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ASP.net实现页面跳转的方法

程序员文章站 2024-03-02 15:51:16
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, eventa...

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;
            }
        }