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!
此网站无法打开
(对于第三个网址-应该说-成功!,但跳过)
最佳答案
try
except
一次只能在其主体或块内部捕获一个异常。
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")
关于python - 无法使用Try,Except进入第三URL进行查询,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/56242599/