what's wrong with this code

gkt2310

hello, any help would be much appreciated! I am trying to display some details from a database table, but this code is not working.

Error is:
Parse error: syntax error, unexpected $end in D:\devserver\admin\clubs\outputfunctions.php on line 91

This is the part of the page that should display the details from the database table:

$sql = "SELECT club_id FROM clubs " .
       "ORDER BY club_name DESC";

$result = mysql_query($sql, $conn);

if (mysql_num_rows($result) == 0) {
  echo "  <br>\n";
  echo "  There are currently no clubs to view.\n";
} else {
  while ($row = mysql_fetch_array($result)) {
    outputClubs($row['club_id']);
  }
}

And it calls the outputClubs function which is here:

function outputClubs($club) {
  global $conn;

  if ($club) {
    $sql = "SELECT *" .
           "FROM clubs" .
           "WHERE club_id = " . $club;
    $result = mysql_query($sql,$conn);

if ($row = mysql_fetch_array($result)) {
  echo "<h3>" . htmlspecialchars($row['club_name']) . "</h3>";
  echo nl2br(htmlspecialchars($row['club_desc']));
}
}

Any suggestions as to where I am going wrong?

Thanks - Gareth

leatherback

Please post line 85 to 95 in outputfunctions.php, so we can see what is happening at the place of the error.

gkt2310

hiya,

the error is referring simply to the end of the page. I know nothing is wrong (technically) down that far, because I have used that part of the script on other pages and it works. It must be something in what I changed (the bit I posted). Maybe I haven't closed something off properly there?

leatherback

well, it encountered a var ( unexpected $end ).. So it must have some code.

Unless you copied it, and is just end.

THen yes. Check the brackets.

gkt2310

cheers! added an extra curly bracket, now it displays the includes that surround it. still getting an error though:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in D:\devserver\admin\clubs\outputfunctions.php on line 36

leatherback

It means that the query doesn't give you the results you'd expected. See my sig.

gkt2310

i don't understand what you mean

leatherback

Use: or die(mysql_error()) after each query and safe a headache

Probably $club is not defined in

$sql = "SELECT *" .
       "FROM clubs" .
       "WHERE club_id = " . $club;