how to open a web page in form of a popup web page
Hello, I have two web pages, let say.
Page A : With a Button on it.
PAge B: with some info.
Now I need to: When I click on the Button of page A.
Page B must open in a pop up window.
I can do this but problem is this when I clicked on the button for first time it does not show me any thing and when I clicked on the button for second time then It shows me Page B .
My code is as follow:
1 protected void Page_Load(object sender, EventArgs e) 2 { 3 if (!this.Page.IsPostBack) 4 { 5 6 } 7 else 8 { 9 //this.btn_ManageEvents_Click(); 10 } 11 } 12 13 14 protected void btn_ManageEvents_Click(object sender, EventArgs e) 15 { 16 btn_ManageEvents.Attributes.Add("OnClick", "PopWin();return false;"); 17 }
<asp:Button ID="btn_ManageEvents" runat="server" Text="Manage Events" Width="100%" OnClick="btn_ManageEvents_Click" /><br />
<script>
function PopWin()
{
window.open("Event_Management.aspx","ScheduleJobs","toolbar=no, directories=no, location=no, status=yes, menubar=no, resizable=no, scrollbars=no, width=850, height=340", true);
}
</script>
How to fix this issue? Please help me to sort out this issue.
Thanks in advance,
junior
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issuemaster
3/16/2007 1:44:57 PM
asp.net.web-forms
93655 articles. 
6 followers.
2 Replies
2449 Views
Remove the below line of code from your button click event and add it on your page_load event
btn_ManageEvents.Attributes.Add("OnClick", "PopWin();return false;");
HC
----
Mark this post as answered if it helped you
Haissam Abdul Malak
MCAD.NET
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Haissam
3/16/2007 3:41:40 PM
Move your code around so that you're adding the attribute to the button in page_load. You shouldn't need any server postbacks or methods to run then. The javascript should open the pop up window.
protected void Page_Load(object sender, EventArgs e)
2 {
3 if (!this.Page.IsPostBack)
4 {
5
6 }
7 else
8 {
9 //this.btn_ManageEvents_Click();
10 }
11
12 btn_ManageEvents.Attributes.Add("OnClick", "PopWin();return false;");
13 }
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richiej
3/16/2007 3:45:17 PM
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