InvalidArgumentException：消息：无效参数：'url' 必须是使用 get() 调用 url 的字符串

Question

首先，我得到了所有页面的总数 url。但是，当我想进入每一页（一页一页）时，它失败了。我怎样才能进入每个页面？


!pip install selenium
from selenium import webdriver
from selenium.webdriver.common.by import By
from selenium.webdriver.support.wait import WebDriverWait
from selenium.webdriver.support import expected_conditions as EC
import urllib as url
from urllib.parse import urlparse
import time

browser = webdriver.Chrome(executable_path='./chromedriver.exe')
wait = WebDriverWait(browser,5)
output = []
for i in range(1,2): # Iterate from page 1 to the last page
    browser.get("https://tw.mall.yahoo.com/search/product?p=%E5%B1%88%E8%87%A3%E6%B0%8F&pg={}".format(i))
    
    # Wait Until the product appear
    wait.until(EC.presence_of_element_located((By.XPATH,"//ul[@class='gridList']")))

    # Get the products
    product_links = browser.find_elements(By.XPATH,"//ul[@class='gridList']/li/a")
    
      # Iterate over 'product_links' to get all the 'href' values
    for link in (product_links):
        print(f"{link.get_attribute('href')}")
        output.append([link.get_attribute('href')])
        for b in output:
            browser.get(b)

输出

InvalidArgumentException: Message: invalid argument: 'url' must be a string
  (Session info: chrome=96.0.4664.45)

Answer 1

i 属于类型 int。您需要将 int 转换为 string 以应用使用 str()函数如下：

format(str(i))

解决方案

所以有效的代码行将是：

for i in range(1,2): # Iterate from page 1 to the last page
    browser.get("https://tw.mall.yahoo.com/search/product?p=%E5%B1%88%E8%87%A3%E6%B0%8F&pg={}".format(str(i)))

InvalidArgumentException：消息：无效参数：'url' 必须是使用 get() 调用 url 的字符串

InvalidArgumentException: Message: invalid argument: 'url' must be a string invoking url using get()

python

url

selenium

urlparse

invalidargumentexception

解决方案