Question

我有多个元素和属性。我只想读几列。从下面的代码，我只想阅读值和评论。排除类型，mimetype和空格。

   <xsd:element name="data">
        <xsd:complexType>
          <xsd:sequence>
            <xsd:element name="value" type="xsd:string" minOccurs="0" msdata:Ordinal="1" />
            <xsd:element name="comment" type="xsd:string" minOccurs="0" msdata:Ordinal="2" />
          </xsd:sequence>
          <xsd:attribute name="name" type="xsd:string" use="required" msdata:Ordinal="1" />
          <xsd:attribute name="type" type="xsd:string" msdata:Ordinal="3" />
          <xsd:attribute name="mimetype" type="xsd:string" msdata:Ordinal="4" />
          <xsd:attribute ref="xml:space" />
        </xsd:complexType>
      </xsd:element>

这是我用来读取文件的代码，但它显示的类型，mimetype和空格我不想要这些额外的列。我该怎么做

C＃代码

   #region this is the method that load the file to the grid
    public void load()
    {
        try
        {
            oDataSet = new DataSet();
            oDataTable = new DataTable();

            //there must always be a file selected
            if (txtInputfile.Text.Length > 0)
            {
                PathSelection = txtInputfile.Text;
            }
            //now adding the rows
            foreach (DataGridViewColumn C in Gridview_Input.Columns)
            {
                oDataTable.Columns.Add(C.DataPropertyName, C.ValueType);
            }

            foreach (DataGridView _Row in Gridview_Input.Rows)
            {
                DataRow row1 = oDataTable.NewRow();
                for (int i = 0; i < Gridview_Input.ColumnCount; i++)
                {
                    oDataTable.Rows.Add(row1);
                }
            }
            //here am transfering the datatable to the dataset
            oDataSet.Tables.Add(oDataTable);
        }

浏览要加载的文件的按钮

  private void btnBrowse1_Click(object sender, EventArgs e)
    {
        try
        {
            BrowseFile();
            //this method loads the file
            load();

            oDataSet = new DataSet();
            //now am reading the files fro the path that is selected
            oDataSet.ReadXml(PathSelection);
            Gridview_Input.DataSource = oDataSet;
            Gridview_Input.DataMember = "Data";
        }
        catch (Exception ex)
        {
            MessageBox.Show("Error occured.\nError message: " + ex.Message, "Error Occured", MessageBoxButtons.OK, MessageBoxIcon.Error);
        }
    }

感谢您的帮助！

Answer 1

您可以使用您想要使用的变量创建一个对象，然后从xml中读取并在阅读时只使用您需要的值填充List

修改

例如：我想你知道如何创建一个类，我用来从xml读取值到List的方法是这样的：

List<YourClass> objectForValuesAndComments=new List<YourClass>(); XmlDocument xmlDoc = new XmlDocument(); xmlDoc.Load(@"C:\Location To XML\File.xml"); foreach (XmlNode subnode in xmlDoc.DocumentElement.ChildNodes) { YourClass objectToInsert=new YourClass(); foreacht(XmlNode SubSubNode in subnode.ChildNodes) { if (SubSubNode.Name == "value") { objectToInsert.Value = SubSubNode.InnerText; } else if (SubSubNode.Name == "comment") { objectToInsert.Comment= SubSubNode.InnerText; } } objectForValuesAndComments.Add(objectToInsert); }

您可以在调试时进行修改，看看哪种方式最适合您:) 希望它有所帮助。

如何在xml中读取多个元素（列）

1 个答案: