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Select Selenium、Python 中的一个元素,而不是使用 ZB6454D4986313710A1891ZED1CEA

[英]Select an element in Selenium, Python, not by using XPath

I was trying to scrape a website, and I need to select only the ul element inside the div with a class "Slider__SliderWrapper-sc-143uniy-0 jrPmnS", however, since there are many div tags with the same class, the only way I have to select just the ul I need is by looking at the href of the a tag, the one inside the h2. I was trying to scrape a website, and I need to select only the ul element inside the div with a class "Slider__SliderWrapper-sc-143uniy-0 jrPmnS", however, since there are many div tags with the same class, the only way我必须 select 只是我需要的 ul 是通过查看 h2 内的 a 标签的 href。 I can't use xpath, because div tags always change position.我不能使用 xpath,因为 div 标签总是改变 position。

<div>
   <h2><a class="slider-components__SectionLink-sc-1r2bduf-3 jchpWs" href="rightOne">Right!</a></h2>
   <div class="Slider__SliderWrapper-sc-143uniy-0 jrPmnS">
      <ul class="Slider__List-sc-143uniy-1 MTYOL">
      the right ul
      </ul>
   </div>
</div>
<div>
   <h2><a class="slider-components__SectionLink-sc-1r2bduf-3 jchpWs" href="wrongOne">Something else</a></h2>
   <div class="Slider__SliderWrapper-sc-143uniy-0 jrPmnS">
      <ul class="Slider__List-sc-143uniy-1 MTYOL">
      the wrong ul
      </ul>
   </div>
</div>

I thought about using css selector but I don't know how to, any help?我想过使用css选择器,但我不知道怎么做,有什么帮助吗?

You definitely CAN use xpath to access the href attribute AND it's contents:您绝对可以使用 xpath 访问 href 属性及其内容:

//a[contains(@href,'rightOne')]

and for the ul:对于 ul:

//h2/a[contains(@href,'rightOne')]/../following-sibling::div/ul

try xpath试试xpath

//a[@href='rightOne']/../following-sibling::div/ul

Explanation:解释:

You cannot use css_selector or any other locator since you are depending on a tag and you have to traverse upwards in DOM first, we are using /.. for that, alternatively you can use /parent::h2 and the next following-sibling using /following-sibling::div and then finally ul child您不能使用css_selector或任何其他locator ,因为您依赖a标签并且您必须首先在 DOM 中向上遍历,我们使用/..为此,或者您可以使用/parent::h2和下一个following-sibling使用/following-sibling::div然后最后是ul child

You cannot get a parent element with css selector, as it's not possible.您无法使用 css 选择器获取父元素,因为这是不可能的。 Check here Is there a CSS parent selector?检查这里是否有 CSS 父选择器?

In your case you would need to get the parent of a[href=rightOne] and get the ul of the following sibling.在您的情况下,您需要获取a[href=rightOne]的父级并获取以下兄弟的ul

With css you could use one of these locators:使用 css 您可以使用以下定位器之一:

div:nth-child(1) .Slider__SliderWrapper-sc-143uniy-0.jrPmnS>.Slider__List-sc-143uniy-1.MTYOL

Or或者

div:nth-child(1) .Slider__SliderWrapper-sc-143uniy-0.jrPmnS>ul

I would select any of XPaths proposed in other two answers if there are not restrictions on selectors.如果对选择器没有限制,我会 select 在其他两个答案中提出的任何 XPaths。

But, if you are using such libraries as BeautfulSoup , you will have to use css selectors, as it does not support XPath.但是,如果您使用BeautfulSoup等库,则必须使用 css 选择器,因为它不支持 XPath。 So, use the ones I proposed.所以,使用我建议的那些。

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