如何在目录中打开多个文件

Question

我需要在python3中构建一个简单的脚本，在目录中打开更多文件，并查看这些文件内部是否为关键字。

所有目录中的文件是这样的： “F * .formatoffile”（*为休闲号码保持）

例子：

f9993546.txt 
f10916138.txt 
f6325802.txt 

Obviusly我只需要打开txt文件那些。

在此先感谢！

最终脚本：

import os 

Path = "path of files" 
filelist = os.listdir(Path) 
for x in filelist: 
    if x.endswith(".txt"): 
     try: 
      with open(Path + x, "r", encoding="ascii") as y: 
       for line in y: 
        if "firefox" in line: 
         print ("Found in %s !" % (x)) 
     except: 
      pass 

Answer 1

这应该做的伎俩：

import os 
Path = "path of the txt files" 
filelist = os.listdir(Path) 
for i in filelist: 
    if i.endswith(".txt"): # You could also add "and i.startswith('f') 
     with open(Path + i, 'r') as f: 
      for line in f: 
       # Here you can check (with regex, if, or whatever if the keyword is in the document.)