1

我开始 .net 并开始我将动态创建多个文本框。

我写了这个：

<%@ Page Language="C#" AutoEventWireup="true" CodeBehind="Default.aspx.cs" Inherits="WebApplication1._Default" %>

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">

<html xmlns="http://www.w3.org/1999/xhtml" >
<head runat="server">
    <title>Page sans titre</title>
</head>
<body>
    <form id="form1" runat="server">
    <div>
        <asp:Button ID="Button1" runat="server" Text="Button" onclick="OnclickButton" />
    </div>
    <asp:TextBox ID="TextBox1" runat="server"></asp:TextBox>
    <br />
    </form>
</body>
</html>

和这个

namespace WebApplication1
{
    public partial class _Default : System.Web.UI.Page
    {    
        protected void Page_Load(object sender, EventArgs e)
        {

        }
        protected void OnclickButton(object sender, EventArgs ea)
        {
            Random random = new Random();
            int randomNumber = random.Next(0, 100);

            Button btnSomeButton = sender as Button;
            btnSomeButton.Text = "I was clicked!" + randomNumber;

            TextBox txt = new TextBox();
            txt.ID = "txt_" + randomNumber;
            form1.Controls.Add(txt);
        }    
    }
}

我不明白为什么当我在 Button1 上单击 2 次时，只出现 1 个文本框。

为什么会有这种行为？做我想做的事的好方法是什么？先感谢您

4

2 回答 2

4

Everytime you click the button it does a post back. Read this on postbacks

The system can only handle one postback at any one time. it needs to wait for the the postback to return before it can process another action.

Also read about the Page lifecycle enter image description here

于 2013-05-21T15:06:19.167 回答
2

Web 是少状态的。换句话说，服务器不知道动态添加了多少控件，除非您存储信息以在回发时持久保存。

因此，您希望将数据保存到ViewState（或 SessionState）中，并在回发时使用相同的 id重新加载它们。

<asp:Button ID="Button1" runat="server" Text="Button" 
  OnClick="Button1_Click"/>
<asp:TextBox ID="TextBox1" runat="server"></asp:TextBox>

private List<int> ControlIds
{
    get { return (List<int>) ViewState["ControlIds"] ?? new List<int>(); }
    set { ViewState["ControlIds"] = value; }
}

protected void Page_Load(object sender, EventArgs e)
{
    if (IsPostBack)
    {
        // Reload the controls on post back
        var ids = ControlIds;
        foreach (var id in ids)
        {
            var txt = new TextBox {ID = "txt_" + id};
            form1.Controls.Add(txt);
        }
    }
}

protected void Button1_Click(object sender, EventArgs e)
{
    var random = new Random();
    int randomNumber = random.Next(0, 100);

    var ids = ControlIds;
    if (ids.Contains(randomNumber))
    {
        // Randam number already exists.
        // We cannot add controls with same ID.
        return;
    }

    ids.Add(randomNumber);
    ControlIds = ids;

    var btnSomeButton = sender as Button;
    btnSomeButton.Text = "I was clicked!" + randomNumber;

    var txt = new TextBox
        {
            ID = "txt_" + randomNumber,
            Text = randomNumber.ToString() // Just for testing
        };
    form1.Controls.Add(txt);
}
于 2013-05-21T15:24:47.447 回答