Protractor 4 : how to locate a parent at 'n' level above by class

Question

例如使用以下 html:

<div class='parent'>
  <p>
    <span class='child'></span>
  </p>
</div>

要从 .child 中找到 .parent，我正在尝试类似的方法:

let child = element(by.css('.child'));
let parent = child.element(by.xpath('../..')); // not working
let parent = child.element(by.xpath('ancestor::.parent')); // not working

实现该目标的最佳方法是什么？

Answer 1

您提到了不正确的定位器组合。这是在 xpath 表达式中包含的 css 值 '.parent'。正确的做法是:

   let parent = child.element(by.xpath('ancestor::div')); 
OR 
   let parent = child.element(by.xpath('ancestor::div[1]'));