在python中调用URL的正确方法是什么

我有csv文件并且传递了csv数据有参数给python代码。在csv文件中有URL数据。在python中调用URL的正确方法是什么？收到错误Cannot navigate to invalid URL在python中调用URL的正确方法是什么

CSV文件

ID,category,link 
sports_shoes,sports-shoes,https://www.flipkart.com/mens-footwear/sports-shoes/pr?otracker=categorytree&page=1&sid=osp%2Ccil%2C1cu

代码：

from selenium import webdriver 
from selenium.webdriver.support.ui import WebDriverWait 
from selenium.webdriver.support import expected_conditions as EC 
from selenium.common.exceptions import TimeoutException 
from selenium.webdriver.common.by import By 
import time 
import csv 

with open('mydata.csv') as csvfile: 
    reader = csv.DictReader(csvfile) 
    for row in reader: 
     #print(row['ID'] ,row['category'],row['link']) 
     url = row['link'] 
     print(url) 
     chrome_path = r"C:\Users\Venkatesh\AppData\Local\Programs\Python\Python35\chromedriver.exe" 
     driver = webdriver.Chrome(chrome_path) 
     RegionIDArray = url 
     data_list=[] 
     data = [] 
     mobile_details_data = [] 
     delay = 30 # seconds 
     for reg in RegionIDArray: 
      driver.get(reg) 
driver.quit()

错误：

Traceback (most recent call last): 
    File ".\input_file.py", line 24, in <module> 
    driver.get(reg) 
    File "C:\Users\Venkatesh\AppData\Local\Programs\Python\Python35\Lib\site-packages\selenium\webdriver\remote\webdriver. 
py", line 250, in get 
    self.execute(Command.GET, {'url': url}) 
    File "C:\Users\Venkatesh\AppData\Local\Programs\Python\Python35\Lib\site-packages\selenium\webdriver\remote\webdriver. 
py", line 238, in execute 
    self.error_handler.check_response(response) 
    File "C:\Users\Venkatesh\AppData\Local\Programs\Python\Python35\Lib\site-packages\selenium\webdriver\remote\errorhandl 
er.py", line 193, in check_response 
    raise exception_class(message, screen, stacktrace) 
selenium.common.exceptions.WebDriverException: Message: unknown error: unhandled inspector error: {"code":-32000,"messag 
e":"Cannot navigate to invalid URL"} 
    (Session info: chrome=57.0.2987.133) 
    (Driver info: chromedriver=2.28.455520 (cc17746adff54984afff480136733114c6b3704b),platform=Windows NT 6.2.9200 x86_64)

来源

2017-04-12 Venkatesh Panabaka

您的url变量包含您想要访问的链接。你的代码正在循环一个字符串，并对每个字符进行一次driver.get()调用。这基本上解释了错误。

因为您已经在循环访问for row in reader:中的数据，所以不需要内部循环。只需使用driver.get(url)即可。

来源

2017-04-12 04:25:30 JRodDynamite

如果mydata.csv有两行，它会打开两个不同的浏览器。我想打开一个浏览器，意思是每次URL将被相同的浏览器取代。 –

@VenkateshPanabaka - 然后在循环之前实例化Chrome驱动程序driver = webdriver.Chrome（chrome_path）。 – JRodDynamite

非常感谢您帮助解决我的问题。 –

在python中调用URL的正确方法是什么

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