PHP:
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title></title>
</head>
<body>
<?php
function Connect () {
mysql_connect ("localhost","root","") or die (mysql_error() );
echo "connected to MySQL";
function Select_db ($database) {
mysql_select_db("$database") or die (mysql_error () );
echo "connected to database";
}
function Create_table () {
mysql_query("CREATE TABLE Tabledatacab
(
id INT NOT NULL AUTO_INCREMENT,
PRIMARY KEY(id),
website name VARCHAR(30),
website url VARCHAR(60),
referral commission VARCHAR (30),
referral levels INT,
payout minimum FLOAT,
payout methods VARCHAR (30),
website info VARCHAR (45)
)") or die(mysql_error());
echo "Table Created with succes";
$database = 'tabledata';
Connect ();
Select_db ($database);
if ( $_POST["create"] == "YES" ) {
Create_table ();
}
else {
echo "no table created!";
}
echo "Job done without problems.";
?>
<form method="post" action="<?php echo $PHP_SELF;?>">
create table? : <input type="text" size="12" maxlength="4" name="create"> <br >
<input type="submit" value="submit" name="submit"><br />
</form>
</body>
</html>
PHP:
Parse error: syntax error, unexpected $end in D:\development\xamp 1.6\xampp\htdocs\MP.php on line 59ik heb geen idee what er fout is.. heb al vanalles geprobeerd!
