How to properly search xml document using LINQ C#
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 ofToList()
; - Element names are case sensitive so pay attention to Firstname for example.
- You can use
XElement.Load()
to load from file for exampleXElement 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())
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