c# XML基于属性获取节点

I have the following xml:

我有以下xml:

<root ...>
  <Tables>
    <Table content="..">
    </Table>
    <Table content="interesting">
      <Item ...></Item>
      <Item ...></Item>
      <Item ...></Item>
    </Table>
    ...etc...
  </Tables>
</root>

I'm using the following code to get the items from the 'interesting' node:

我使用以下代码从“有趣”节点获取项目:

XElement xel = XElement.Parse(resp);

var nodes = from n in xel.Elements("Tables").Elements("Table")
            where n.Attribute("content").Value == "interesting"
            select n;

var items = from i in nodes.Elements()
            select i;

Is there a simpler, cleaner way to achieve this?

有没有更简单、更干净的方法来实现这个目标?

4 个解决方案

#1

Well there's no point in using a query expression for items, and you can wrap the whole thing up very easily in a single statement. I wouldn't even bother with a query expression for that:

对项使用查询表达式是没有意义的，您可以很容易地将整个内容封装在一个语句中。我甚至不需要查询表达式:

var items = XElement.Parse(resp)
                    .Elements("Tables")
                    .Elements("Table")
                    .Where(n => n.Attribute("content").Value == "interesting")
                    .Elements();

Note that this (and your current query) will throw an exception for any Table element without a content attribute. If you'd rather just skip it, you can use:

注意，这(以及您当前的查询)将为任何没有content属性的表元素抛出异常。如果你宁愿跳过它，你可以使用:

.Where(n => (string) n.Attribute("content") == "interesting")

instead.

代替。

#2

You can use XPath (extension is in System.Xml.XPath namespace) to select all items in one line:

可以使用XPath(扩展名在System.Xml中。选择一行中的所有项:

var items = xel.XPathSelectElements("//Table[@content='interesting']/Item");

#3

If you don't need nodes outside of your query for items, you can just do this:

如果在查询项之外不需要节点，可以这样做:

var items = from n in xel.Elements("Tables").Elements("Table")
            where n.Attribute("content").Value == "interesting"
            from i in n.Elements()
            select i;

#4

using xml document
XmlDocument xdoc = new XmlDocument();

使用xml文档XmlDocument xdoc = new XmlDocument();

var item= xdoc.GetElementsByTagName("Table[@content='interesting']/Item");

var = xdoc.GetElementsByTagName项(“表[@content = '有趣']/项目”);

#1