c# - DataGridView 反复重新创建列

Question

我在为 DataGridView 控件分配 DataSource 属性时遇到问题。我的 DataSource 是 DataTable 的 DefaultView 并且，正如预期的那样，在 DataGridView 中自动创建列以匹配当我分配它时 DataTable 中的那些。

接下来发生的是列似乎被 DataGridView 自动删除并重新创建了 2 次。为什么会这样？

在表单的构造函数中:

//A DataTable is created with 5 columns
//The DataTable is populated with some rows.

myDgv.AutoGenerateColumns = true;
myDgv.DataSource = myDataTable.DefaultView;
// myDgv.ColumnAdded event is fired 5 times.
// WHY: myDgv.ColumnRemoved event is fired 5 times.
// WHY: myDgv.ColumnAdded event is fired 5 times.
// WHY: myDgv.ColumnRemoved event is fired 5 times.
// WHY: myDgv.ColumnAdded event is fired 5 times.

---

编辑:添加了一个(希望如此)自包含示例。如果我在事件处理程序中设置断点，我会点击“添加”一个 6 次，然后点击“删除”一个 4 次。 DataTable 包含 2 列，我从不要求在我的代码中删除任何列。

using System;
using System.Collections.Generic;
using System.ComponentModel;
using System.Data;
using System.Drawing;
using System.Linq;
using System.Text;
using System.Windows.Forms;

namespace asdasdgf
{
    public class Form1 : Form
    {
        public Form1()
        {
            InitializeComponent();

            DataTable dt = new DataTable();
            dt.Columns.Add("Col1", typeof(int));
            dt.Columns.Add("Col2", typeof(string));

            foreach (int i in Enumerable.Range(0, 10))
            {
                var row = dt.NewRow();
                row["Col1"] = i;
                row["Col2"] = "stackoverflow";
                dt.Rows.Add(row);
            }

            dataGridView1.ColumnAdded += new DataGridViewColumnEventHandler(dataGridView1_ColumnAdded);
            dataGridView1.ColumnRemoved += new DataGridViewColumnEventHandler(dataGridView1_ColumnRemoved);

            dataGridView1.DataSource = dt.DefaultView;
        }

        void dataGridView1_ColumnRemoved(object sender, DataGridViewColumnEventArgs e)
        {
            // Break here
        }

        void dataGridView1_ColumnAdded(object sender, DataGridViewColumnEventArgs e)
        {
            // Break here
        }

        // Form1.Designer.cs contents:
        #region Windows Form Designer generated code
        private System.ComponentModel.IContainer components = null;
        private System.Windows.Forms.DataGridView dataGridView1;

        protected override void Dispose(bool disposing)
        {
            if (disposing && (components != null))
            {
                components.Dispose();
            }
            base.Dispose(disposing);
        }



        /// <summary>
        /// Required method for Designer support - do not modify
        /// the contents of this method with the code editor.
        /// </summary>
        private void InitializeComponent()
        {
            this.dataGridView1 = new System.Windows.Forms.DataGridView();
            ((System.ComponentModel.ISupportInitialize)(this.dataGridView1)).BeginInit();
            this.SuspendLayout();
            // 
            // dataGridView1
            // 
            this.dataGridView1.ColumnHeadersHeightSizeMode = System.Windows.Forms.DataGridViewColumnHeadersHeightSizeMode.AutoSize;
            this.dataGridView1.Location = new System.Drawing.Point(12, 41);
            this.dataGridView1.Name = "dataGridView1";
            this.dataGridView1.Size = new System.Drawing.Size(240, 150);
            this.dataGridView1.TabIndex = 0;
            // 
            // Form1
            // 
            this.AutoScaleDimensions = new System.Drawing.SizeF(6F, 13F);
            this.AutoScaleMode = System.Windows.Forms.AutoScaleMode.Font;
            this.ClientSize = new System.Drawing.Size(284, 264);
            this.Controls.Add(this.dataGridView1);
            this.Name = "Form1";
            this.Text = "Form1";
            ((System.ComponentModel.ISupportInitialize)(this.dataGridView1)).EndInit();
            this.ResumeLayout(false);

        }

        #endregion
    }
}

Answer 1

我有一个应用程序使用 DataGridView 来显示 DataTable 的内容，但我是这样链接它们的:

dataGridView.DataSource = dataTable;

你能试试这个吗？

编辑:

我在自动生成的代码中也有以下内容:

this.dataGridView.EditMode = System.Windows.Forms.DataGridViewEditMode.EditProgrammatically;

但不确定为什么这会产生影响。