从目录打开文件，并保留文件名和内容

Question

我想从目录中读取所有文件，并将文件名和内容导入到列表中。我写了下面的代码，但它只导入目录中的最后一个文件。

import os 

fileList = [] 

for dirname, dirnames, filenames in os.walk('.'): 
    for filename in filenames: 
     file = os.path.join(dirname, filename) 
    f = open(file, 'r') 
    contents = f.read() 
    newItem = (file, contents) 
    fileList.append(newItem) 


print(fileList) 

Answer 1

似乎是缩进问题。试试这个：

import os 

fileList = [] 

for dirname, dirnames, filenames in os.walk('.'): 
    for filename in filenames: 
     file = os.path.join(dirname, filename) 
     f = open(file, 'r') 
     contents = f.read() 
     newItem = (file, contents) 
     fileList.append(newItem) 

print(fileList) 

此外，你不关闭你的程序中的文件是不好的风格。这是更好地运用到处理打开和关闭文件所谓的上下文管理器：

import os 

fileList = [] 

for dirname, dirnames, filenames in os.walk('.'): 
    for filename in filenames: 
     file = os.path.join(dirname, filename) 
     with open(file, 'r') as f: 
      contents = f.read() 
      newItem = (file, contents) 
      fileList.append(newItem) 

print(fileList)