C#LINQ with XML,无法将具有相同名称的多个字段提取到对象中

Question

尝试使用LINQ XML将以下XML文件(可以更改)读入对象.

<?xml version='1.0'?>
<config>
    <report>
        <name>Adjustment Report</name>
        <extension>pdf</extension>
        <filetype>adobe_pdf</Filetype>
        <field name="total" type="currency" />
        <field name="adjust" type="currency" />
        <field name="monthly" type="currency" />
        <output>
            <format>Excel</format>
            <AutoFormat>True</AutoFormat>
        </output>
        <reportstart>adjustment report</reportstart>
        <reportend></reportend>
        <linebegins>
            <regex>(?&lt;ssn&gt;\d{3}-\d{2}-\d{4})</Regex>
        </linebegins>
        <regex>"(?&lt;last&gt;\s{0,1}[A-Z-'.]+\s{0,1}[A-Z-'.]+),(?&lt;first&gt;\s[A-Z-'.]+\s{0,1})(?&lt;middle&gt;[A-Z][.]|\s{0,1})"></Regex>
        <regex>"(?&lt;ssn&gt;\d{3}-\d{2}-\d{4})"</Regex>
        <regex>"(?&lt;total&gt;\$[,/d]+)(?&lt;adjust&gt;\$[,/d]+)(?&lt;monthly&gt;\$[,/d]+)"</Regex>
    </report>
</config>

什么是无效的是将多个元素读入对象.我只能读第一个.显然,持有该字段的Object需要是一个数组？这是我到目前为止的代码.

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

namespace ElementDemo
{
 class Program
 {
  static void Main(string[] args)
  {

   XElement xml = XElement.Load("C:\\TEMP\\test.xml");

   var reports = from report in xml.Descendants("report")
    where report.Element("name").Value.Contains("Adjustment Report")
    select new
    {
    Name = report.Element("name").Value,
    Extension = report.Element("extension").Value,
    FileType = report.Element("filetype").Value,
    // Fields : How is this done?
   };

   foreach(var obj in reports)
   {
    Console.WriteLine("Name: " + obj.Name );
   };
   Console.ReadLine();
     }
 }
}

提前致谢.

Answer 1

使用该Elements方法获取所有field元素,然后调用Select将它们转换为对象.

例如:

var reports = from report in xml.Descendants("report")
    where report.Element("name").Value.Contains("Adjustment Report")
    select new {
        Name = report.Element("name").Value,
        Extension = report.Element("extension").Value,
        FileType = report.Element("filetype").Value,
        Fields = report.Elements("field")
            .Select(f => new {
                Name = f.Attribute("name").Value, 
                Type = f.Attribute("type").Value 
            }).ToArray()
    };