Paul, This all started because when I try this: <?php echo $rs->Fields(22);?> It work fine, as long as there is a non-null value there, otherwise it produces an error. Also, I'm working with a Microsoft SQL 2000 database, not MySQL....not sure if that matters.... But "echo $rs->Fields(22)" works perfectly for dumping values out of my $rs recordset...that is, unless the value is NULL is the database - then I get: Catchable fatal error: Object of class variant could not be converted to string in D:\Inetpub\wwwroot\evaluations\lookup2.php on line 176 -----Original Message----- From: Paul M Foster [mailto:paulf@xxxxxxxxxxxxxxxxx] Sent: Tuesday, August 25, 2009 4:39 PM To: php-general@xxxxxxxxxxxxx Subject: Re: How to output a NULL field? On Tue, Aug 25, 2009 at 02:00:04PM -0400, David Stoltz 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! Just a thought... do you really mean $rs->Fields(22) or do you mean $rs->Fields[22]? The former is a function call and the latter is an array variable. Paul -- Paul M. Foster -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
