擅长:python、mysql、java
<p>您可以尝试<a href="http://lxml.de/tutorial.html" rel="nofollow noreferrer">lxml.etree</a>如下-我使用循环来选择具有相同位置的所有节点。在</p>
<p><strong>所需xpath的示例是</strong>-我使用了<a href="https://stackoverflow.com/questions/27183353/what-is-the-difference-between-absolute-and-relative-xpaths-which-is-preferred">relative xpath</a>,因为它在长节点路径的情况下非常有用。在</p>
<p><code>.//hudson.model.StringParameterDefinition/name[contains(text(),'project_name')]/following-sibling::defaultValue</code></p>
<p>或者</p>
<pre><code>.//hudson.model.StringParameterDefinition/name[contains(text(),'project_name')]/following::defaultValue[1]
</code></pre>
<hr/>
^{pr2}$
<p>输出-</p>
<pre><code>['my_customer', '*****']
['*****']
['*****']
</code></pre>