首页 > 解决方案 > Selenium 在 Webelement 中通过 xpath 查找

问题描述

我有这个页面:

<html>
<head>
    <title>Document Example</title>
</head>
<body>
    <div id="container">
        <div class="section">
            <h1>Section 1</h1>
            <li class="links">
                <ul><a href="link.com/1"></a></ul>
                <ul><a href="link.com/2"></a></ul>
                <ul><a href="link.com/3"></a></ul>
            </li>
        </div>
        <div class="section">
            <h1>Section 2</h1>
            <li class="links">
                <ul><a href="link.com/4"></a></ul>
                <ul><a href="link.com/5"></a></ul>
                <ul><a href="link.com/6"></a></ul>
            </li>
        </div>
        <div class="section">
            <h1>Section 3</h1>
            <li class="links">
                <ul><a href="link.com/7"></a></ul>
                <ul><a href="link.com/8"></a></ul>
                <ul><a href="link.com/9"></a></ul>
            </li>
        </div>
    </div>
</body>
</html>

我想要按部分分组链接:

driver.get("mypage.com")

sections = driver.find_elements_by_xpath("//div[@class='section']")
for section in sections:
    section_name = section.find_element_by_xpath("//h1[@class='name']").get_attribute('innerHTML') #This fails

    links = section.find_elements_by_xpath("//ul/a") #This find all links in page, not only links in section

问题是,我通过 xpath 在所有页面中找到元素列表(WebElement),现在我想通过 xpath 查找元素部分中的所有特定元素,而不是在所有页面中。

怎么了?

编辑 1

这解决了问题(现在 xpath 开始 whit dot . ): driver.get("mypage.com")

sections = driver.find_elements_by_xpath("//div[@class='section']")
for section in sections:
    section_name = section.find_element_by_xpath(".//h1[@class='name']").get_attribute('innerHTML') #This fails

    links = section.find_elements_by_xpath(".//ul/a") #This find all links in page, not only links in section

在这里找到的解释是:http: //selenium-python.readthedocs.io/api.html#selenium.webdriver.remote.webelement.WebElement.find_element_by_xpath

标签: pythonseleniumxpathselenium-chromedriver

解决方案


可能有用

sections = driver.find_elements_by_xpath("//div[@class='section']")
for i, section in enumerate(sections):
    section_name = section.text
    links = section.find_elements_by_xpath("//div[@class='section'][%s]//a" % str(i+1))

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