Logo Questions Linux Laravel Mysql Ubuntu Git Menu
 

XDocument get XML element by the value of its name attribute

I have an XML result like this

<response>
  <lst name="responseHeader">
    <int name="status">0</int>
    <int name="QTime">16</int>
  </lst>
  <result name="response" numFound="3" start="0" maxScore="1.0">
    <doc>
      <str name="ContaFirstname">
        firstname1                                                   
      </str>
      <str name="ContaId">6557</str>
      <str name="ContaJobTitle">Manager</str>
      <str name="ContaSurname">surname1
      </str>
    </doc>
    <doc>
      <str name="ContaFirstname">firstname2</str>
      <str name="ContaId">6203</str>
      <str name="ContaJobTitle">Director</str>
      <str name="ContaSurname">surname2</str>
    </doc>
  </result>
</response>

I want to get a list of objects, and every object will contain the value of ContaFirstname, ContaId, ContaJobTitle and ContaSurname

I tried something like this, but that's not right because I get them all NULL

var test = from c in xml.Descendants("doc")
                    select new 
                    {
                        firstname = c.Element("ContaFirstname"),
                        surnmane = c.Element("ContaSurname")
                    }; 

So how can access these elements by name?

like image 404
kbaccouche Avatar asked Jan 24 '13 14:01

kbaccouche


2 Answers

You don't want to access the elements by name as most people would interpret that statement. You want to access the elements by the value of their name attribute:

firstname = (string) c.Elements("str")
                      .First(x => x.Attribute("name").Value == "ContaFirstname");
//etc

You may well want to abstract that into a separate method, as it's going to be a pain to do it multiple times. For example:

public static XElement ElementByNameAttribute(this XContainer container,
                                              string name)
{
    return container.Elements("str")
                    .First(x => x.Attribute("name").Value == name);
}

Then:

var test = from c in xml.Descendants("doc")
           select new 
           { 
               firstname = c.ElementByNameAttribute("ContaFirstname").Value,
               surnmane = c.ElementByNameAttribute("ContaSurname").Value
           }; 

If you have any chance to give your document a more sensible structure, that would be preferable...

like image 96
Jon Skeet Avatar answered Oct 25 '22 18:10

Jon Skeet


Does this solve your problem:

var test = from c in xml.Descendants("doc")
           select new 
           {
               firstname = c.Elements("str").First(element => element.Attribute("name").Value == "ContaFirstname"),
               surnmane = c.Elements("str").First(element => element.Attribute("name").Value == "ContaSurname")
           }; 

or, if you want the values (instead of XElement:

var test = from c in xml.Descendants("doc")
           select new 
           {
               firstname = c.Elements("str").First(element => element.Attribute("name").Value == "ContaFirstname").Value,
               surnmane = c.Elements("str").First(element => element.Attribute("name").Value == "ContaSurname").Value
           }; 
like image 25
Cédric Bignon Avatar answered Oct 25 '22 18:10

Cédric Bignon