Get attribute values from XML

Question

This is my XML string.

<?xml version="1.0" encoding="UTF-8"?>
<Menu TextField="Menu" NavigateUrlField="/#" MenuID="74">
   <Menu TextField="Assets" NavigateUrlField="/Assets" MenuID="1870" />
   <Menu TextField="Asset Category" NavigateUrlField="/AssetCategory" MenuID="1871" />
   <Menu TextField="Asset Location" NavigateUrlField="/AssetLocation" MenuID="1872" />
</Menu>

I need to get the TextField and MenuId from this XML. Only from Menu tag that is below root Menu tag.

Text Field               MenuId
-------------------------------
Assets                   1870
Asset Category           1871
Asset Location           1872

I tried below code but not working:

var xmlDoc = new XmlDocument();
xmlDoc.LoadXml(xml);

XmlNodeList elemList = xmlDoc.GetElementsByTagName("Menu");
for (int i = 0; i < elemList.Count; i++)
{
    string menuId = elemList[i].Attributes["MenuID"].Value;
    string textField = elemList[i].Attributes["TextField"].Value;
}

Please someone help me.

Possible duplicate of [What is a NullReferenceException, and how do I fix it?](http://stackoverflow.com/questions/4660142/what-is-a-nullreferenceexception-and-how-do-i-fix-it) — Thomas Weller, Nov 30 '16 at 07:59

score 1 · Answer 1 · answered Nov 30 '16 at 07:43

1

Please pay attention, to you attribute name "MenuID", not "MenuId"

XmlNode root = xmlDoc.DocumentElement;
XmlNodeList elemList = root.SelectNodes("Menu");
for (int i = 0; i < elemList.Count; i++)
{       
    string menuId = elemList[i].Attributes["MenuID"].Value;
    string textField = elemList[i].Attributes["TextField"].Value;
}

answered Nov 30 '16 at 07:43

Maksim Simkin

9,561
4
36
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Holy cow, only 1 out of 6 answers that explains what is actually wrong and not rewriting the whole code. Keep on doing good work! – Thomas Weller Nov 30 '16 at 08:03

score 0 · Accepted Answer · answered Nov 30 '16 at 07:39

Using 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);
            XElement menu = doc.Element("Menu");
            var results = menu.Descendants("Menu").Select(x => new {
                textField = (string)x.Attribute("TextField"),
                menuID = (int)x.Attribute("MenuID")
            }).ToList();
        }
    }
}

Thanks. This helped me out. – good-to-know Nov 30 '16 at 07:40 — good-to-know, Nov 30 '16 at 07:40

Peter B · Answer 3 · 2016-11-30T07:44:24.463

-1

You can use a simple XPath query to find the Menu at parent level and then select all Menu nodes below it:

XmlNodeList elemList = xmlDoc.SelectNodes("Menu/Menu");

Then also change Attributes["MenuId"].Value to Attributes["MenuID"].Value because all names are case sensitive in XML.

edited Nov 30 '16 at 07:44

answered Nov 30 '16 at 07:38

Peter B

22,460
5
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Did someone by accident lean on the downvote button? My solution works, I actually tested it, so why? – Peter B Nov 30 '16 at 13:06

Get attribute values from XML

3 Answers3