在 Python 中使用 Selenium 从某个 div 获取链接

2022-01-15 00:00:00 python selenium selenium-chromedriver selenium-webdriver

问题描述

我有以下 HTML 页面.我想获取特定 div 中的所有链接.这是我的 HTML 代码:

I have the following HTML page. I want to get all the links inside a specific div. Here is my HTML code:

<div class="rec_view">
    <a href='www.xyz.com/firstlink.html'>
        <img src='imga.png'>
    </a>
    <a href='www.xyz.com/seclink.html'>
        <img src='imgb.png'>
    </a>
    <a href='www.xyz.com/thrdlink.html'>
        <img src='imgc.png'>
    </a>
</div>

我想获取 rec_view div 上的所有链接.所以我想要的那些链接是，

I want to get all the links that are present on the rec_view div. So those links that I want are,

www.xyz.com/firstlink.html
www.xyz.com/seclink.html
www.xyz.com/thrdlink.html

这是我尝试过的 Python 代码

Here is the Python code which I tried with

from selenium import webdriver;
webpage = r"https://www.testurl.com/page/123/"
driver = webdriver.Chrome("C:chromedriver_win32chromedriver.exe")
driver.get(webpage)
element = driver.find_element_by_css_selector("div[class='rec_view']>a")
link = element.get_attribute("href")
print(link)

如何在 Python 上使用 selenium 获取这些链接?

How can I get those links using selenium on Python?

解决方案

根据您共享的 HTML 来获取 rec_view div 上存在的所有链接的列表，您可以使用以下代码块:

As per the HTML you have shared to get the list of all the links that are present on the rec_view div you can use the following code block :

from selenium import webdriver

driver = webdriver.Chrome(executable_path=r'C:chromedriver_win32chromedriver.exe')
driver.get('https://www.testurl.com/page/123/')
elements = driver.find_elements_by_css_selector("div.rec_view a")
for element in elements:
    print(element.get_attribute("href"))

注意 :因为您需要从 div 标签中收集所有 href 属性，所以而不是 find_element_* 你需要使用 find_elements_*.此外，> 指的是直接 <a> 子节点，因为您需要遍历所有 <a> 子节点，因此所需的 css_selector 将是 div.rec_view a

Note : As you need to collect all the href attributes from the div tag so instead of find_element_* you need to use find_elements_*. Additionally, > refers to immediate <a> child node where as you need to traverse all the <a> child nodes so the desired css_selector will be div.rec_view a

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