Håller nu på att bygga ett "bättre" adminsystem än det jag gjort tidigare,
och kör nu med en array.
Functionen ser ut såhär (inc_functions.php):
function info () {
$fraga = mysql_query("SELECT * FROM episodes") or exit(mysql_error()); }
while ($r = mysql_fetch_array($fraga)) {
$fraga[] = array( "title" => $r[title], "season" => $r[s], "episode" => $r[e] );
}
Och jag hämtar ut infon via (episodes.php):
<?PHP
foreach (info as $r) { ?>
<?php echo $r['title']; ?>
<?php echo $r['season']; ?>
<?php echo $r['episode']; ?>
<?php } ?>
Men jag får en bunt med felmeddelanden ..
Undefined variable: fraga in C:\EasyPHP\www\simpsonslife\include\inc_functions.php on line 6
mysql_fetch_array(): supplied argument is not a valid MySQL result resource in C:\EasyPHP\www\simpsonslife\include\inc_functions.php on line 6
Use of undefined constant info - assumed 'info' in C:\EasyPHP\www\simpsonslife\episodes.php on line 4
Invalid argument supplied for foreach() in C:\EasyPHP\www\simpsonslife\episodes.php on line 4