$1 is a bad variable. You can't have variables that start with a number.
Try $a or $foo or $bar.
On Thu, 12 Dec 2002, mike karthauser wrote:
> If i try the below in my query i get no results.
>
> Ideally i would like to do this (not valid sql)
>
> printf("\n");
> while ($myrow = mysql_fetch_ar
If i try the below in my query i get no results.
Ideally i would like to do this (not valid sql)
\n");
while ($myrow = mysql_fetch_array($result)) {
$l=$myrow['coursecode'];
?>>\n");
?>
on 11/12/02 5:26 pm, Peter Beckman at [EMAIL PROTECTED] wrote:
> LEFT(str,len)
> Returns the leftmost len c
LEFT(str,len)
Returns the leftmost len characters from the string str:
mysql> SELECT LEFT('foobarbar', 5);
-> 'fooba'
This function is multi-byte safe.
select left(coursecode,40) from courses
Peter
On Wed, 11 Dec 2002, mike karthauser wrote:
> I am using
>
