XML to LINQ element selection

Question

I have an xml document in the following format.

<?xml version="1.0" encoding="UTF-8" ?>

<Rows>
<Row>    
    <Field Name='PhysicalLocation'>11;#West</Field>
    <Field Name='ID'>3327</Field>
</Row>
</Rows>

And am attempting to do a linq selection with it.

I have tried the following.

XDocument xmlDoc = XDocument.Load("C:\\manifest.xml");

var query = from item in xmlDoc.Descendants("Rows").Elements()
                    select new { ID = item.Attribute("ID").Value, Value = item.Attribute("PhysicalLocation").Value };

And also

XDocument xmlDoc = XDocument.Load("C:\\manifest.xml");

var query = from item in xmlDoc.Descendants("Rows").Elements()
                    select new { ID = item.Element("ID"), Value = item.Element("PhysicalLocation") };

And in both cases I seem to be coming up short. It is generating the expected amount of rows but the values are not being filled.

Could anyone point me in the right direction? What am I missing?

Enigmativity · Accepted Answer · 2012-02-06 23:34:10Z

1

How about trying a query like this:

var query =
    from item in xmlDoc.Descendants("Rows").Elements()
    let values = item.Elements("Field")
        .ToDictionary(x => x.Attribute("Name").Value, x => x.Value)
    select new
    {
        ID = values["ID"],
        Value = values["PhysicalLocation"],
    };

answered Feb 6, 2012 at 23:34

Enigmativity

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3 Comments

Maxim Gershkovich Over a year ago

Thank you for your input but unfortunately with the above code I'm receiving the following error: {"The given key was not present in the dictionary."}

Enigmativity Over a year ago

@MaximGershkovich - I used your example XML provided in your question and it worked fine for me. Let me know if you have bad XML or if you have slightly more complicated requirements.

Maxim Gershkovich Over a year ago

My bad, I had some bad data (I had a 40mb XML dataset). You're correct, your code works great. Thank you...

Antony Scott · Accepted Answer · 2012-02-06 23:30:04Z

1

you don't have an attribute called 'PhysicalLocation' or 'ID'. you only have attributes called 'Name'.

You need to add a where clause on the value of the 'name' attribute to find your ID and PhysicalLocation

answered Feb 6, 2012 at 23:30

Antony Scott

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Comments

Pawel · Accepted Answer · 2012-02-07 00:34:00Z

0

I think this should do the trick:

var xDoc = XDocument.Parse(
@"<?xml version='1.0' encoding='UTF-8' ?> 

<Rows> 
    <Row>     
        <Field Name='PhysicalLocation'>11;#West</Field> 
        <Field Name='ID'>3327</Field> 
    </Row> 
</Rows>");

var res = from row in xDoc.Root.Elements("Row")
          select new
          {
              ID = (int)row.Elements("Field").Single(e => (string)e.Attribute("Name") == "ID"),
              PhysicalLocaltion = (string)row.Elements("Field").Single(e => (string)e.Attribute("Name") == "PhysicalLocation"),
          };

foreach (var r in res)
{
    Console.WriteLine(r);
}

Here is the result printed by the loop:

{ ID = 3327, PhysicalLocaltion = 11;#West }

answered Feb 7, 2012 at 0:34

Pawel

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Comments

mnsr · Accepted Answer · 2012-02-07 00:50:08Z

Try this:

        var xd = XDocument.Load("C:\\manifest.xml");
        var query = xd.Root.Descendants("Row").Elements("Field")
            .Select(s => new
            {
                Name = (string)s.Attribute("Name"),
                Value = s.Value
            });

The above code, loops through each "Row" element, then reads the "Field" element data. It will return the following anonymous list:

Name = PhysicalLocation
Value = 11;#West
Name = ID
Value = 3327

To loop through the query, you can use the following code:

        var sb = new StringBuilder();
        foreach (var i in query)
        {
            sb.Append("\n Name = ").Append(i.Name).Append("\n Value = ").Append(i.Value);
        }

Finally, to find the Field element value by Name, you can use the following query:

var query2 = query.Where(w => w.Name == "ID").Single().Value;

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