按文件名通配符打开文件 [英] Open file by filename wildcard

问题描述

我有一个文本文件目录，所有文件的扩展名都为.txt.我的目标是打印文本文件的内容.我希望能够使用通配符*.txt指定要打开的文件名(我正在考虑类似F:\text\*.txt这样的行?)，分割文本文件的行，然后打印输出.

I have a directory of text files that all have the extension .txt. My goal is to print the contents of the text file. I wish to be able use the wildcard *.txt to specify the file name I wish to open (I'm thinking along the lines of something like F:\text\*.txt?), split the lines of the text file, then print the output.

这是我要执行的操作的一个示例，但是我希望能够在执行命令时更改somefile.

Here is an example of what I want to do, but I want to be able to change somefile when executing my command.

f = open('F:\text\somefile.txt', 'r')
for line in f:
    print line,

我之前已经签出了glob模块，但是我不知道如何对文件做任何实际的事情.这是我想出的，不起作用.

I had checked out the glob module earlier, but I couldn't figure out how to actually do anything to the files. Here is what I came up with, not working.

filepath = "F:\irc\as\*.txt"
txt = glob.glob(filepath)

lines = string.split(txt, '\n') #AttributeError: 'list' object has no attribute 'split'
print lines

推荐答案

import os
import re
path = "/home/mypath"
for filename in os.listdir(path):
    if re.match("text\d+.txt", filename):
        with open(os.path.join(path, filename), 'r') as f:
            for line in f:
                print line,

尽管您忽略了我的完美解决方案，但您可以执行以下操作:

Although you ignored my perfectly fine solution, here you go:

import glob
path = "/home/mydir/*.txt"
for filename in glob.glob(path):
    with open(filename, 'r') as f:
        for line in f:
            print line,

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