l_starter_l Posted December 6, 2008 Share Posted December 6, 2008 I need help to create a dynamic drop down menu from a database This is where I select the manufacturer of a car and the the corresponding models load in the second list. I wrote this code but it doesnt seem to work and anyone correct it for me <?php $conn=odbc_connect("cisproject","" ,"");// database connection $make = $_POST['manuf_name']; if ($make){ ///////////////////////////////////////////////// $query = sprintf("SELECT * FROM Models where manufacture_id='$make'"); $result = @mysql_query($query); $rowModel = mysql_fetch_array($result); ///////////////////////////////////////////////// } ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" /> <title>Untitled Document</title> </head> <body> <form id="form1" name="form1" method="post" action="drop.php" > <select name="make" onChange="document.forms[0].submit()"> <option value="">Select Make</option> <option value="1" <?php if(!(strcmp(1, $make))){echo "selected";}?>>LEXUS</option> <option value="2" <?php if(!(strcmp(2, $make))){echo "selected";}?>>TOYOTA</option> </select> <select name="model"> <option value="">Select Model</option> <?php do { ?> <option value="<?php echo $rowModel['manufactures_id']; ?>"><?php echo $rowModel['model_name']; ?></option> <?php }while ($rowModel = mysql_fetch_array($result)); ?> </select> </form> </body> </html> Can someone pls help me with my code Link to comment https://forums.phpfreaks.com/topic/135737-dynamic-drop-down-list/ Share on other sites More sharing options...
timmah1 Posted December 6, 2008 Share Posted December 6, 2008 http://dynamicdrive.com/dynamicindex16/chainedselects/index.htm Link to comment https://forums.phpfreaks.com/topic/135737-dynamic-drop-down-list/#findComment-707388 Share on other sites More sharing options...
fanfavorite Posted December 6, 2008 Share Posted December 6, 2008 $query = sprintf("SELECT * FROM Models where manufacture_id='$make'"); <option value="<?php echo $rowModel['manufactures_id']; ?>"><?php echo $rowModel ['model_name']; ?></option> Watch the naming, they must match exactly. Link to comment https://forums.phpfreaks.com/topic/135737-dynamic-drop-down-list/#findComment-707436 Share on other sites More sharing options...
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