On Thursday 29 November 2001 01:29, Kevin Ruiz wrote:
> I've come across yet another problem.
[snip]
> $ci = "select contactid from users where username='$username' and
> password='$password'";
> $cir = mysql_query($ci)
> or die("Couldn't execute");
> $query = "select * from my_contacts where contactid='$cir'";
>
> $result = mysql_query($query) or
> die( mysql_error() );
>
> When I try to print $cir to see if anything's getting passed I keep getting
> something that reads "resource id #2".
[snip]
> Does anyone know what this means and how I can work around it.
$cir does NOT contain the actual results of your query. It is only a
*pointer* to your results. To get at the actual result(s):
print "<table>\n";
while ($line = mysql_fetch_array($cir)) {
print "\t<tr>\n";
while(list($col_name, $col_value) = each($line)) {
print "\t\t<td>$col_value</td>\n";
}
print "\t</tr>\n";
}
print "</table>\n";
Read the manual for full details.
hth
--
Jason Wong -> Gremlins Associates -> www.gremlins.com.hk
/*
Total strangers need love, too; and I'm stranger than most.
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