Javascript Popup in an If Statement

jerseyralph

I am modifing a php quiz script so that if the quiz taker achieves a score of 70 that a new window pops up using javascript.

Here is the begining of the section I what to modify:

# display the final score ?
if ($score >= 70)
{
OpenTable();
echo "<font class=pn-title>"._QUIZZYOURSCORE." : $score</font><br>";
echo "<script language=\"JavaScript\">\n";

The javascript looks like this:

What is the easiest way to do this?

Thanks

Ralph

andrewtayloruk

I'm not exactly sure what you want to do, but I think you're after something like this -

$body = '';

if($score > 70) {

// do something in php

$body = 'onLoad="doJavascriptFunction()"';
}

?>
<html>
<head>
</head>
<body <? echo $body; ?>>

</body>

This way body is normally blank, however, if the score is above 70 an onLoad flag is added to the body tag forcing the JS to pop a window up

jerseyralph

Hmmm....

That may be it. But where does the javascript go? Is it declared somewhere else?

I am just too new at this, but it is fun trying to figure it out!!

coditoergosum

There's a few different ways to do what you're wanting to do. If it were me, I'd define the javascript in a function somewhere in the head of the HTML, then in your if clause just call the javascript function.

Something like this:

<html>
<head> 
<script type="text/javascript">
function doit() {
// javascript code goes here
}
</script>
</head>
// whatever other stuff here
<?php
if ($score >= 70) 
{ 
OpenTable(); 
echo "<font class=pn-title>"._QUIZZYOURSCORE." : $score</font><br>"; 
echo "<script language=\"JavaScript\">\ndoit();\n</script>";
} 
?>

jerseyralph

Thanks. I will play around with that one.