Question

我有文件列表，我需要检查它们是否为空如果它们是非空的打印文件名和文件内容，则不执行任何操作

例如：file 1.html content：a，2.html content：b，3.html -empty

需要创建包含两个文件内容的结果文件：

output.txt的：

1.html

a

2.html

b

我有这段代码：

import os


files = ["1.html", "2.html", "3.html"];

for i in range(len(files)):

 with open(files) as file:
     first = file.read(1)
     if not first:
        print('') #nothing to print
     else:
        print file #print file name
        print file.read() #print file content

得到：

with open(files) as file:
TypeError: coercing to Unicode: need string or buffer, list found

Answer 1

你太复杂了，只是先加载文件内容 - 如果有的话打印出来，如果没有则忽略：

files = ["1.html", "2.html", "3.html"]

for filename in files:
    with open(filename, "r") as f:
        contents = f.read()
        if contents:
            print(filename)
            print(contents)

Answer 2

因为打开初始数组而不是文件[i]，所以你的with语句没有用。处理此问题的更好方法是：

files = ["1.html", "2.html", "3.html"];

for f in files:

 with open(f) as file:
     first = file.read(1)
     if not first:
        print('') #nothing to print
     else:
        print f #print file name
        print file.read() #print file content

Answer 3

for file in files:
    with open(file) as fin:
        if fin.read():
            print file
            print file.read()

Python，检查多个文件是否为空打印结果以分隔文件

3 个答案: