If i try the below in my query i get no results.
Ideally i would like to do this (not valid sql)
? $result = mysql_query(SELECT * FROM courses,$db);
printf(select name=\coursecode\\n);
while ($myrow = mysql_fetch_array($result)) {
$l=$myrow['coursecode'];
?option
$1 is a bad variable. You can't have variables that start with a number.
Try $a or $foo or $bar.
On Thu, 12 Dec 2002, mike karthauser wrote:
If i try the below in my query i get no results.
Ideally i would like to do this (not valid sql)
? $result = mysql_query(SELECT * FROM courses,$db);
I am using
? $result = mysql_query(SELECT * FROM courses,$db);
printf(select name=\coursecode\\n);
while ($myrow = mysql_fetch_array($result)) {
$l=$myrow['coursecode'];
?option value=?=$l;??=($_POST['coursecode']==$l)?'
selected=selected':;??=$myrow['title'];?/option?
}
printf(/select\n);
?
to