Sorry - I don't know what you mean by DB class?
I'm using Microsoft SQL 2000....with this code:
<?php
//create an instance of the ADO connection object
$conn = new COM ("ADODB.Connection")
or die("Cannot start ADO");
//define connection string, specify database driver
$connStr = "PROVIDER=SQLOLEDB;SERVER=xxxx;UID=xxx;PWD=xxxx;DATABASE=xxxx";
$conn->open($connStr); //Open the connection to the database
$query = "SELECT * FROM eval_evaluations WHERE id = ".$_POST["eval"];
$rs = $conn->execute($query);
echo $rs->Fields(22); //this is where that particular field is NULL, and
produces the error
....
-----Original Message-----
From: hack988 hack988 [mailto:hack...@dev.htwap.com]
Sent: Wednesday, August 26, 2009 8:08 AM
To: David Stoltz
Cc: php-general@lists.php.net
Subject: Re: [PHP] How to output a NULL field?
Could you post your database's class to here?
I'm use mssql with php for several years and read NULL Fields is never
appear your case.
2009/8/26 David Stoltz <dsto...@shh.org>:
> I tried that -it's in the first part of my message
>
>
> -----Original Message-----
> From: hack988 hack988 [mailto:hack...@dev.htwap.com]
> Sent: Wednesday, August 26, 2009 7:39 AM
> To: David Stoltz
> Cc: Paul M Foster; php-general@lists.php.net
> Subject: Re: [PHP] How to output a NULL field?
>
> use is_null() check it
>
> 2009/8/26 David Stoltz <dsto...@shh.org>:
>> 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:pa...@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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>
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