this.dataGrid1 = new System.Windows.Forms.DataGrid();
this.dataGrid1.DataMember = "";
this.dataGrid1.Location = new System.Drawing.Point(36, 50);
this.dataGrid1.Name = "dataGrid1";
this.dataGrid1.Size = new System.Drawing.Size(464, 432);
this.dataGrid1.TabIndex = 0;
//
this.AutoScaleBaseSize = new System.Drawing.Size(35, 13);
this.ClientSize = new System.Drawing.Size(592, 573);
this.Controls.AddRange(new System.Windows.Forms.Control[] { this.dataGrid1 });
((System.ComponentModel.ISupportInitialize)(this.dataGrid1)).EndInit();
this.ResumeLayout(false);
XmlDataDocument xmlDatadoc = new XmlDataDocument();
xmlDatadoc.DataSet.ReadXml("abcd.xml");
DataSet ds = new DataSet("abc");
ds = xmlDatadoc.DataSet;
dataGrid1.DataSource = ds.Tables[0];
上面的代码读取DataGrid中的XML文件并显示。有人可以告诉我如何编辑DataGrid,以便可以编辑XML中相应的值吗?
文件正在被修改,但不会反映更改。 – akanki
@akanki回答修改。 – swiftgp
ds.Tables [0] =(DataTable)(dataGrid1.DataSource)...这行正确地在你身边正常工作,因为dis显示错误无法分配ErrorProperty或索引器'System.Data.DataTableCollection.this [int]' - 它是只读的 – akanki