XML attributes c#

Question

I need to display the processingStatus and every field node value along with it's corresponding name of the following XML.

<?xml version="1.0" encoding="UTF-8"?>
<T24 xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns="http://www.temenos.com/T24/OFSML/130" xsi:schemaLocation="http://www.temenos.com/T24/OFSML/130 ofsml13.xsd">
  <serviceResponse>
    <ofsTransactionProcessed application="FUNDS.TRANSFER" function="INPUT" operation="VALIDATE" processingStatus="OK" version="BOOK.TRAN.VAL.MCB">
      <transactionId>FT15056T2QLP</transactionId>
      <field mv="1" name="TRANSACTION.TYPE" sv="1">AC</field>
      <field mv="1" name="CURRENCY.MKT.DR" sv="1">1</field>
      <field mv="1" name="DEBIT.CURRENCY" sv="1">USD</field>
      <field mv="1" name="DEBIT.AMOUNT" sv="1">125.00</field>
      <field mv="1" name="DEBIT.VALUE.DATE" sv="1">20150225</field>

    </ofsTransactionProcessed>
  </serviceResponse>
</T24>

Can anyone help me please? my code: EDITED CODE:

XmlTextReader reader = new XmlTextReader("XML03.xml");
            Console.WriteLine(reader.Attributes["processingStatus"].Value);
            while (reader.Read())
            {
                Console.WriteLine(reader.Value);

            }
            Console.ReadLine();

now please describe whats wrong with the code along with output/exception you are getting — Steve
– Steve, Commented Aug 15, 2016 at 21:10
I usually use XElement msdn.microsoft.com/en-us/library/… for retrieving values from XML. You may want to look into it for this. If you know the names of the attributes, you should have no problems extracting the values and printing them out. — Zack
– Zack, Commented Aug 15, 2016 at 21:12

Codor · Accepted Answer · 2016-08-15 21:10:54Z

1

One possibility would be to define a type or several types which reflect the structure of the Xml; the members would have to be endowed with the attributes XmlElementAttribute which is documented here, and the attribute XmlAttributeAttribute, which is documented here; if done suitably, the file contents can be deserialized with the XmlDeserializer class, which is documented here. A detailed explanation on how to use the deserialization facilities can be found in this official tutorial.

answered Aug 15, 2016 at 21:10

Codor

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jdweng · Accepted Answer · 2016-08-15 21:12:10Z

1

Try xml linq

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Xml;
using System.Xml.Linq;

namespace ConsoleApplication1
{
    class Program
    {
        const string FILENAME = @"c:\temp\test.xml";
        static void Main(string[] args)
        {
            XDocument doc = XDocument.Load(FILENAME);
            var results = doc.Descendants().Where(x => x.Name.LocalName == "ofsTransactionProcessed").Select(x => new {
                application = (string)x.Attribute("application"),
                function = (string)x.Attribute("function"),
                operation = (string)x.Attribute("operation"),
                processingStatus = (string)x.Attribute("processingStatus"),
                version = (string)x.Attribute("version"),
                transactionId = x.Elements().Where(y => y.Name.LocalName == "transactionId").Select(z => (string)z).FirstOrDefault(),
                fields = x.Elements().Where(y => y.Name.LocalName == "field").Select(z => new {
                    mv = (string)z.Attribute("mv"),
                    name = (string)z.Attribute("name"),
                    sv = (string)z.Attribute("sv"),
                    value = (string)z
                }).ToList()
            }).FirstOrDefault();
        }
    }
}

answered Aug 15, 2016 at 21:12

jdweng

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3 Comments

jdweng Over a year ago

How big is the xml file? XmlReader isn't needed for small xml files. What tags repeat? I can modify code to work with xml reader but need the elements that repeat. If your input is a string then simply replace Load(FILENAME) with Parse(your xml string).

velvt Over a year ago

How to I output all the results? i.e processingStatus and fields?

jdweng Over a year ago

What format? You can get each property using results.application, results.function, .... You can get the fields using a foreach like : foreach(var field in fields){ field.mn, field.name, field.sv, and field.vaue}

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