ASP.net实现页面跳转的方法

  主要是使用response的属性,代码如下:

  

复制代码 代码如下:

  protected void LinkButton1_Click(object sender, EventArgs e)

  {

  string url = "InfoShow.aspx";

  Response.Redirect(url);

  }

  protected void LinkButton1_Click(object sender, EventArgs e)

  {

  string url = "InfoShow.aspx";

  Response.Redirect(url);

  }//当然我们可以在页面跳转的时候进行参数传递,代码如下:

  protected void Menu1_MenuItemClick(object sender, MenuEventArgs e)

  {

  string url, s;

  s = e.Item.Value.ToString();

  url = "InfoRelease.aspx?UserName=" + s.Trim();

  Response.Redirect(url);

  }

  protected void Menu1_MenuItemClick(object sender, MenuEventArgs e)

  {

  string url, s;

  s = e.Item.Value.ToString();

  url = "InfoRelease.aspx?UserName=" + s.Trim();

  Response.Redirect(url);

  }

  上面是我的一个项目中的代码所以有item,大家可以根据自己的情况进行设定。

  那么既然传递了参数,在跳转的页面怎样获取呢???

  代码如下:

  

复制代码 代码如下:

  protected void Page_Load(object sender, EventArgs e)

  {

  if (Request["UserName"] != null)

  {

  string s = Request["UserName"].ToString();

  //           form1.Visible = true;

  txtInfoclass.Text = s;

  }

  }

  protected void Page_Load(object sender, EventArgs e)

  {

  if (Request["UserName"] != null)

  {

  string s = Request["UserName"].ToString();

  //           form1.Visible = true;

  txtInfoclass.Text = s;

  }

  }

  

复制代码 代码如下:

  protected void Page_Load(object sender, EventArgs e)

  {

  txtAlert.Text = "";

  if (Request["UserName"] != null)

  {

  string s = Request["UserName"].ToString();

  //           form1.Visible = true;

  txtInfoclass.Text = s;

  }

  }

  protected void Page_Load(object sender, EventArgs e)

  {

  txtAlert.Text = "";

  if (Request["UserName"] != null)

  {

  string s = Request["UserName"].ToString();

  //           form1.Visible = true;

  txtInfoclass.Text = s;

  }

  }