De PHPhulp kroeg
Variable oproepen voor in de head
Voorbeeld
index.php
Code (php)
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<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=windows-1252">
<title>Nieuwe pagina 1</title>
</head>
<body>
<?php
include ('include.php');
?>
</body>
</html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=windows-1252">
<title>Nieuwe pagina 1</title>
</head>
<body>
<?php
include ('include.php');
?>
</body>
</html>
En include.php
Code (php)
En dan moeten de java-scripts bovenin komen, resultaat:
Code (php)
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<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=windows-1252">
<title>Nieuwe pagina 1</title>
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.4.2/jquery.min.js"></script>
<script type="text/javascript" src="script.js"></script>
</head>
<body>
some stuffsome stuff
</body>
</html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=windows-1252">
<title>Nieuwe pagina 1</title>
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.4.2/jquery.min.js"></script>
<script type="text/javascript" src="script.js"></script>
</head>
<body>
some stuffsome stuff
</body>
</html>
Heb hier het e.e.a. gevonden, maar lukt niet helemaal.
Groet,
Dennis
Gewijzigd op 12/07/2011 23:56:35 door Dennis van Dijk
Code (php)
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<?php
$a = 'some stuff';
$b = '<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.4.2/jquery.min.js"></script>';
$c = '<script type="text/javascript" src="script.js"></script>';
$d = 'some stuff';
?>
$a = 'some stuff';
$b = '<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.4.2/jquery.min.js"></script>';
$c = '<script type="text/javascript" src="script.js"></script>';
$d = 'some stuff';
?>