在特定属性处加入 3 个 XML 文档并使用 ZD7EFA19FBE20239D74Z5ADB6 中的 LINQ 获取单个 XML 文档

Question

1. I have 3 xml document data.
2. Each have similar attribute "ID"
3. I tried to join 2 xmls but not getting any joy so far as it result into error saying object null blaa..blaa.
4. Is there anyone who can help me to create a 4th XML output as below

<--INPUT XML file #1-->
<name_info>
    <person id = "101">
        <name>"Thomas Edison"<name>
        <age>"35"</age>
    </person>
        <person id = "102">
        <name>"Louis Pasture"<name>
        <age>"40"</age>
    </person>
</name_info>

<--INPUT XML file #2-->
<phone_info>
    <person id = "101">
        <phone>"7777777777"</phone>
    </person>
    <person id = "102">
        <phone>"99999999"</phone>
    </person>
</phone_info>

<--INPUT XML file #3-->
<country_info>
    <person id = "101">
        <country>"England"</country>
    </person>
    <person id = "102">
        <country>"Paris"</country>
    </person>
</country_info>

<--OUTPUT XML file #4-->
<person>
    <person id = "101">
        <name>"Thomas Edison"<name>
        <age>"35"</age>
        <phone>"7777777777</phone>
        <country>"England"</country>
        </person>
    <person id = "102">
        <name>"Louis Pasture"<name>
        <age>"40"</age>
        <phone>"99999999</phonephone>
        <country>"Paris"</country>
    </person>
<person/>

Answer 1

Xml Attributes value need double quotes around the values like

See xml linq solution below:

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


namespace ConsoleApplication4
{
    class Program
    {
        const string FILENAME = @"c:\temp\test.xml";
        const string FILENAME1 = @"c:\temp\test1.xml";
        const string FILENAME2 = @"c:\temp\test2.xml";
        static void Main(string[] args)
        {
            XDocument doc = XDocument.Load(FILENAME);
            XDocument doc1 = XDocument.Load(FILENAME1);
            XDocument doc2 = XDocument.Load(FILENAME2);


            var joins = (from d in doc.Descendants("person")
                         join d1 in doc1.Descendants("person") on (int)d.Attribute("id") equals (int)d1.Attribute("id")
                         join d2 in doc2.Descendants("person") on (int)d.Attribute("id") equals (int)d2.Attribute("id")
                         select new { d = d, d1 = d1, d2 = d2 }
                         ).ToList();

            string ident = "<?xml version=\"1.0\" encoding=\"utf-8\" ?> <person></person>";

            XDocument outputXml = XDocument.Parse(ident);
            XElement persons = outputXml.Root;

            foreach (var join in joins)
            {
                XElement person = XElement.Parse(join.d.ToString());
                person.Add(XElement.Parse(join.d1.Element("phone").ToString()));
                person.Add(XElement.Parse(join.d2.Element("country").ToString()));
                persons.Add(person);


            }

        }
    }


}