problem with arrays
Posted: Tue Sep 06, 2005 1:53 am
value of
$ans = 1;
but when i printed my query it gave me result some thing like this
Result
SELECT GameAnswer FROM `games` WHERE GameId = '7'
SELECT GameAnswer FROM `games` WHERE GameId = '6'
SELECT GameAnswer FROM `games` WHERE GameId = '3'
SELECT GameAnswer FROM `games` WHERE GameId = '2'
Thanx.
Mannan.
$ans = 1;
Code: Select all
$b[$ans] = $GameId;
//what i think is it shoud replace $b[1] 's old value with value of $GameId.
for($i=0; $i<4; $i++)
{
$SkyRes1 = mysql_query("SELECT GameAnswer FROM `games` WHERE GameId = '$b[$i]'") or die(mysql_error());
$Num1 = mysql_num_rows($SkyRes1);
if($Num1 > 0)
{
if($SkyRow1 = mysql_fetch_object($SkyRes1))
{
$GameAnswer1 = $SkyRow1 -> GameAnswer;
}
}
}Result
SELECT GameAnswer FROM `games` WHERE GameId = '7'
SELECT GameAnswer FROM `games` WHERE GameId = '6'
SELECT GameAnswer FROM `games` WHERE GameId = '3'
SELECT GameAnswer FROM `games` WHERE GameId = '2'
Thanx.
Mannan.