Case, when "document.ready" obstruct - php

Question

I have php file, in this file I have this code:
<script language="JavaScript" type="text/javascript" src="jquery.js"></script>
<script language="JavaScript">
$(document).ready( function () {
var myvar = <?php echo json_encode($myvar); ?> ;
});
</script>
<script language="JavaScript" type="text/javascript" src="costum.js"> </script>
and in costum.js file I have code:
$(document).ready( function () {
alert(myvar );
});
this not working, error consol returns "myvar is undefined"
if in php file I write this (that is, without "document.ready")
<script language="JavaScript">
var myvar = <?php echo json_encode($myvar); ?> ;
</script>
in costum.js file, code alredy is working. Please tell why this happened?

Answer 1

try with
<script>
var myvar;
$(document).ready( function () {
myvar = <?php echo json_encode($myvar); ?> ;
});
</script>
your variable has to be declared as global (or in other words, in the outer scope) to be viewed from both document.ready functions.
As a side note language attribute is not necessary. Even type is not necessary (if you're using html5 doctype)

Answer 2

Your myvar is in the local scope of the ready-function. Move the var declaration outside to make it global and available to the other script.
However, as you just assign to a variable, you won't need to wait for DOMready anyway. Just use
<script type="text/javascript">
var myvar = <?php echo json_encode($myvar); ?>;
</script>
BTW, the language attribute is deprecated.

Answer 3

local variable inside the function is only visible in the function scope.
when you declare variable in the global scope, then it is the global variable.
You could expose it to global scope by:
$(document).ready( function () {
var myvar = <?php echo json_encode($myvar); ?>;
window['myvar'] = myvar;
});