Python 无法使用Try访问查询的第三个URL,除非
它不会转到第三个url-Python 无法使用Try访问查询的第三个URL,除非,python,Python,它不会转到第三个url- try: "code here..." except requests.exceptions.ConnectionError: pass # doesn't pass to try: 这是密码- import requests try: for url in ['google.com', 'skypeassets.com', 'yahoo.com']: http = ("http://") url2
try:
"code here..."
except requests.exceptions.ConnectionError:
pass # doesn't pass to try:
这是密码-
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-
成功无法访问此网站
(对于第三个url-应该说-Success!,但是跳过)
除了之外的
一次只能在其主体或块内捕获一个异常。
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")
你的尝试应该在循环中。