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 $q4;?>
>
> What am I doing wrong?
>
> Thanks!
First off, if the value is NULL in the database then in PHP it will be
the string "NULL" and not a null value as far as I remember. Second,
you cant use a function/method in empty(). Thirdly, the string "NULL"
is not empty. I'm not sure what DB class you're using here or what the
Fields() method actually returns. You should do a
var_dump($rs->Fields(22)) to see. If it returns a string (especially
"NULL"), then this or some variation should work:
$q4 = $rs->Fields(22);
if($q4 == "NULL"){
$q4 = "";
}
If it returns an empty string or false then you may have nothing to do.
--
Thanks!
-Shawn
http://www.spidean.com
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