页面跳转 ASP.net实现页面跳转的方法
人气:0主要是使用response的属性,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:
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;
}
}
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;
}
}
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