首次加载时如何从弹出窗口返回对父窗口的关注

Question

I am working on a website. Where while loading website first time opening index page with one pop window. First my problem is I want to minimize this pop window automatically while loading first time and focus return to website without any click in website.

Here is my code in index page I write this function:

myFunction();
 function myFunction() {
   //var myWindow = window.open("music.php", "", "width=320,height=60");

   var myWindow = window.open("music.php", "", "directories=0,titlebar=0,toolbar=0,location=0,status=0,menubar=0,scrollbars=no,resizable=no,left=-100000000, top=100000, width=320, height=1, display=none","visibility=hidden");
    //myWindow.moveTo(0, 0);
}

And pop window having page extension music.php which should be minimize automatically and focus return to index page without any click on website.

I have already try with set the focus of a popup window to website everytime , but it was also not working.

Answer 1

Well, The window object has an opener property as well. When we open the browser, that property is null, and when we open a popup , it points to the window the popup has been opened through.

I think something like this should suffice.

myFunction();
 function myFunction() {

   var myWindow = window.open("music.php", "", "directories=0,titlebar=0,toolbar=0,location=0,status=0,menubar=0,scrollbars=no,resizable=no,left=-100000000, top=100000, width=320, height=1, display=none","visibility=hidden");

   myWindow.opener.focus()   // can also do window.focus()

}

The execution of JS should not stop, but if it does, you have to add the focus code to your child page/popup page.(Depending on varying browsers)

like

popupHTMLJS.js

focusParent();
function focusParent(){
    parent = window.opener;
    parent.focus()
}