c# - 单击按钮时如何创建弹出窗口

Question

这是我为按钮准备的代码。我希望配置按钮在单击时显示一个弹出窗口。我查阅并尝试了很多不同的代码，但没有一个有效。我觉得我缺少一个关键组件，但我不确定那是什么。感谢帮助!

tr = new TableRow();
// Create the cell
tc = new TableCell();
tc.Width = Unit.Point(300);
tc.BorderWidth = 0;
tc.BorderStyle = BorderStyle.None;

Button ConfigButton = new Button();
ConfigButton.Text = "Configuration";
ConfigButton.Visible = true;

tc.Controls.Add(ConfigButton);
tr.Cells.Add(tc);
tbl.Controls.Add(tr);

Answer 1

Using JavaScript Along with ASP.NET您将执行以下操作:

// when you create the button, you can add attributes
Button ConfigButton = new Button();

// this example will display alert dialog box
ConfigButton.Attributes.Add("onclick", "javascript:alert('ALERT ALERT!!!')");

// to get you popup windows
// you would use window.open with the BLANK target

ConfigButton.Text = "Configuration";
ConfigButton.Visible = true;