### {len(open_files)} OPEN FILES: [{", ".join(f.name for f in open_files)}]', file=sys.stderr)
    for file in open_files:
        print(f'Open file {file.name}:
{"".join(traceback.format_list(file.stack))}', file=sys.stderr)


open_files = set()
io.open = opener(io.open)
builtins.open = opener(builtins.open)

这篇关于检查在 Python 中打开了哪些文件的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持编程学习网！