python實現檔案名稱批量替換和內容替換

指定檔案夾，指定檔案類型，替換該檔案夾下全部檔案的內容。

注意在window下的讀寫內容需要指定編碼，還需要在檔案頭指定#coding:utf-8 編碼，避免出現編碼問題。

複製代碼 代碼如下:

#coding:utf-8
import os
import os.path

path='.'
oldStr='.php'
newStr='.html'

for (dirpath, dirnames, filenames) in os.walk(path):
for file in filenames:
if os.path.splitext(file)[1]=='.html':
print(file)
filepath=os.path.join(dirpath,file)
try:
text_file = open(filepath, "r")
lines = text_file.readlines()
text_file.close()
output = open(filepath,'w',encoding= 'utf-8')
for line in lines:
#print(line)
if not line:
break
if(oldStr in line):
tmp = line.split(oldStr)
temp = tmp[0] + newStr + tmp[1]
output.write(temp)
else:
output.write(line)
output.close()
except Exception:
print(Exception)
break

這個樣本可以批量替換檔案名稱和內容

複製代碼 代碼如下:

#!/usr/bin/env python
# -*- coding: utf-8 -*-

import os, re

def multi_replace(text, adict):
rx = re.compile('|'.join(map(re.escape, adict)))
def xlat(match):
return adict[match.group(0)]
return rx.sub(xlat, text)

def batrename(curdir, pairs):
for fn in os.listdir(curdir):
newfn = multi_replace(fn, pairs)
if newfn != fn:
print("Renames %s to %s in %s." % (fn, newfn, curdir))
os.rename(os.path.join(curdir, fn), os.path.join(curdir, newfn))
file = os.path.join(curdir, newfn)

if os.path.isdir(file):
batrename(file, pairs)
continue

text = open(file).read()
newtext = multi_replace(text, pairs)
if newtext != text:
print("Renames %s." % (file,))
open(file, 'w').write(newtext)

if __name__=="__main__":
while True:
oldname = raw_input("Old name: ")
newname = raw_input("New name: ")
if oldname and newname:
batrename(os.path.abspath('.'), {oldname:newname})
else: break

