Need help in JavaScript , openWindow ();

Question

I am trying to run the following JavaScript program,

function myFunction() {
 var x;
 if (confirm("Press a button!") == true) {
 x = "You pressed Ok!";
 } else {
 x = "You pressed Cancel!";
 }
 document.getElementById("demo").innerHTML = x;
}

<p>Click the button to display a confirm box.</p>
<button onclick="myFunction()">Try it</button>
<p id="demo"></p>

But I really need is not just saying "You pressed Ok" instead of that it need to open a link. Help me please.

I hope

openWindow("http://fb.com")

this line will help, but don't know where to place it and make a working program.

• DCruz22: You should use window.open("http://www.google.com") and just place it where your x = "You pressed Ok!"; is.
• Heretic Monkey: A simple search of the internet for "how to open a window in javascript" would have found the answer to this.

Answer 1

if (confirm("Press a button!")) {
 window.open('www.google.com')
}

• John Weisz: FYI, if (confirm(...)) is identical to if (confirm(...) == true), because the condition statement inside an if is evaluated as a boolean expression, and checking whether any given boolean expression e for e == true is essentially the same as the expression itself.

Answer 2

You just need to change your function and place your openWindow() call inside it. Because I don't know what your openWindow() function does, I would use window.open():

function myFunction() {
 window.open("http://fb.com");
}

Answer 3

function myFunction() {
var x;
if (confirm("Press a button!") == true) {
location.href = "http://www.google.com";
} else {
x = "You pressed Cancel!";
}
document.getElementById("demo").innerHTML = x;
}

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<p>Click the button to display a confirm box.</p>
<button onclick="myFunction()">Try it</button>
<p id="demo"></p>