Re: How to output a NULL field?

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On Wed, Aug 26, 2009 at 9:51 AM, David Stoltz<Dstoltz@xxxxxxx> 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:

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