点击在asp.net按钮 [英] click on button in asp.net
本文介绍了点击在asp.net按钮的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我动态创建一个按钮,并将其放置在一个占位符,如下
I am creating a button dynamically and place it in a placeholder as below
<asp:Button ID="generateTableSchema" runat="server" Text="Generate Table" OnClick="generate_Click" />
protected void generate_Click(object sender, EventArgs e)
{
Button button = new Button();
button.Text = "Generate Table";
button.ID = "generateTable";
button.OnClick = hello();
PlaceHolder1.Controls.Add(button);
}
但onclick事件不点火。
but onclick event is not firing.
这是我收到的错误
System.Web.UI.WebControls.Button.OnClick(System.EventArgs)' is inaccessible due to its protection level
您好是如下...
public void hello()
{
Label1.Text = "heellllllllllo";
}
什么是错在这里????
What's wrong here????
protected void Page_Load(object sender, EventArgs e)
{
if (!Page.IsPostBack)
{
}
else
{
button.Click += ButtonClick;
}
}
@Darenü意味着像这样...
@Daren u mean like this...
推荐答案
由于您要添加的按钮编程的盟友必须添加事件处理程序。
Because you are adding the button programmatic ally you have to add the event handler.
所以这会工作。
修改
包裹内按钮的Page_Load
EDIT Wrapped the button INSIDE Page_Load
protected void Page_Load(object sender, EventArgs e)
{
Button button = new Button();
button.Text = "Generate Table";
button.ID = "generateTable";
button.Click += hello; /// THIS is the handler
PlaceHolder1.Controls.Add(button);
}
ButtonClick将是你的方法的名称。
ButtonClick would be the name of your method.
protected void hello(Object sender, EventArgs e)
{
// ...
}
另外,你在运行时产生这个需要makesure这个被调用的回传也。
Also, as you're generating this at runtime you need to makesure this gets called on postbacks too.
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