dc_jt Posted January 31, 2008 Share Posted January 31, 2008 Hi Im using this tutorial to create two drop downs (the second depending on the whats selected from the first) using this page: http://www.ajax-tutorials.com/tutorial-list/resource/AJAX_Dynamic_Drop_Down_List_PHP_xajax/view.php I have got the two drop downs working however I want to be able to use the value of the second drop down for something else but not sure how I can retrieve this. E.g Ive got drop down 1 consisting of regions - 'North West','North East','South' and if I select 'North West' I get the options of 'Liverpool','Manchester','Preston'. Now say I select Liverpool, how can I use this value to do something else? I dont want another dropdown I basically want to use this Id to collect info about Liverpool. Please see my code below: require($_SERVER['DOCUMENT_ROOT'].'/assets/js/xajax_core/xajax.inc.php'); $xajax = new xajax(); //$xajax->configure('debug',true); class myXajaxResponse extends xajaxResponse { function addCreateOptions($sSelectId, $options) { $this->script("document.getElementById('".$sSelectId."').length=0"); if (sizeof($options) >0) { foreach ($options as $option) { $this->script("addOption('".$sSelectId."','".$option."','".$option."');"); } } } } //get regions $aRegion = array(); foreach($aRegions as $oRegion){ array_push($aRegion, $oRegion['region_id']); } //get stores $aStore = array(); foreach ($aRegion as $rRegion){ $aStores = $oTblStores->GetAjaxStoresByLocation($rRegion); $aStore[$rRegion] = array(); if($aStores) foreach ($aStores as $oStore){ array_push($aStore[$rRegion], $oStore['Store']); } } // adds an option to the select function addStores($selectId, $aRegion) { global $aStore; $objResponse = new myXajaxResponse(); $data = $aStore[$aRegion]; $objResponse->addCreateOptions($selectId, $data); //return $objResponse->getXML(); return $objResponse; } function addHours($selectId, $transport) { global $aHour; $objResponse = new myXajaxResponse(); $data = $aHour[$transport]; $objResponse->addCreateOptions($selectId, $data); //return $objResponse->getXML(); return $objResponse; } $xajax->registerFunction("addStores"); $xajax->registerFunction("addHours"); $xajax->processRequest(); ?> <? if (isset($_POST['Submit'])) { print_r($_POST); } Javascript <? $xajax->printJavascript("/assets/js/"); ?> <script type="text/javascript"> function addOption(selectId, val, txt) { var objOption = new Option(txt, val); document.getElementById(selectId).options.add(objOption); } </script> Drop down html code <div class="ordering-step"> <div class="step-col step-col-border-right"><h2>Select store for collection</h2> <form name="form1" method="POST" action=""> <input name="sMode" type="hidden" value="save"> <input name="region" id="region" type="hidden" value=""> <label for="region">Choose Region</label> <select name="region_id" id="region_id" onchange="xajax_addStores('store', document.form1.region_id.value)"> <option value="">Select Region</option> <?php while ($oRegion = mysql_fetch_object($rRegions)) {?> <option value="<?= $oRegion->region_id?>"><?= $oRegion->name?></option> <? } ?> </select><br clear="all" /> <label for="store">Choose Store</label> <select name="store" id="store" onchange="xajax_addHours('hours', document.form1.store.value)"> <option value="">Select Store</option> </select> </form> Any idea how I get the Id of the second drop down to use somewhere else? Thanks Quote Link to comment Share on other sites More sharing options...
haku Posted January 31, 2008 Share Posted January 31, 2008 I replied to this for you in the other forum. Quote Link to comment Share on other sites More sharing options...
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