如何使用python删除功能从具有404状态代码的文件中删除网址？

Posted 2023-02-21

我必须使用python remove功能从状态为404的文件中删除网址。但是我不确定为什么它不起作用。

代码：

#!/usr/bin/python

import requests



url_lines = open('url.txt').read().splitlines()
for url in url_lines:
    remove_url = requests.get(url)
    if remove_url.status_code == 404:
       print remove_url.status_code
       url_lines.remove(url)

url.txt文件包含以下几行：

https://www.amazon.co.uk/jksdkkhsdhk
http://www.google.com

应该从url.txt文件中删除https://www.amazon.co.uk/jksdkkhsdhk行。

非常感谢您的帮助。

答案

我用自我解释的评论写了这篇文章希望对您有帮助

import requests

with open('url.txt') as file:  # better file handling using context manager
    good_guys = []
    for line in file:  # simpler way rather than for line in file.read().splitlines()
        url = line.strip()  # get a rid of line endings and spaces

        # the pythonic way is to use try ... execpt block
        try:
            if not url:
                continue  # skip empty url
            resp = requests.get(url)
            if resp.status_code != 404:
                good_guys.append(line)
        except Exception as e:
            print('error while handling url,', e)

# save your new urls to new file named url2.txt, you still can overwrite same input file but 
# this is not a good practice, if you made a mistake you will still have your original file
with open('url2.txt', 'w') as file:
    file.writelines(good_guys)

另一答案

您可以跳过它：

if remove_url.status_code == 404:
    continue

您不应该在for循环中尝试删除它。而是将其添加到另一个列表remove_from_urls，然后在for循环之后，删除新列表中的所有索引。这可以通过以下方式完成：

remove_from_urls = []

for url in url_lines:
    remove_url = requests.get(url)
    if remove_url.status_code == 404:
        remove_from_urls.append(remove_url)
        continue
    # Code for handling non-404 requests

url_lines = [url for url in url_lines if url not in remove_from_urls]

# Save urls example
with open('urls.txt', 'w+') as file:
    for item in url_lines:
        file.write(item + '\n')