把弹出窗口写成通用方法

1.添加System.Web.Extensions引用

2.

 　　　　/// <summary>
/// 弹出信息
/// </summary>
/// <param name="page"></param>
/// <param name="msg"></param>
        public void showMessage(Control page,string msg)
        {
string str = "alert('"+msg+"');";
            ScriptManager.RegisterStartupScript(page, Type.GetType("System.String"), str, str, true);
        }