Programming

I'm putting my club's membership database online.
I have a Member table in the database with fields such as FirstName, LastName etc. I have a web form with matching input variables Member_FirstName, Member_LastName that's going to be used for inserting a new member into the database, so I'll be populating an SQL statement such as

Code:

insert Member
(FirstName, LastName .... )
values
("$_POST['Member_FirstName']", "$_POST['Member_LastName']" ...)

My plan was to have a set of intermediate variables

Code:

$Member_FirstName = $_POST['Member_FirstName'];
$Member_LastName = $_POST['Member_LastName'];

so that I could have a pre built SQL statement that would simply substitute the intermediate variables.

OK - that's the background...now the question is, how do I generate the intermediate variables in a loop of column names.
So in pseudo code...

Code:

for each $col (select * from Member)
    $var = "Member_" . $col;                 ($var="MemberFirstName")
    ${$var} = $_POST[$var];                  Create the intermediate variable
                                                This is the problem line!

I've been trying to use 'eval()' but can't get the syntax correct, so any views on correct syntax or a better/alternative way to do this appreciated.

Jerry

eval is evil. You'd spend more code on trying to prevent security problems than anything else, and even then there'd probably be (more than) a few left.

Check the PHP Manual for Variable variables.

Brilliant - thanks pludi. I thought I'd tried this previously and couldn't get it working, so eval was a clutch at straws. The description and example on the link were invaluable.

Jerry