I wan to echo a random swf variable into html, but I can not seem to get the right result.
By checking with firebug I can see that it outputs my php code php echo '$randomImage'
What I have so far:
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML+RDFa 1.0//EN" "http://www.w3.org/MarkUp/DTD/xhtml-rdfa-1.dtd">
<html>
<head>
<style media="screen" type="text/css">
.content {
position: absolute;
height: 200px;
width: 400px;
margin: -100px 0 0 -200px;
top: 25%;
left: 50%;
}
</style>
</head>
<body>
<?php
$imagesDir = '/';
$images = glob($imagesDir . '*.{swf}', GLOB_BRACE);
$randomImage = $images[array_rand($images)];
?>
<div class="content">
<center><object width="675" height="517"><param value="<?php echo '$randomImage' ?>" name="movie"><embed width="475" height="317" type="application/x-shockwave-flash" src="<?php echo '$randomImage' ?>"></object></center><center></center>
</div>
</body>
</html>
Remove the quotes.
<?php echo $randomImage; ?>
PHP interprets that as a string rather than the variable $randomImage. You can use variables in quotes but they have to be double quotes like so.
echo "$randomImage";
There is no reason to do this however because you can just echo the variable. This is useful in situations when you want to format a string with multiple variables without concatenation operations.
echo "Hello, my name is $myName!, I am a $species.";
Which is more simple and readable than the alternative.
echo "Hello, my name is ", $myName ,"!, I am a ", $species ,".";
You either:
By placing your variable in single quotes you are telling PHP that your dollar sign is to be taken literally and not used to indicate that there is variable.
So use
<?php echo $randomImage ?>
Or
<?php echo "$randomImage" ?>
To echo properly you should remove the quotes '$randomImage' -> $randomImage or you can wrap them in "" (double quotes) "$randomImage"