Question

我有以下DataTable：

<Data>
    <A>
        <lable>A1</lable>
        <B>
            <lable>B1</lable>
            <C>
                <lable>C1</lable>
                <lable>C2</lable>
                <lable>C3</lable>
            </C>
            <lable>B2</lable>
            <C>
                <lable>C1</lable>
                <lable>C2</lable>
            </C>
        </B>
    </A>
</Data>

我尝试使用C＃将其转换为XML格式，如下所示：

    using System;
    using System.Collections.Generic;
    using System.ComponentModel;
    using System.Data;
    using System.Drawing;
    using System.Linq;
    using System.Text;
    using System.Threading.Tasks;
    using System.Windows.Forms;
    using System.Xml.Linq;
    using System.Data.SqlClient;

    SqlConnection con = new SqlConnection("Integrated Security=SSPI;Persist Security Info=False;Initial Catalog=Files;Data Source=localhost");
    con.Open();
    SqlCommand cmd = new SqlCommand("select * from MyTable",con);
    SqlDataAdapter da = new SqlDataAdapter(cmd);
    DataTable DT = new DataTable("Data");
    da.Fill(DT);
    dataGridView1.DataSource = DT;
    con.Close();
    string XML = ToXmlFormat(DT, 0);
    Console.WriteLine(XML);


public string ToXmlFormat(this DataTable table, int metaIndex = 0)
{
    XDocument xdoc = new XDocument(
        new XElement(table.TableName,
            from column in table.Columns.Cast<DataColumn>()
            where column != table.Columns[metaIndex]
            select new XElement(column.ColumnName,
                from row in table.AsEnumerable()
                select new XElement(row.Field<string>(metaIndex), row[column])
                )
            )
        );

    return xdoc.ToString();
}

我进行了深入搜索，我在网上发现了一些有用的内容，例如this。但那个使用关系，我的数据只是一个表。我也试过这个C＃代码：

var li = document.createElement("li");

li.classList.add('my-li-class')

结果是另一种格式，不是我所解释的嵌套XML格式。 see image

如何将表格转换为我需要的XML格式？

Answer 1

您需要使用GroupBy对行进行分组，然后在XElements中选择所需的部分。
下面的代码应该做你想要的：

    var xml = new XElement(table.TableName, table.Rows.Cast<DataRow>()
        .GroupBy(row => (string)row[0])
        .Select(g =>
            new XElement(table.Columns[0].ColumnName,
                new XElement("label", g.Key),
                g.GroupBy(row => (string)row[1])
                 .Select(g1 =>
                    new XElement(table.Columns[1].ColumnName,
                        new XElement("label", g1.Key),
                        new XElement(table.Columns[2].ColumnName,
                            g1.Select(row =>
                                new XElement("label", (string)row[2])
                            )
                        )
                    )
                )
            )
        )
    ).ToString();

小提琴：https://dotnetfiddle.net/qEWNvj

Answer 2

我的风格与其他答案略有不同。代码经过测试并符合您的预期结果。

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

namespace ConsoleApplication1
{
    class Program
    {
        static void Main(string[] args)
        {
            DataTable dt = new DataTable();
            dt.Columns.Add("A", typeof(string));
            dt.Columns.Add("B", typeof(string));
            dt.Columns.Add("C", typeof(string));

            dt.Rows.Add(new object[] {"A1", "B1","C1"});
            dt.Rows.Add(new object[] {"A1", "B1","C2"});
            dt.Rows.Add(new object[] {"A1", "B1","C3"});
            dt.Rows.Add(new object[] {"A1", "B2","C1"});
            dt.Rows.Add(new object[] {"A1", "B2","C2"});

            dt = dt.AsEnumerable()
                .OrderBy(x => x.Field<string>("A"))
                .ThenBy(x => x.Field<string>("B"))
                .ThenBy(x => x.Field<string>("C"))
                .CopyToDataTable();

            XElement data = new XElement("Data", new XElement("A", dt.AsEnumerable()
                .GroupBy(g1 => g1.Field<string>("A")).Select(g1a =>  new object[] { 
                    new XElement("lable",(string)g1a.Key),
                    new XElement("B",
                        g1a.GroupBy(g2 => g2.Field<string>("B")).Select(g2b => new object[] {
                            new XElement("lable", (string)g2b.Key),
                            new XElement("C",
                                g2b.Select(g3c => new XElement("lable", g3c.Field<string>("C"))
                            ))}
                    ))}
            )));

        }
    }
}

在C＃中将DataTable转换为XML层次结构

2 个答案: