python - 无法使用 Try、Except 访问第三个 URL 进行查询

Question

它不会转到第三个网址 -

try:
        "code here..."
except requests.exceptions.ConnectionError:
    pass  # doesn't pass to try:

这是代码-

import requests

try:
    for url in ['google.com', 'skypeassets.com', 'yahoo.com']:

        http = ("http://")
        url2 = (http + url)
        page = requests.get(url2)

        if page.status_code == 200:
            print('Success!')
        elif page.status_code == 404:
            print('Not Found.')

except requests.exceptions.ConnectionError:
    print("This site cannot be reached")
        pass

输出-

成功!
无法访问该网站
(对于第三个网址 - 应该说 - 成功!，但跳过)

Answer 1

try except 一次只能捕获其主体或 block 内的一个异常。这意味着您必须在 for 循环中使用它。

import requests

for url in ['google.com', 'skypeassets.com', 'yahoo.com']:
    try:
        http = "http://"
        url2 = http + url
        page = requests.get(url2)

        if page.status_code == 200:
            print('Success!')
        elif page.status_code == 404:
            print('Not Found.')

    except requests.exceptions.ConnectionError:
        print("This site cannot be reached")