页面无刷新的PostBack事件
TextBox的onchange事件会引起页面重新刷新，这使得输入很不方便，于是我想了以下的解决办法。
思路：1、调用简单的前台脚本打开一个新页， 2、把相关参数传递给该新页处理（该新页的前后台均可处理）， 3、处理后所得数据再用前台脚本返回opener， 4、数据处理完毕后被打开的页面自动关闭。代码例：要处理的页面send.aspx：
string windowAttribs = "width=10px," + "height=10px," + "left='+((screen.width +" + "10" + ") * 2)+'," + "top='+ (screen.height + " + "10" + ") * 2+'";
TextBox2.Attributes["onchange"] = "var aaa = this.value;window.open('open_XML.aspx?get_id=TextBox2&get_value='+aaa,'open_test','"+windowAttribs+"')";
TextBox3.Attributes["onchange"] = "var aaa = this.value;window.open('open_page.aspx?get_id=TextBox3&get_value='+aaa,'open_test','"+windowAttribs+"')";
打开的页面open_page.aspx：
private void Page_Load(object sender, System.EventArgs e)
{
object ls_o1 = Request.QueryString["get_id"].ToString();
object ls_o2 = Request.QueryString["get_value"].ToString();
if(ls_o1 != null && ls_o2 != null)
{
get_id.Value = (string)ls_o1;
get_value.Value = (string)ls_o2;
}
send_value_back();
}
private void send_value_back()
{
if(get_id.Value != "" && get_value.Value != null)
{
this.Response.Write("window.opener.Form1."+get_id.Value+".value='aaaaaaaaaaaa'"+"");
this.Response.Write("window.opener=null;window.close();");
}
}