Re: How to output a NULL field?

On Wed, Aug 26, 2009 at 9:51 AM, David Stoltz<Dstoltz@shh.org> wrote: > 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 $rs->Fields(22); //this is where that particular field is NULL, and produces the error > > .... > Because you are using COM, you can't use PHP's empty(), isset(), or is_null() in your if(...) statement. I've not used COM for much in PHP, but I think you'll have to do something like this: switch (variant_get_type($rs->Fields(22)) { case VT_EMPTY: case VT_NULL: $q4 = ''; break; case VT_UI1: