Re: How to output a NULL field?

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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

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