弹出对话框后,跳到另一个界面(转)
写个通用的工具类,加入如下方法,以后方便调用:
/// <summary>
/// 显示错误并跳转
/// </summary>
/// <param name="err">错误信息 </param>
/// <param name="url">url地址 </param>
public static void throwErrorGotoURL(string err, string url)
{
System.Text.StringBuilder builder = new System.Text.StringBuilder();
builder.Append(" <script language=\"javascript\">");
builder.Append("alert(\"" + err + "\");");
builder.Append("location='" + url + "';");
builder.Append(" </script>");
HttpContext.Current.Response.Write(builder.ToString());
HttpContext.Current.Response.End();
}
/// <summary>
/// 显示错误并跳转
/// </summary>
/// <param name="err">错误信息 </param>
/// <param name="url">url地址 </param>
public static void throwErrorGotoURL(string err, string url)
{
System.Text.StringBuilder builder = new System.Text.StringBuilder();
builder.Append(" <script language=\"javascript\">");
builder.Append("alert(\"" + err + "\");");
builder.Append("location='" + url + "';");
builder.Append(" </script>");
HttpContext.Current.Response.Write(builder.ToString());
HttpContext.Current.Response.End();
}