Python遍历目录查找文件内容

04/12/2017

以前要修改wordpress模板上的具体内容时，总是把模板文件夹下载下来，通过sublime text打开文件夹，才能具体的查找到相关内容
当然linux通过grep也可以实现，但还是达不到自己想要的效果
于是用所学的pythoon生成器写了一个小程序
该程序能够查找文件夹下各个文件，并返回关键词所在行号、该行内容、文件所在地址

具体实现

#!/bin/python3

# -*- coding:utf-8 -*-

#modules

import os,sys

#functions

def init(func):

''' initialization generator functions '''

def wrapper(*args,**kwargs):

res = func(*args,**kwargs)

next(res)

return res

return wrapper

def get_filepath(keyword,dir_path):

'''get file path'''

path_list = os.walk(dir_path)

file_path = ('%s/%s'%(x[0],y) for x in path_list for y in x[-1])

for x in file_path:

open_file(cat_file(print_file())).send((x,keyword))

@init

def open_file(target):

'''open file send to cat_file'''

while True:

path,keyword = yield

with open(path,'r') as file_read:

target.send((path,keyword,file_read))

@init

def cat_file(target):

'''cat file content , grep by keyword,get keywork in line and send to print'''

while True:

count = 1

path,keyword,file_read = yield

for x in file_read:

if keyword in x:

target.send((path,x.strip(),count))

count += 1

@init

def print_file():

'''print keywork in which line\line content\file path'''

while True:

path,line,count = yield

print("file:%s in line:\033[1;43m%s\033[0m found(%s)" %(path,count,line))

#main program

def main():

if len(sys.argv) == 3:

if os.path.isdir(sys.argv[2]):

get_filepath(sys.argv[1],sys.argv[2])

else:

print("path must directory!")

else:

print("find.py [keyword] [directory path]")

#program entry

if __name__ == '__main__':

main()

Don't settle for good. Demand Great

Python遍历目录查找文件内容

具体实现

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