这里Xpath的蟒蛇发现节点是HTML代码:后特定的文本
<div id="someid">
<h2>Specific text 1</h2>
<a class="hyperlinks" href="link"> link1 inside specific text 1</a>
<a class="hyperlinks" href="link"> link2 inside specific text 1</a>
<a class="hyperlinks" href="link"> link3 inside specific text 1</a>
<h2>Specific text 2</h2>
<a class="hyperlinks" href="link"> link1 inside specific text 2</a>
<a class="hyperlinks" href="link"> link2 inside specific text 2</a>
<a class="hyperlinks" href="link"> link3 inside specific text 2</a>
<a class="hyperlinks" href="link"> link4 inside specific text 2</a>
<h2>Specific text 3</h2>
<a class="hyperlinks" href="link"> link1 inside specific text 3</a>
<a class="hyperlinks" href="link"> link2 inside specific text 3</a>
</div>
我必须清楚地找到下的每个“特定文字”链接。问题是,如果我用Python语言编写如下代码:
links = root.xpath("//div[@id='someid']//a")
for link in links:
print link.attrib['href']
它打印出的所有环节,不论“特定文字X”,而我想是这样的:
print "link under Specific text:"+specific+" link:"+link.attrib['href']
请建议
那么,基于提供的XML文档,您想要的确切输出是什么?这不清楚。请编辑您的问题并添加此要求。 –