RE: [PHP] How to output a NULL field?
| From: | David Stoltz | Date: | Wed, 26 Aug 2009 11:29:53 +0000 |
| Subject: | RE: [PHP] How to output a NULL field? | ||
| References: | 1 2 3 4 5 | Groups: | php.general |
| Request: | Send a blank email to php-general+get-297252@lists.php.net to get a copy of this message | ||
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@quillandmouse.com]
Sent: Tuesday, August 25, 2009 4:39 PM
To: php-general@lists.php.net
Subject: Re: [PHP] 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
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