if(empty($rs->Fields(22))){
$q4 = '';
}else{
$q4 = ''.$rs->Fields(22);
}
Still produces errors, whether using "empty" or "is_null"....with single
quotes....
???
-----Original Message-----
From: Bastien Koert [mailto:phps...@gmail.com]
Sent: Tuesday, August 25, 2009 2:17 PM
To: David Stoltz
Cc: php-general@lists.php.net
Subject: Re: [PHP] How to output a NULL field?
On Tue, Aug 25, 2009 at 2:00 PM, David Stoltz<dsto...@shh.org> wrote:
> $rs->Fields(22) equals a NULL in the database
>
> My Code:
>
> if(empty($rs->Fields(22))){
> $q4 = "";
> }else{
> $q4 = $rs->Fields(22);
> }
>
> Produces this error:
> Fatal error: Can't use method return value in write context in
> D:\Inetpub\wwwroot\evaluations\lookup2.php on line 32
>
> Line 32 is the "if" line...
>
> If I switch the code to (using is_null):
> if(is_null($rs->Fields(22))){
> $q4 = "";
> }else{
> $q4 = $rs->Fields(22);
> }
>
> It produces this error:
> Catchable fatal error: Object of class variant could not be converted to
> string in D:\Inetpub\wwwroot\evaluations\lookup2.php on line 196
>
> Line 196 is: <?php echo $q4;?>
>
> What am I doing wrong?
>
> Thanks!
>
> --
> PHP General Mailing List (http://www.php.net/)
> To unsubscribe, visit: http://www.php.net/unsub.php
>
>
$q4 = '' . $rs->Fields(22);
Note that it's two single quotes
--
Bastien
Cat, the other other white meat
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