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I'm trying to create a variable background, where the image changes based on the time of day. This code I have USED to work, but I did something somewhere along the line and didn't notice that the functionality had broken. Can someone explain to me why this doesn't work?
<html>
<?php
function day()
{
if ( $hour >= 6 && $hour <= 18 )
{
return 1;
} else { return 0; }
}
?>
<style type="text/css">
body
{
background-image: url('<?php echo (day() ? 'images/day_sheep.jpg'
: 'images/night_sheep.jpg'); ?>');
background-position: 50% 50%;
background-repeat: no-repeat;
background-color: silver
}
a {text-decoration:none;}
a:link {color:#ff0000;}
a:visited {color:#0000FF;}
a:hover {text-decoration:underline;}
</style>
</html>
785
Inside your function day(), $hour is unset. It will be treated as 0 in a numerical context, and if you enable reporting of notices, you will see notices warning you of an unset variable. Did it used to be a global variable? Did you remove code that set its value or declared it as global?
Edit: Also, on a point of style, I feel it would look neater to have an external CSS file like this:
body {
background-position: 50% 50%;
background-repeat: no-repeat;
background-color: silver
}
body.day {
background-image: url('images/day_sheep.jpg');
}
body.night {
background-image: url('images/night_sheep.jpg');
}
and then you can get rid of the CSS section of your php script, but include the above CSS file, and you need only have the following:
<body class="<?php echo day() ? 'day' : 'night'; ?>">
You never declare what $hour is.
<html>
<?php
function day()
{
$hour = date('G');
if ( $hour >= 6 && $hour <= 18 )
{
return 1;
} else { return 0; }
}
?>
... snip ...
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