文章目录
- 问题
- 解决
问题
写了一个小工具,会检索给定目录下所有文件以及内容,目录的指定的字符,并替换想要的字符
同时会自动判断文档的格式是utf-8,gbk,或者其他类型,保证不读取出错 比如 check_word=[“111”,“222”,“333”,“4444”,“55”,“66”]
解决
代码语言:javascript复制import os,shutil
import re
import time
import datetime
import chardet
allFileNum = 0
myfile=[]
mydir=[]
def get_encode(file_my):
file_my = open(file_my,mode='rb') # 以二进制模式读取文件
data = file_my.read() # 获取文件内容
file_my.close() # 关闭文件
result = chardet.detect(data) # 检测文件内容
return result["encoding"]
def printPath(level, path):
global allFileNum
''''' 打印一个目录下的所有文件夹和文件 '''
# 所有文件夹,第一个字段是次目录的级别
dirList = []
# 所有文件
fileList = []
# 返回一个列表,其中包含在目录条目的名称(google翻译)
files = os.listdir(path)
# 先添加目录级别
dirList.append(str(level))
for f in files:
if (os.path.isdir(path '/' f)):
# 排除隐藏文件夹。因为隐藏文件夹过多
if (f[0] == '.'):
pass
else:
# 添加非隐藏文件夹
dirList.append(f)
mydir.append(path '/' f)
if (os.path.isfile(path '/' f)):
# 添加文件
fileList.append(f)
myfile.append(path '/' f)
# 当一个标志使用,文件夹列表第一个级别不打印
i_dl = 0
for dl in dirList:
if (i_dl == 0):
i_dl = i_dl 1
else:
# print("得到的文件夹",'-' * (int(dirList[0])), dl)
# 打印目录下的所有文件夹和文件,目录级别 1
printPath((int(dirList[0]) 1), path '/' dl)
for fl in fileList:
# print("得到的文件路径",'-' * (int(dirList[0])), fl)
# 随便计算一下有多少个文件
allFileNum = allFileNum 1
def update_file(file_my):
print("开始处理文件",file_my)
#修改文件内容
check_word=["111","222","333","4444","55","66"]
change_word=["Learn_Java","learn01","learn02","learn03","learn04","learn05"]
delete_word=["class"]
old_name = file_my.split("/")[-1]
if old_name.split(".")[1] in delete_word:
os.remove(file_my)
else:
f = open(file_my, 'r')
alllines = f.readlines()
f.close()
type_encode=get_encode(file_my)
f = open(file_my, 'w ',encoding=type_encode)
for eachline in alllines:
for i in range(0,len(check_word)):
eachline =eachline.replace(check_word[i],change_word[i])
f.writelines(eachline)
f.close()
#修改文件名字
base_path = file_my[:-len(old_name)]
temp_old_name=old_name
for i in range(0, len(check_word)):
old_name = old_name.replace(check_word[i], change_word[i])
os.rename(os.path.join(base_path,temp_old_name), os.path.join(base_path,old_name))
def update_folder(folder_my):
print("开始处理文件夹",folder_my)
#有一个规律,只有最后一个不同路径才是正确的,可修改的,否则就是之前修改过的,因此应该是逆序倒着修改
charc_list=[]
temp_folder=folder_my
judge_name=folder_my.split("/")[-1]
base_path=folder_my[:-len(judge_name)]
temp_name=judge_name
check_word = ["11","22"]
change_word = ["Learn_Java","Learn_Java01"]
for i in range(0, len(check_word)):
judge_name = judge_name.replace(check_word[i], change_word[i])
if temp_name!=judge_name:#如果名字变化说明不同了,应该修改
os.rename(temp_folder,base_path judge_name)
this_folder = input("需要更新信息的代码路径:").replace("\",'/')
printPath(1, this_folder)
for i in myfile:
print(i)
i.replace("\","/")
update_file(i)
mydir=sorted(mydir,key=lambda x: len(x),reverse=True)
for i in mydir:
print(i)
update_folder(i)
input("已经处理完成,按任意键结束")
发布者:全栈程序员栈长,转载请注明出处:https://javaforall.cn/100236.html原文链接: