Return Results Not Found In A Query

[tomtimms]

I am trying to do a search against my mysql database for names that are "NOT" found.  I have a search form that is submitted and I would like to return the names that were not found in the database.  Of course I am able to retrieve the names that are found but not the ones that are not.  Anyone know the best approach to this?  I was thinking to run a comparison of my query to the search variable however I want to know if there is an easier way.  Thanks to anyone who an assist. 

[MadTechie]

Find records with the Name MadTechie

SELECT * FROM table WHERE name = 'MadTechie'

 

Find records without the Name MadTechie

SELECT * FROM table WHERE name != 'MadTechie'

[tomtimms]

Ahh I don't think I explained right, When I perform the query, I want it to look for all the values in my array and send back the values that were not found.

[MadTechie]

okay,

as this is in the PHP section and not the SQL one 

try this logic find all items in the database and store them in an array and then use arraydiff to generate a new array of unfound items

 

try this untested script (update table and field of course)

<?php
$findme = array("One","Two","Three","Four");
$query = sprintf("SELECT field FROM table WHERE field in (%s)",implode(",", $findme));
$result = mysql_query($query);
$found = array();
while($row = mysql_fetch_assoc($result)){
    $found[] = $row['field'];
}
$notFound = array_diff($findme, $found);
echo "<pre>";var_dump($notFound);echo "</pre>";
?>