Python 显示 No such file or directory 错误，尽管文件存在

问题描述

我想从多个文件中搜索一个字符串

I want to search a string from multiple files

我的尝试:

import os
path= 'sample1/nvram2/logs'
all_files=os.listdir(path)
for my_file1 in all_files:
    print(my_file1)
    with open(my_file1, 'r') as my_file2:
        print(my_file2)
        for line in my_file2:
            if 'string' in line:
                print(my_file2)

输出:

C:\Users\user1\scripts>python search_string_3.py
abcd.txt
Traceback (most recent call last):
  File "search_string_3.py", line 6, in <module>
    with open(my_file1, 'r') as my_file2:
FileNotFoundError: [Errno 2] No such file or directory: 'abcd.txt'

但是文件 abcd.txt 存在于 C:\Users\user1\scripts\sample1\nvram2\logs

But file abcd.txt is present in C:\Users\user1\scripts\sample1\nvram2\logs

为什么错误显示没有这样的文件或目录?

Why the Error shows that No such file or directory?

使用 glob:

当我使用 all_files=glob.glob(path) 而不是 all_files=os.listdir(path)

C:\Users\user1\scripts>python search_string_3.py
sample1/nvram2/logs
Traceback (most recent call last):
  File "search_string_3.py", line 7, in <module>
    with open(my_file1, 'r') as my_file2:
PermissionError: [Errno 13] Permission denied: 'sample1/nvram2/logs'

推荐答案

你发现/猜到了第一个问题.用文件名加入目录可以解决它.经典:

you figured out/guessed the first issue. Joining the directory with the filename solves it. A classic:

with open(os.path.join(path,my_file1), 'r') as my_file2:

如果您不尝试使用 glob 进行某些操作，我就不会在意回答.现在:

I wouldn't have cared to answer if you didn't attempt something with glob. Now:

for x in glob.glob(path):

因为 path 是一个目录，glob 将它作为自身进行评估(你得到一个包含一个元素的列表:[path]).您需要添加通配符:

since path is a directory, glob evaluates it as itself (you get a list with one element: [path]). You need to add a wildcard:

for x in glob.glob(os.path.join(path,"*")):

glob 的另一个问题是，如果目录(或模式)与任何内容都不匹配，您不会收到任何错误.它什么都不做……os.listdir 版本至少崩溃了.

The other issue with glob is that if the directory (or the pattern) doesn't match anything you're not getting any error. It just does nothing... The os.listdir version crashes at least.

并在打开之前测试它是否是文件(在两种情况下)，因为尝试打开目录会导致 I/O 异常:

and also test if it's a file before opening (in both cases) because attempting to open a directory results in an I/O exception:

if os.path.isfile(x):
  with open ...

简而言之 os.path 包是您操作文件时的朋友.或者 pathlib 如果你喜欢面向对象的路径操作.

In a nutshell os.path package is your friend when manipulating files. Or pathlib if you like object-oriented path manipulations.

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