i am trying to write a php program n ends up giving me parse error on line 21

shygvy · 05-07-2012, 09:26 AM

Parse error: parse error in C:\xampp\htdocs\corp\shared\function.php on line 21
function get_pages_for_subject($subject_id){
$query ="SELECT*
FROM pages WHERE subject_id = {$subject_id};
ORDER by position ASC";
$page_set = mysql_query($query);
confirm_query($page_set);
return $pages_set;
}
ma line 21 is function get_pages_for_subject($subject_id){

klurik · 05-07-2012, 10:24 AM

Try remove semicolon before ORDER?

shygvy · 05-08-2012, 06:48 AM

it worked

klurik · 05-08-2012, 03:35 PM

Thats great

shygvy · 05-09-2012, 06:49 AM

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in C:\content.php on line 40

while($page = mysql_fetch_array($page_set)){
echo "<li";
if ($page ["id"]==$sel_page){
echo "class=\"selected\"";
}
echo"><a href=\"content.php? page=" . urlencode($page["id"]).
"\">{$pages["name"]}</a></li>";
}

this is my line 40

eantoranz · 05-09-2012, 08:32 AM

How are you setting $page_set? Cause you are returning it so you must be saving it somewhere, right?

shygvy · 05-09-2012, 10:11 AM

eantoranz

you are heaven sent...thanks a million.