maybe somebody else will be able to come up with a
much more graceful solution to this, but i think you
need to forget unset()-- all it does is destroy a var
within a script. your db data doesn't become a var
until you fetch it, but unset() isn't a mysql function
and doesn't interact with the db at all. your example
code is a little hard to read, but is there a reason
why you can't just...
$query = "UPDATE tablename set password='$n_password'
WHERE id = '$id'";
--- [EMAIL PROTECTED] wrote:
> Good day everyone,
>
> I'm writing because the username/password database
> table we had written for
> us does not seem to be accepting changes to a
> password. When I make a
> change, not only does the new password not take
> effect but it disables the
> entire account making it useless)
>
> Below is the code from the change.php3 file. I
> think the problem is
> possibly being caused with the unset() function but
> I'm not sure....
>
> Any ideas?
>
> Thanks again for your help. Shawna
>
>
> <? php3
>
> if($n_password1 != $n_password2){ print "passwords
> do not match. try again."; exit;
> };$password1=$n_password1;$username =
> addslashes($n_username);$name_prefix =
> addslashes($name_prefix);$name_first =
> addslashes($name_first);$name_middle =
> addslashes($name_middle);$name_last =
> addslashes($name_last);$name_suffix =
> addslashes($name_suffix);$email =
> addslashes($email);$designation =
> addslashes($designation);$title =
> addslashes($title);$company =
> addslashes($company);if($newsletter != "y"){
> $newsletter = "n"; };
> (....not sure what is going on here.... )if($id){
> unset($password); if($password1){ $password = ",
> password = password('$password1')"; };
>
> mysql_db_query("$db", "update members set username
> = '$username',
> name_prefix = '$name_prefix', name_first =
> '$name_first', name_middle =
> '$name_middle', name_last = '$name_last',
> name_suffix = '$name_suffix',
> email = '$email', designation = '$designation',
> title = '$title',
> foodchain='$foodchain', company = '$company',
> newsletter = '$newsletter'
> $password where id = $id") || print
> "FAILED.\n\n";}else{ mysql_db_query("$db", "insert
> into members (id, username, password,
> foodchain, name_prefix, name_first, name_middle,
> name_last, name_suffix,
> email, designation, title, company, newsletter)
> values(null, '$username',
> password('$password1'), $foodchain, '$name_prefix',
> '$name_first',
> '$name_middle', '$name_last', '$name_suffix',
> '$email', '$designation',
> '$title', '$company', '$newsletter')"); };?>
>
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