On Friday 10 May 2002 10:52, Steve Buehler wrote:
> No....keep writing to the list. Next time, you might actually
Good idea.
[snip]
> >What baffles me, is
> >
> > while ($VerifyPassword = mysql_fetch_array($PasswordCheckSQL)) {
> > // Compare UserIDs
> > if ($VerifyPasword[ContactID] = $_SESSION['ContactID']) {
> > $_SESSION['Authenticated'] = "True";
> > $AuthenticationFailure = "False";
> > } else {
> > $_SESSION['Authenticated'] = "False";
> > $AuthenticationFailure = "True";
> > }
> >
> >Works ($VerifiyPassword[ContactID] comes out of a MySQL Lookup,
> >$_SESSION['ContactID'] was previously, also looked up via a MySQL Query).
> >However, from what I understand in the documentation, in this case, the if
> >statement should be '==' in which case, it doesn't work ?!?!?!?!?!?!?
> > This is exactly what I am talking about, and why it is so confusing.
> > Everywhere, I use either a double =, or triple = in the if statements,
> > with a ! to use the is "not" true... Only in this specific statement,
> > the only way I can get it to work, was with a single =. Now what makes
> > that if statement so special over the others, that this one requires a
> > single = and not a double like all the other hundreds I have in my code?
What is confusing you is that the single equal sign /seems/ to work and is
masking the real error(s) in your code.
if ($VerifyPasword[ContactID] = $_SESSION['ContactID'])
This statement is /always/ true if $_SESSION['ContactID'] is non-zero, or
non-empty.
If your code is copy-and-paste then the error is most likely because you used
a single 's' in $VerifyPasword[ContactID]. And also is better to single-quote
your array indices (see manual on arrays for reason):
$VerifyPassword['ContactID']
--
Jason Wong -> Gremlins Associates -> www.gremlins.com.hk
Open Source Software Systems Integrators
* Web Design & Hosting * Internet & Intranet Applications Development *
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