I am trying to open jquery popup box in asp.net
i have four pages in array
------------------------------------
i.e.
var i=0;
var forms;
this.aForms = new Array("Page1.asp",
"Page2.asp",
"Page3.asp",
"Page4.asp");
and open dialog code
----------------------
var url = form ;
$.showModalDialog({
url: url,
dialogArguments: window,
height: 470,
width: 400,
scrollable: false,
onClose: function () {
var returnedValue = this.returnValue;}
});
while ((Btnstat =="next"))
{
objWizard.showDialog();
form = this.aForms[i];
i++
}
and i next button on popup window and i am setting value of Btnstat variable= next and at the end finish
when i am running the code the dailog is not opening until loop fineshes its process and at the end only one page get display
so how can i display my first page in popup and click of next button another page should display in popup
Plz help me....