我必须在所有文件中递归递归找到所有行(以字符串" excel"开头((目录和子目录中(。我需要每个文件名所找到的行(例如::例如:: filename1:Line1建立...filename2:
line2建立...输出结果中的文件称为" logfile"如果没有建立的行,则未保存在logfile中。
import os
word="excel"
from os.path import join
for (dirname, dirs, files) in os.walk('/batch/'):
for filename in files:
thefile = os.path.join(dirname,filename)
for line in files:
if line.startswith(word):
print (line)
print (thefile)
谢谢
您的代码只有一个小问题:最大的是您在文件名上循环而不是文件内容。
import os
word="excel"
from os.path import join
for (dirname, dirs, files) in os.walk('/batch/'):
for filename in files:
thefile = os.path.join(dirname, filename)
with open(thefile) as f:
for line in f:
if line.startswith(word):
print (line)
print (thefile)
编辑:
import os
word="excel"
from os.path import join
with open('log_result.txt', 'w') as log_file:
for (dirname, dirs, files) in os.walk('/tmp/toto'):
for filename in files:
thefile = os.path.join(dirname, filename)
with open(thefile) as f:
lines = [line for line in f if line.startswith(word)]
if lines:
log_file.write("File {}:n".format(thefile))
log_file.writelines(lines)
这是固定的代码。您无需重新浏览相同的文件列表。OS.Walk((将返回目录中的所有子目录,您需要做的就是循环所有目录。
示例代码
import glob
import os
word="excel"
for (dirname, dirs, files) in os.walk("/batch/"):
for file_ in files :
if file_.startswith(word):
print(file_)
print(os.path.join(dirname, file_))
for dir_ in dirs :
myfiles = glob.glob(os.path.join(dirname,dir_))
for myfile in myfiles:
if myfile.startswith(word):
print(myfile)
print(os.path.join(dirname,myfiles))
希望这有帮助