Xml节点的Xml后代

时间:2018-05-18 01:02:52

标签: c# xml xpath

这是一个简化的mxl结构'xml',

<store>

<book_1>
   <author_1><name>Alice</name><age>30</age></author_1>
   <author_2><name>Bob</name><age>31</age></author_2>
<book_1>
   <author_1><name>Charley</name><age>29</age></author_1>
   <author_2><name>Dory</name><age>25</age></author_2> 
<book_1>    
</store>

这是我尝试过的;

 XmlDocument submission = new XmlDocument();
   submission.LoadXml(xml);
    var bookNodes = submission.SelectNodes("//*[starts-with(local-name(),'book_')]");

这给了我一本书的清单。

 foreach (XmlNode book in bookNodes)
  {
    //I want to do something like to find book authors for the book in context e.g. for the first book I just want nodes for Alice and Bob.
   // var bookAuthors = book.SelectNodes("decendants::[starts-with(local-name(),'author_')"); 

  }

我怎样才能开始检查后代元素?

编辑: 好像这是一个错字......

var bookAuthors = book.SelectNodes(“ descendant :: * [starts-with(local-name(),'MeritCriterion _')] ”);

1 个答案:

答案 0 :(得分:1)

您可以使用以下XPath语法访问后代节点:

XmlDocument submission = new XmlDocument();
submission.LoadXml(xml);

var bookNodes = submission.SelectNodes("//*[starts-with(local-name(),'book_')]");

foreach (XmlNode book in bookNodes)
{
    var author = book.SelectNodes("descendant::*[starts-with(local-name(),'author_')]");

    foreach (XmlNode authorInfo in author)
    {
        Console.WriteLine(authorInfo .InnerText);
    }
}

简而言之,您需要访问descendant::(all)[starts-with],否则您只是尝试在XPath中不访问任何后代。 :)