altinnovation Posted September 19, 2011 Share Posted September 19, 2011 Hello, I have been able to turn a multiple select option form into a php array in this form: <?php if(isset($_POST['select3'])) { $aVenues = $_POST['select3']; if(!isset($aVenues)) { echo("<p>You didn't select any venues!</p>\n"); } else { $nVenues = count($aVenues); echo("<p>You selected $nVenues venues: "); for($i=0; $i < $nVenues; $i++) { echo($aVenues[$i] . " "); } echo("</p>"); } } ?> $aVenues is the individual numbers received from the form and $nVenues is the number of those received. Basically, what you get with this code is this: You selected 2 venues: 2 6 The numbers 2 and 6 are id's sent from the form, and I would like to use these numbers to request the ID of a row in a database and make a simple table. How would I go about doing that? My Mysql is basic and all I came up with is this: $sql = "SELECT * FROM some_database WHERE id IN ($aVenues)"; Thanks everyone )) Quote Link to comment https://forums.phpfreaks.com/topic/247448-use-variable-as-id-and-request-table-from-mysql/ Share on other sites More sharing options...
AyKay47 Posted September 19, 2011 Share Posted September 19, 2011 your query will work for what you want to do.. are you not sure how to make a table with the info? or are you asking if your query is correct for what you want to accomplish? Quote Link to comment https://forums.phpfreaks.com/topic/247448-use-variable-as-id-and-request-table-from-mysql/#findComment-1270712 Share on other sites More sharing options...
altinnovation Posted September 19, 2011 Author Share Posted September 19, 2011 Hello AyKay47, you are correct I know that I should request the permissions at the beginning, but I'm not sure where exactly to put this sql request and to draw and I'm also confused about the table :/ By the way, that query I made is correct ? Quote Link to comment https://forums.phpfreaks.com/topic/247448-use-variable-as-id-and-request-table-from-mysql/#findComment-1270731 Share on other sites More sharing options...
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