Question

My sql script is not returning a value ‘sid’ when I use it with php, but when I run it against the actual database it returns the value. I cannot figure out where my error is. My tables are as follows:

Song
aid (1)
sid (999)
title (Moonlight Sonata)
surl

Album
aid (1)
artist (Beethoven)
genre (Classic)
album

Here is my code:

$sql="SELECT S.title, S.surl, S.sid, A.aid, A.album, A.artist FROM challenge C, 
    song S, album A WHERE C.fid = '$uid' AND uid = '$user' AND date = '$date' 
    AND C.sid = S.sid AND S.aid = A.aid";

$result=mysql_query($sql);

if (mysql_num_rows($result) != 0){
    while($row=mysql_fetch_array($result)){
    $title = $row[title];
    $album = $row[album];
    $surl = $row[surl];
    $artist = $row[artist];
    $aid = $row[aid];
    $sid = $row[sid];

echo "Album: $album Artist: $artist Title: $title Song: $surl SongID: $sid";
    }
}

Everything is printing except the value ‘sid’ – it should be printing 999. Any help would be greatly appreciated!!

Answer 1

$row[title] is not the syntax you’re looking for. If you’d enable error reporting you’d see warnings about undefined constants. Just writing title makes PHP look for a constant called “title”, which does not exist. Only then does it fall back onto strings and tries if you maybe meant the string 'title'. SID happens to be a predefined constant, so that’s the only time it’s using the value of the constant, and there’s no index in the array with the value of that constant.

Quote your strings! Write $row['title'] and $row['sid'].

Please read “Why is $foo[bar] wrong?”.
Furthermore read the article about SQL injection and escape your input!