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How to properly search xml document using LINQ C#

Tags:

c#

xml

linq

I have a hard time time figuring out how to properly search an xml document. I have been reading the other forms like crazy today but just can't seem to understand it. Was hopeing someone could give me a little more detailed information on how to do this the correct way and why using LINQ. Here is the XML file.

<?xml version="1.0" encoding="utf-8"?>
<body>
  <Customers>
    <Client>
      <Firstname Value="someguy" />
      <LastName Value="test" />
      <PhoneNumber Value="541555555" />
      <Address Value="55 nowhere" />
      <City Value="sometown" />
      <State Value="somestate" />
    </Client>
  </Customers>
</body>

What I am tyring to accomplish is to return all of the values of each element that matches by a name of a customer. Here is my code.

       IEnumerable<XElement> test = doc.Root.Descendants()
            .Where(nodeName => nodeName.Name == "Client"
            && nodeName.Descendants().Any(x => x.Name == "Firstname"
            && x.Value == "someguy"));

        foreach (XElement m in test)
        {
            MessageBox.Show(m.ToString());
        }

Would really appreciate the help. Please also if possible explain what the idea of using LINQ is like the format if you will. Not really sure how to explain what I am asking but for the most part just more understanding of the way it works or format etc...

EDIT

I have tried the solution given and still nothing is seeming to work. Please show me what I am doing wrong here.

        private void button2_Click(object sender, EventArgs e)
    {
        string seach = txtSearch.Text;

        XDocument doc = XDocument.Load(@"C:\users\tim\desktop\test.xml");

        var result = doc.Elements("Customers")
            .Elements("Client")
            .Where(x => x.Elements("Firstname")
                         .Where(c => c.Attribute("Value").Value == "someguy")
                         .Any())
            .ToList();

        foreach (var m in result)
        {
            MessageBox.Show(m.ToString());
        }
    }

EDIT:

Okay so I have gotten it to work now and output the data that I was looking for. Can someone please tell me if there is a more efficent way of doing what I have posted below this edit.

        private void button2_Click(object sender, EventArgs e)
    {
        string seach = txtSearch.Text;

        XDocument doc = XDocument.Load(@"C:\users\tim\desktop\test.xml");

        var result = (from clientNode in doc.Root.Descendants("Client")
                      from name in clientNode.Descendants("Firstname")
                      where name.Attribute("Value").Value == "someguy"
                      select new
                      {
                          Fname = clientNode.Element("Firstname").Attribute("Value").Value,
                          Lname = clientNode.Element("LastName").Attribute("Value").Value,
                          Phone = clientNode.Element("PhoneNumber").Attribute("Value").Value,
                          Address = clientNode.Element("Address").Attribute("Value").Value,
                          City = clientNode.Element("City").Attribute("Value").Value,
                          State = clientNode.Element("State").Attribute("Value").Value
                      });

        foreach (var m in result)
        {
            MessageBox.Show(m.Fname + "\n" +
                            m.Lname + "\n" +
                            m.Phone + "\n" +
                            m.Address + "\n" +
                            m.City + "\n" +
                            m.State);
        }
like image 683
Timg Avatar asked Sep 29 '15 20:09

Timg


2 Answers

You can use XElement and search using a LINQ query like these:

XElement doc = XElement.Parse(xml);
var result = doc.Elements("Customers")
                .Elements("Client")
                .Where(x => x.Elements("Firstname")
                             .Where(c => c.Attribute("Value").Value == "someguy")
                             .Any())
                .ToList();

So with input:

var xml =
@"<?xml version=""1.0"" encoding=""utf-8""?>
<body>
    <Customers>
    <Client>
        <Firstname Value=""someguy"" />
        <LastName Value=""some last name"" />
        <PhoneNumber Value=""123456"" />
        <Address Value=""some where"" />
        <City Value=""some town"" />
        <State Value=""some state"" />
    </Client>
    <Client>
        <Firstname Value=""someotherguy"" />
        <LastName Value=""some other last name"" />
        <PhoneNumber Value=""123456"" />
        <Address Value=""some other where"" />
        <City Value=""some other town"" />
        <State Value=""some other state"" />
    </Client>
    </Customers>
</body>";

XElement doc = XElement.Parse(xml);
var result = doc.Elements("Customers")
                .Elements("Client")
                .Where(x => x.Elements("Firstname")
                             .Where(c => c.Attribute("Value").Value == "someguy")
                             .Any())
                .ToList();

The result will be:

<Client>
    <Firstname Value=""someguy"" />
    <LastName Value=""some last name"" />
    <PhoneNumber Value=""123456"" />
    <Address Value=""some where"" />
    <City Value=""some town"" />
    <State Value=""some state"" />
</Client>

And you can show values for example:

MessageBox.Show(string.Format("Firstname: {0}\nLastName: {1}\nPhoneNumber: {2}\nAddress: {3}\nCity: {4}\nState: {5}",
                result[0].Element("Firstname").Attribute("Value").Value,
                result[0].Element("LastName").Attribute("Value").Value,
                result[0].Element("PhoneNumber").Attribute("Value").Value,
                result[0].Element("Address").Attribute("Value").Value,
                result[0].Element("City").Attribute("Value").Value,
                result[0].Element("State").Attribute("Value").Value));

Note:

  • If you know the result should contain 0 or 1 elements, you can use FirstOrDefault() instead of ToList();
  • Element names are case sensitive so pay attention to Firstname for example.
  • You can use XElement.Load() to load from file for example XElement doc = XElement.Load(@"d:\file.xml");
  • The query will be more fault tolerant if you select elements this way .Where(c => c.Name.ToString().ToLower() == "Customers".ToLower())
  • The query will be more fault-tolerant if you select attributes this way .Where(c => c.Attributes("Value").Where(a=>a.Value == "someguy").Any())
like image 199
Reza Aghaei Avatar answered Oct 17 '22 11:10

Reza Aghaei


As an alternative to @Reza Aghaei's solution, XPath is also a solution

var xml =
@"<?xml version=""1.0"" encoding=""utf-8""?>
<body>
    <Customers>
    <Client>
        <Firstname Value=""someguy"" />
        <LastName Value=""some last name"" />
        <PhoneNumber Value=""123456"" />
        <Address Value=""some where"" />
        <City Value=""some town"" />
        <State Value=""some state"" />
    </Client>
    <Client>
        <Firstname Value=""someotherguy"" />
        <LastName Value=""some other last name"" />
        <PhoneNumber Value=""123456"" />
        <Address Value=""some other where"" />
        <City Value=""some other town"" />
        <State Value=""some other state"" />
    </Client>
    <Client>
        <Firstname Value=""someguy"" />
        <LastName Value=""some other last name"" />
        <PhoneNumber Value=""12345634543"" />
        <Address Value=""some other where"" />
        <City Value=""some other town"" />
        <State Value=""some other state"" />
    </Client>
    </Customers>
</body>";
XElement doc = XElement.Parse(xml);

foreach(var client in doc
 .XPathSelectElements("./Customers/Client/Firstname[@Value='someguy']")
 .Select(x => x.Parent))
    Console.WriteLine (client);

If you prefer a Linq To Xml solution :

var results = (from c in doc.Descendants("Client")
            from f in c.Descendants("Firstname")
             where (string)f.Attribute("Value") == "someguy"
             select c).ToList();
foreach(var r in results)
    Console.WriteLine (r);

How to: Write Linq to Xml Queries with Complex Filtering

like image 25
ziddarth Avatar answered Oct 17 '22 13:10

ziddarth