>
Sent: Wednesday, September 19, 2001 4:34 PM
Subject: RE: [PHP-DB] Storing code in a mysql database
> I've now progressed to the point where I'm not getting eval errors. The
next
> problem is that is doesn't substitute the variable names: so the code
>
> $company_cod
I've now progressed to the point where I'm not getting eval errors. The next
problem is that is doesn't substitute the variable names: so the code
$company_code=1;
eval($sql='select * from vehicle where company=$company_code';");
ends up $sql equal to 'select * from vehicle where company=$comp
