不要使用后退按钮弹出弹出窗口

Question

I used pop up in my site.The problem is, when i open a pop up and go to the other page and then go back to previous page, a pop up will refresh and open again!! without any pressed! How can i fix it? I want that this pop up wont come up again by coming back to that page.

here is my Pop up code:

.......
  <script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jqueryui/1.5.3/jquery-ui.min.js" ></script>  
<script type='text/javascript' src='jquery-1.2.3.min.js'></script>
<script type='text/javascript' src='menu.js'></script>

  <script type="text/javascript">
      function pop_up(url) {
          newwindow = window.open(url, 'name', 'height=517,width=885,scrollbars=yes,toolbar=no,menubar=no,resizable=yes,location=no,directories=no,status=no,titlebar=no,left=400,top=120');
          if (window.focus) { newwindow.focus() }
          return false;
      }
    </script>


<link href="style.css" rel="stylesheet" type="text/css" />

    <style type="text/css">
        .style2
        {
.......

and behind code:

Page.ClientScript.RegisterStartupScript(GetType(), "popup", "pop_up('" + "PopUpEmailing.aspx" + "');", true);

Answer 1

One solution would be to use the window.name property. I know this solution can lead to conflicts if another library also tries to use this. But I prefer this more then using a cookie. (You could also think over using local storage)

function displayPopUp() {
    if( window.name != "popup-displayed" ) {
        window.name = "popup-displayed";
        ... popup code ...
    }
}

Answer 2

Instead of using Page.ClientScript.RegisterStartupScript try the following.

<asp:Button runat="server" ID="btnTest" Text="My Button" OnClientClick="pop_up('" + "PopUpEmailing.aspx" + "')"><asp:Button>

The above will create a button and when the button will be click a client side event will be called and it will run your pop_up javascript code.