A
Andy
Hi,
I want to create a div ajax popup:
var popup = document.createElement('div');
popup.id = 'my_id';
popup.innerHTML = popuptext;
popup.style.clip = 'auto';
popup.style.overflow = 'scroll';
popup.style.position = 'absolute';
popup.style.width = popupwidth;
popup.style.height = popupheight;
popup.style.top = ymouse + "px";
popup.style.left = xmouse + "px";
popup.onmouseout = 'alert(this)';
document.body.appendChild(popup);
I cannot get the alert set via setting popup.onmouseout to come up.
Is it possible to set onmouseout attribute of a div element created
this way?
Andy
I want to create a div ajax popup:
var popup = document.createElement('div');
popup.id = 'my_id';
popup.innerHTML = popuptext;
popup.style.clip = 'auto';
popup.style.overflow = 'scroll';
popup.style.position = 'absolute';
popup.style.width = popupwidth;
popup.style.height = popupheight;
popup.style.top = ymouse + "px";
popup.style.left = xmouse + "px";
popup.onmouseout = 'alert(this)';
document.body.appendChild(popup);
I cannot get the alert set via setting popup.onmouseout to come up.
Is it possible to set onmouseout attribute of a div element created
this way?
Andy