我有一些这样的 HTML:
I have some HTML like this:
<h4 class="box_header clearfix"> <span> <a rel="dialog" href="www.google/?q=word">Search</a> </span> <small> <span> <a rel="dialog" href="www.google/?q=word">Search</a> </span> </h4>我正在尝试使用 Selenium 在 Java 中获取 href.我尝试了以下方法:
I am trying to get the href here in Java using Selenium. I have tried the following:
selenium.getText("xpath=/descendant::h4[@class='box_header clearfix']/"); selenium.getAttribute("xpath=/descendant::h4[@class='box_header clearfix']/");但这些都不起作用.它一直抱怨我的 xpath 无效.谁能告诉我我做错了什么?
But none of these work. It keeps complaining that my xpath is invalid. Can someone tell me what mistake I am doing?
推荐答案你应该使用getAttribute来获取链接的href.您的 XPath 需要对最终节点的引用以及所需的属性.以下应该有效:
You should use getAttribute to get the href of the link. Your XPath needs a reference to the final node, plus the required attribute. The following should work:
selenium.getAttribute("xpath=/descendant::h4[@class='box_header clearfix']/a@href");您还可以修改您的 XPath,使其更改更加灵活,甚至使用 CSS 来定位元素:
You could also modify your XPath so that it's a bit more flexible to change, or even use CSS to locate the element:
//modified xpath selenium.getAttribute("//h4[contains(@class,'box_header')]/a@href"); //css locator selenium.getAttribute("css=.box_header a@href");更多推荐
Selenium:无法理解 xPath
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