如何将 csv 转换为 xml，其中 csv 中的每一列 = xml 标记的属性

Question

如何通过 C# 将 csv 转换为 xml，其中 csv 中的每一列都是 xml 标记中的属性？让我告诉你我的意思：这是 CSV：

1;A;a;b;c;d;
2;B;e;f;g;h;

这是Xml:

<section>
  <row code="1" s1="A">
    <col code="5">a</col>
    <col code="6">b</col>
    <col code="7">c</col>
    <col code="8">d</col>
  </row>
  <row code="2" s1="B">
    <col code="5">e</col>
    <col code="6">f</col>
    <col code="7">g</col>
    <col code="8">h</col>
  </row>
</section>

其中 "row code" = 第 1 列，"s1"=第 2 列，"col code=5"=第 3 列，"col code=6"=第 4 列等。这就是任务。

现在我有 xmlSchema class、csv class，但我在映射 csv 时遇到一些问题 - xml。这是 XMLSchema.cs 的一部分：

public partial class reportSectionsSectionRowCol
{
    private string codeField;
    private string valueField;
    /// <remarks/>
    [System.Xml.Serialization.XmlAttributeAttribute()]
    public string code
    {   get {return this.codeField;}
        set {this.codeField = value;}
    }
    [System.Xml.Serialization.XmlTextAttribute()]
    public string Value
    {
        get{return this.valueField;}
        set{this.valueField = value;}
    }
}

col - XML 的元素具有属性代码和值。

<row code="1" s1="A">
    <col code="5">a</col>
    <col code="6">b</col>
    <col code="7">c</col>
    <col code="8">d</col>
</row>

这是映射：

static void MappingCSVtoXML(report report, CSVData csvData)
        {
            if (csvData != null)
            {
                var col = new reportSectionsSectionRowCol[]{   new reportSectionsSectionRowCol(),
                                                               new reportSectionsSectionRowCol(),
                                                               new reportSectionsSectionRowCol(),
                                                               new reportSectionsSectionRowCol()};                


                var row = new reportSectionsSectionRow[2] { new reportSectionsSectionRow(), new reportSectionsSectionRow() };
                for (int i = 0; i < 2; i++)
                {
                    col[0].code = "5"; col[0].Value = csvData.rows[i].depreciationGroupNumber;
                    col[1].code = "6"; col[1].Value = csvData.rows[i].commissioningYear;
                    col[2].code = "7"; col[2].Value = csvData.rows[i].startYear;
                    col[3].code = "8"; col[3].Value = csvData.rows[i].ageAtLiquidation;


                    row[i].code = csvData.rows[i].okof.Substring(0, 3); // 1,2
                    row[i].col = col; // Reference type
                    row[i].s1 = csvData.rows[i].rowNumber; //A,B                               
                }

                report.sections = new reportSectionsSection[1]{new reportSectionsSection(){code = "1", row = row}};

            }
        }

问题是 col 是引用类型并且只记住最后一个值。总的结果是

<section>
  <row code="1" s1="A">
    <col code="5">e</col>
    <col code="6">f</col>
    <col code="7">g</col>
    <col code="8">h</col>
  </row>
  <row code="2" s1="B">
    <col code="5">e</col>
    <col code="6">f</col>
    <col code="7">g</col>
    <col code="8">h</col>
  </row>
</section>

我该如何解决？

Answer 1

这是我的class结构

的映射代码

static Section Map( IEnumerable<Csv1Data> source )
{
    return new Section
    {
        Rows = source
            .Select( e => new SectionRow
            {
                Code = e.Code,
                Value = e.S1,
                Columns = new List<SectionRowCol>
                {
                    new SectionRowCol{ Code = 5, Value = e.Col5, },
                    new SectionRowCol{ Code = 6, Value = e.Col6, },
                    new SectionRowCol{ Code = 7, Value = e.Col7, },
                    new SectionRowCol{ Code = 8, Value = e.Col8, },
                }
            } )
            .ToList(),
    };
}

简单的检查和用例

static void Main( string[] args )
{
    var source = new List<Csv1Data> {
        new Csv1Data { Code = 1, S1 = "A", Col5 = "a", Col6 = "b", Col7 = "c", Col8 = "d" },
        new Csv1Data { Code = 2, S1 = "B", Col5 = "a", Col6 = "b", Col7 = "c", Col8 = "d" },
    };

    var output = Map( source );

    XmlSerializer ser = new XmlSerializer( typeof( ExportModels.Xml1.Section ) );

    using ( var stream = new MemoryStream() )
    {
        using ( var writer = new StreamWriter( stream, Encoding.UTF8 ) )
        {
            ser.Serialize( writer, output );
        }

        var bytes = stream.ToArray();
        var str = Encoding.UTF8.GetString( bytes );
        Console.WriteLine( str );
    }
}

XML-Data-Classes

// HINWEIS: Für den generierten Code ist möglicherweise mindestens .NET Framework 4.5 oder .NET Core/Standard 2.0 erforderlich.
/// <remarks/>
[System.SerializableAttribute()]
[System.ComponentModel.DesignerCategoryAttribute( "code" )]
[System.Xml.Serialization.XmlTypeAttribute( AnonymousType = true )]
[System.Xml.Serialization.XmlRootAttribute( Namespace = "", IsNullable = false, ElementName = "section" )]
public partial class Section
{

    private List<SectionRow> _rowField;

    /// <remarks/>
    [System.Xml.Serialization.XmlElementAttribute( "row" )]
    public List<SectionRow> Rows
    {
        get
        {
            return this._rowField;
        }
        set
        {
            this._rowField = value;
        }
    }
}

/// <remarks/>
[System.SerializableAttribute()]
[System.ComponentModel.DesignerCategoryAttribute( "code" )]
[System.Xml.Serialization.XmlTypeAttribute( AnonymousType = true )]
public partial class SectionRow
{

    private List<SectionRowCol> _columnsField;

    private int _codeField;

    private string _valueField;

    /// <remarks/>
    [System.Xml.Serialization.XmlElementAttribute( "col" )]
    public List<SectionRowCol> Columns
    {
        get
        {
            return this._columnsField;
        }
        set
        {
            this._columnsField = value;
        }
    }

    /// <remarks/>
    [System.Xml.Serialization.XmlAttributeAttribute( AttributeName = "code" )]
    public int Code
    {
        get
        {
            return this._codeField;
        }
        set
        {
            this._codeField = value;
        }
    }

    /// <remarks/>
    [System.Xml.Serialization.XmlAttributeAttribute( AttributeName = "s1" )]
    public string Value
    {
        get
        {
            return this._valueField;
        }
        set
        {
            this._valueField = value;
        }
    }
}

/// <remarks/>
[System.SerializableAttribute()]
[System.ComponentModel.DesignerCategoryAttribute( "code" )]
[System.Xml.Serialization.XmlTypeAttribute( AnonymousType = true )]
public partial class SectionRowCol
{

    private byte _codeField;

    private string _valueField;

    /// <remarks/>
    [System.Xml.Serialization.XmlAttributeAttribute( AttributeName = "code" )]
    public byte Code
    {
        get
        {
            return this._codeField;
        }
        set
        {
            this._codeField = value;
        }
    }

    /// <remarks/>
    [System.Xml.Serialization.XmlTextAttribute()]
    public string Value
    {
        get
        {
            return this._valueField;
        }
        set
        {
            this._valueField = value;
        }
    }
}

CSV-Data-Class

class Csv1Data
{
    public int Code { get; set; }
    public string S1 { get; set; }
    public string Col5 { get; set; }
    public string Col6 { get; set; }
    public string Col7 { get; set; }
    public string Col8 { get; set; }
}

如何将 csv 转换为 xml，其中 csv 中的每一列 = xml 标记的属性

How to convert csv into xml where each column in csv = xml attribute of tag

c#

xml

csv

converters