I realized that as soon as I hit Send - doh! :)
Jonathan Hilgeman wrote:
> Actually,
> He should be using three. While one company could use many services, you
> might also have other companies that offer the same service. So there should
> be 3 tables to allow a many-to-many relationship. See m
Actually,
He should be using three. While one company could use many services, you
might also have other companies that offer the same service. So there should
be 3 tables to allow a many-to-many relationship. See my e-mail response to
him.
- Jonathan
"Chris Hobbs" <[EMAIL PROTECTED]> wrote in m
I won't add to the code suggestions already given, but I will share one
thought about your tables - you absolutely _should_ be using two. Any
given company can have from one service to thousands (take GE,
everything from light bulbs to cruise missiles :). The only way to
capture that data effe
\n";
}
echo "".$ar['services']."";
}
}
mysql_free_result($res);
?>
Andrey Hristov
IcyGEN Corporation
http://www.icygen.com
99%
- Original Message -
From: "Andrius Jakutis" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Friday, August 24, 2001 5:04 AM
Subject: [PHP
";
print getServices($row[companyid]) . "";
print "";
}
} else {
print "No Companies";
}
Function getServices ($id) {
$sql = "SELECT services FROM services WHERE companyid = $id";
$result = mysql_query($sql);
$num = mysql_num
