C# XML get nodes based on attribute

Question

I have the following 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:

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?

That's not valid XML... the first Table doesn't have an endtag called Table... — It'sNotALie.
– It'sNotALie., Commented Jun 10, 2013 at 11:17
I really prefer to use XmlSerializer over XElement or XmlDocument. — Fendy
– Fendy, Commented Jun 10, 2013 at 11:17

Jon Skeet · Accepted Answer · 2013-06-10 11:18:28Z

5

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:

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:

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

instead.

answered Jun 10, 2013 at 11:18

Jon Skeet

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1 Comment

Sergey Berezovskiy Over a year ago

Good point on casting attribute instead of accessing Value property

Sergey Berezovskiy · Accepted Answer · 2013-06-10 11:23:06Z

2

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

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

answered Jun 10, 2013 at 11:23

Sergey Berezovskiy

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Comments

p.s.w.g · Accepted Answer · 2013-06-10 11:19:36Z

1

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

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

answered Jun 10, 2013 at 11:19

p.s.w.g

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Comments

Deepak Saralaya · Accepted Answer · 2013-06-10 11:28:09Z

1

using xml document
XmlDocument xdoc = new XmlDocument();

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

answered Jun 10, 2013 at 11:28

Deepak Saralaya

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