获取属性 href returns 通用 link 硒 python

Question

我想得到一个 href link，它应该像这里的 link

<a href="https://www.allmusic.com/album/a-link-in-the-chain-mw0000072661">A Link in the Chain</a>

相反我得到

'https://www.allmusic.com/advanced-search#'

这是代码。我认为问题在于页面是我必须先单击内容才能将其拉起，idk tho。感谢您的帮助

#This gets every album in the page
browser = webdriver.Chrome('c:\Users\16308\Documents\VSCPython\chromedriver')
def get_page(pge_url):
    browser.get(pge_url)
    time.sleep(1)
    add_stuff(pge_url)
    time.sleep(2)
    thing1 = browser.find_element_by_class_name('content-container')
    time.sleep(4)
    thing2 = thing1.find_element_by_class_name('content')
    time.sleep(8)
    thing3 = thing2.find_element_by_class_name('results')
    time.sleep(4)
    thing4 = thing3.find_element_by_class_name('desktop-results')
    time.sleep(4)
    thing5 = browser.find_element_by_class_name('title').find_element_by_css_selector('a').get_attribute('href')
    print("5: ",thing5)
    return thing5

get_page('https://www.allmusic.com/advanced-search')

#clicks range of years
def add_stuff(current_page):
    time.sleep(1)
    browser.get(current_page)
    time.sleep(2)
    selectOne = Select(browser.find_element_by_class_name('start-year'))
    time.sleep(4)
    selectOne.select_by_visible_text('1920')
    time.sleep(8)
    selectTwo = Select(browser.find_element_by_class_name('end-year'))
    time.sleep(8)
    selectTwo.select_by_visible_text('2022')

#IMPORT STATEMENTS
from selenium import webdriver
from selenium.webdriver.common.keys import Keys
from selenium.webdriver.support.ui import WebDriverWait
from selenium.webdriver.common.by import By
from selenium.webdriver.support import expected_conditions as EC
import time
from selenium.webdriver.chrome.options import Options
from selenium import webdriver
from selenium.webdriver.support import expected_conditions 
from selenium.webdriver.support.ui import Select
from selenium.webdriver.support.ui import Select
import dns
import pymongo 
import sys
from pymongo import MongoClient

Answer 1

你可以试试Xpath

driver.find_element(By.XPATH,"*The full XPATH here*")

Answer 2

我假设你有一个 div class，它被称为 "title".因此，在这种情况下，您可以轻松使用 xpath 轻松获取该 href。

thing5 = browser.find_element_by_xpath('.//div[@class="title"]/a').get_attribute('href')

获取属性 href returns 通用 link 硒 python

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