Variable

Senior WebHelper Joined: 21 Apr 2003 Posts: 153

How would I add the number contained in the variable $i to the begining of $wl. Ex: if $i = 1 I want the variable $1wl = 10, if $i = 2 I want the variable $2wl = 10.

Team Member Joined: 06 Jan 2002 Posts: 2564

Last time I tried (a few days ago!), I could use the $$ method in that way. So you'll have to settle for:

$wl[$i] (which will give $wl[1] or $wl[2])

Posted: Sun May 18, 2003 8:39 am (20 years, 11 months ago)

if I've understood what your trying to do this might work:

${$i."wl"} = 10

Senior WebHelper Joined: 21 Apr 2003 Posts: 153

I cant get it to work right, I was trying to make this code shorter, the WHERE needs to go up to 38 so doing it the way I have it means a lot of code. Can someone find a way to make it so I only need 1 and it will loop it or somthing?

Code:

$result = mysql_query( "SELECT * FROM Game_Standings WHERE id = 1)
or die("SELECT Error: ".mysql_error());
$num=mysql_numrows($result);
$i=1;
while ($i < $num) {
$wl1 = mysql_result($result,$i,"wl");
$sc1 = mysql_result($result,$i,"sc");
++$i;
}
$result = mysql_query( "SELECT * FROM Game_Standings WHERE id = 2)
or die("SELECT Error: ".mysql_error());
$num=mysql_numrows($result);
$i=1;
while ($i < $num) {
$wl2 = mysql_result($result,$i,"wl");
$sc2 = mysql_result($result,$i,"sc");
++$i;
}
$result = mysql_query( "SELECT * FROM Game_Standings WHERE id = 3)
or die("SELECT Error: ".mysql_error());
$num=mysql_numrows($result);
$i=1;
while ($i < $num) {
$wl3 = mysql_result($result,$i,"wl");
$sc3 = mysql_result($result,$i,"sc");
++$i;
}
$result = mysql_query( "SELECT * FROM Game_Standings WHERE id = 4)
or die("SELECT Error: ".mysql_error());
$num=mysql_numrows($result);
$i=1;
while ($i < $num) {
$wl4 = mysql_result($result,$i,"wl");
$sc4 = mysql_result($result,$i,"sc");
++$i;
}