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问题描述
我发现这段代码可以读取特定文件的所有行.
I found this piece of code that reads all the lines of a specific file.
我如何编辑它,让它一一读取目录文件夹"中的所有文件(html、文本、php .etc),而不必指定每个文件的路径?我想在目录中的每个文件中搜索一个关键字.
How can I edit it to make it read all the files (html, text, php .etc) in the directory "folder" one by one without me having to specify the path to each file? I want to search each file in the directory for a keyword.
path = '/Users/folder/index.html'
files = glob.glob(path)
for name in files:
try:
with open(name) as f:
sys.stdout.write(f.read())
except IOError as exc:
if exc.errno != errno.EISDIR:
raise
推荐答案
import os
your_path = 'some_path'
files = os.listdir(your_path)
keyword = 'your_keyword'
for file in files:
if os.path.isfile(os.path.join(your_path, file)):
f = open(os.path.join(your_path, file),'r')
for x in f:
if keyword in x:
#do what you want
f.close()
os.listdir('your_path')
将列出目录的所有内容os.path.isfile
是否会检查它的文件
os.listdir('your_path')
will list all content of a directoryos.path.isfile
will check its file or not
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