[PHP-DB] Question on how to test a value

[Lara J. Fabans]

Hi, friends,

Here is how I'm putting the information into the database:
$query7 = isset($price) ? "INSERT INTO price (itemid, price) VALUES ($itemid, 
'$price')" : 0;
$result7 = @pg_Exec($connection, $query7 );

I'm having difficulties with my test in display.
I've tried isset and == 'NULL' and == '0' as such:

    if ($values2 == 'NULL')
    {
       print "<tr><td class=\"display\">&nbsp;</td></tr>";
    } else
    {
    print "<tr><td class=\"display\">US$ $values2->price</td></tr></table></td>";
    }


and


    if (isset($values2))
    {
       print "<tr><td class=\"display\">US$ $values2->price</td></tr></table></td>";
    } else
    {
    print "<tr><td class=\"display\">&nbsp;</td></tr>";
    }


However, in both cases, if the price doesn't exist in the database, the  US$ still 
shows up in the
display.

db=> select * from price where itemid=50;
 priceid | itemid | price 
---------+--------+-------
      48 |     50 | 
(1 row)


I was surprised that the insert put something into the database at all.

I could use some advice.  

I cleaned up the database to remove all the lines where price wasn't set.
Why is it creating a table row?  If the insert form is passing in where they didn't 
set the price,
wouldn't that get picked up by the isset?  

And if there isn't a row in the database, wouldn't testing isset of the result of the 
query be
sufficient, or should I do a test on $values2->price instead?

I've tried lots of things and nothing seems to be doing the right behavior.  So thanks!
Lara