我想进行一些基本的URL验证，如果URL无效，除非用户输入了有效的URL，否则不应继续进行请求。
import time
from selenium import webdriver
from selenium.webdriver.common.keys import Keys
user_url = input('Please enter a valid url:')
driver = webdriver.Chrome('/home/m/Desktop/chromedriver')
driver.get(user_url)
HEADERS = {'user-agent': 'Mozilla/5.0 (X11; Linux x86_64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/83.0.4103.61 Safari/537.36', 'accept': '*/*'}
time.sleep(8)
imagecounter = driver.find_elements_by_css_selector('img')
print('Number of HTML image tags:')
print(len(imagecounter))
您能否修改代码并解释发生了什么？我已经尝试过一些库，但是我认为由于我的编码技巧很差，所以没有运气。
您可以使用请求获取HTTP状态代码
import requests
import time
from selenium import webdriver
from selenium.webdriver.common.keys import Keys
user_url = input('Please enter a valid url:')
# send a get request to the page, and if the status code is not OK
# ask for a different url
req = requests.get(user_url)
while req.status_code != requests.codes['ok']:
user_url = input('Please enter a valid url:')
driver = webdriver.Chrome()
driver.get(user_url)
HEADERS = {'user-agent': 'Mozilla/5.0 (X11; Linux x86_64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/83.0.4103.61 Safari/537.36', 'accept': '*/*'}
time.sleep(8)
imagecounter = driver.find_elements_by_css_selector('img')
print('Number of HTML image tags:')
print(len(imagecounter))
要在继续操作之前验证用户提供的url，可以使用Python's python-requests模块来检查请求状态，并且可以使用以下解决方案：
代码块：
from selenium import webdriver
import requests
while True:
user_url = str(input("Please enter a valid url:"))
req = requests.get(user_url)
if req.status_code != requests.codes['ok']:
print("Not a valid url, please try again...")
continue
else:
break
print("URL was a valid one... Continuing...")
driver = webdriver.Chrome(executable_path=r'C:\WebDrivers\chromedriver.exe')
driver.get(user_url)
# perform your rest of the tasks
控制台输出：
Please enter a valid url:https://www.goodday.com
Not a valid url, please try again...
Please enter a valid url:https://www.goodday.com
Not a valid url, please try again...
Please enter a valid url:https://www.goodday.com
Not a valid url, please try again...
Please enter a valid url:https://www.google.com
URL was a valid one... Continuing...
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