Been down that road. I've used unset($duplicate), $duplicate="" - so I don't
know. I have been going around in circles since 5 this afternoon EST with
this - no matter what I do I get the same results. I'm just at a complete
loss. I've even tried setting it to a letter $duplicate="T", but get the
same results. I'm not a pro, but it has to be the way the variable is being
returned - extra hidden characters, slashes, dots, whatever. When I do echo
$duplicate - it displays 1, but when I do if($duplicate == 1) nothing
happens - I even tried "1". I am so frustrated and tired right now, I just
don't get this.
----- Original Message -----
From: <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>; <[EMAIL PROTECTED]>
Sent: Wednesday, April 02, 2003 2:16 AM
Subject: RE: [PHP-WIN] I really don't understand this?
> I _THINK_ this is to do with scope.
>
> If you say
>
> elseif ($duplicate = ...)
> {
> }
>
> and this is the first use of the variable $dublicate, then $duplicate goes
> out of scope at the end of the right brace.
>
> I could be wrong about that, but it's worth a try. See what happens if you
> declare the variable $duplicate _before_ the first if(...). For example:
>
> $duplicate = "";
> if (...)
> ...
>
> Jill
>
>
> -----Original Message-----
> From: ODCS [mailto:[EMAIL PROTECTED]
> Sent: Wednesday, April 02, 2003 6:26 AM
> To: PHP Windows
> Subject: [PHP-WIN] I really don't understand this?
>
>
> I have spent the last 8 hours (seriously) trying to figure out this stupid
> piece of code, and it's just beyond me why it doesn't work. I hope someone
> here can help, or even follow the code.
>
> I have a form where the user inputs information - the code below is the
> error checking for one of the fields. The first IF statement just checks
> that the filed is not empty.
>
> The first ELSEIF checks a function to see if the user is already in the
> mysql database and returns the letter T, D or S. ** This part works fine.
>
> If the first ELSIF is not = to T then the second ELSEIF is checked to see
if
> the letter returned is D. I have checked over and over and over and over
and
> the letter D is being returned (I have checked the third ELSEIF and it is
> the same). It appears like they are just being ignored.
>
> Either I am just not getting the concept, or there is something seriously
> weird going on.
>
> Any help in solving this mystery is truely appreciated.
>
> if (empty($form["name"])) { $error["name"] = "*"; $message["name"] =
> "Required field!";
> }
> elseif($duplicate = duplicate_name($form["name"]) == "T") {
> $error["name"] = "*"; $message["name"] = "Screen name already used!";
> }
> elseif ($duplicate == "D") {
> $error["name"] = "*"; $message["name"] = "Database error - Please try
> again in a few moments!";
> }
> elseif ($duplicate == "S") {
> $error["name"] = "*"; $message["name"] = "Server error - Please try
> again in a few moments!";
> }
>
>
> function duplicate_name($name) {
>
> if (!($connect = @mysql_connect($serverhost,$serveruser,$serverpass))) {
> **** These are all defined above - not included here.
> $server_error = "S";
> }
>
> if ([EMAIL PROTECTED]($databasename)) {
> $server_error = "D";
> }
>
> if($server_error) { $duplicate = $server_error;
> }
> else
> {
> $query = "SELECT name FROM $tablename WHERE name = '$name'";
>
> if(!$query) { $duplicate = "D"; }
>
> else {
> $result = mysql_query($query);
> $line = mysql_fetch_array($result);
>
> if($line['name'] == $name) { $duplicate = "T"; }
>
> mysql_free_result($result);
> mysql_close($connect);
>
> }
> }
>
> echo $duplicate; ***** This shows that the variable $duplicate contains
> T, D, or S.
>
> return($duplicate);
> }
>
>
>
> --
> PHP Windows Mailing List (http://www.php.net/)
> To unsubscribe, visit: http://www.php.net/unsub.php
>
> --
> PHP Windows Mailing List (http://www.php.net/)
> To unsubscribe, visit: http://www.php.net/unsub.php
>
>
--
PHP Windows Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
