saravanataee Posted December 31, 2012 Share Posted December 31, 2012 Dear all, I am actually trying to have triple dropdown in my php form. i.e 1st drop down initially loads with default values and once selecting a value in it, then second dropdown loads based on the value selected in 1st combo, following 3rd combo based on value selected in 2nd combo. I referred this example : http://roshanbh.com.np/2008/01/populate-triple-drop-down-list-change-options-value-from-database-using-ajax-and-php.html The problem is the Source code itself not working. when i run the source code with changes made as specified by the author, i am able to select the first combo with values usa,Canada. but second and third combo does not work.. My index file is <code><HTML> <HEAD> <TITLE>AJAX DEMO</TITLE> <script language="javascript" type="text/javascript"> function getXMLHTTP() { //fuction to return the xml http object var xmlhttp=false; try{ xmlhttp=new XMLHttpRequest(); } catch(e) { try{ xmlhttp= new ActiveXObject("Microsoft.XMLHTTP"); } catch(e){ try{ xmlhttp = new ActiveXObject("Msxml2.XMLHTTP"); } catch(e1){ xmlhttp=false; } } } return xmlhttp; } function getState(countryId) { alert(req.responseText); var strURL="findState.php?country="+countryId; var req = getXMLHTTP(); if (req) { req.onreadystatechange = function() { if (req.readyState == 4) { // only if "OK" if (req.status == 200) { document.getElementById('statediv').innerHTML=req.responseText; } else { alert("There was a problem while using XMLHTTP:\n" + req.statusText); } } } req.open("GET", strURL, true); req.send(null); } } function getCity(countryId,stateId) { var strURL="findCity.php?country="+countryId+"&state="+stateId; var req = getXMLHTTP(); if (req) { req.onreadystatechange = function() { if (req.readyState == 4) // only if "OK" { if (req.status == 200) { document.getElementById('citydiv').innerHTML=req.responseText; } else { alert("There was a problem while using XMLHTTP:\n" + req.statusText); } } } req.open("GET", strURL, true); req.send(null); } } </script> </HEAD> <BODY> <form method="post" name="form1"> <table border="0" cellpadding="0" cellspacing="0" width="60%"><tbody> <tr> <td width="150">Country</td> <td width="150"><select style="background-color: #ffffa0" name="country" onchange="getState(this.value)"><option>Select Country</option><option value="1">USA</option><option value="2">Canada</option> </select></td> </tr> <tr> <td>State</td> <td> <p id="statediv"> <select style="background-color: #ffffa0" name="state"><option>Select Country First</option> </select></td> </tr> <tr> <td>City</td> <td> <p id="citydiv"> <select style="background-color: #ffffa0" name="city"><option>Select State First</option> </select></td> </tr> </tbody></table> </form> </BODY> </HTML> </code> findCity.php: <code> <!--//---------------------------------+ // Developed by Roshan Bhattarai | // http://roshanbh.com.np | // Contact for custom scripts | // or implementation help. | // email-nepaliboy007@yahoo.com | //---------------------------------+--> <? #### Roshan's Ajax dropdown code with php #### Copyright reserved to Roshan Bhattarai - nepaliboy007@yahoo.com #### if you have any problem contact me at http://roshanbh.com.np #### fell free to visit my blog http://php-ajax-guru.blogspot.com ?> <? $countryId=intval($_GET['country']); $stateId=intval($_GET['state']); $link = mysql_connect('localhost', 'root', ''); //changet the configuration in required if (!$link) { die('Could not connect: ' . mysql_error()); } mysql_select_db('db_ajax'); $query="SELECT id,city FROM city WHERE countryid='$countryId' AND stateid='$stateId'"; $result=mysql_query($query); ?> <select name="city"> <option>Select City</option> <? while($row=mysql_fetch_array($result)) { ?> <option value><?=$row['city']?></option> <? } ?> </select> </code> findState.php: <code> <!--//---------------------------------+ // Developed by Roshan Bhattarai | // http://roshanbh.com.np | // Contact for custom scripts | // or implementation help. | // email-nepaliboy007@yahoo.com | //---------------------------------+--> <? #### Roshan's Ajax dropdown code with php #### Copyright reserved to Roshan Bhattarai - nepaliboy007@yahoo.com #### if you have any problem contact me at http://roshanbh.com.np #### fell free to visit my blog http://roshanbh.com.np ?> <? $country=intval($_GET['country']); $link = mysql_connect('localhost', 'root', ''); //changet the configuration in required if (!$link) { die('Could not connect: ' . mysql_error()); } mysql_select_db('db_ajax'); $query="SELECT id,statename FROM state WHERE countryid='$country'"; $result=mysql_query($query); ?> <select name="state" onchange="getCity(<?=$country?>,this.value)"> <option>Select State</option> <? while($row=mysql_fetch_array($result)) { ?> <option value=<?=$row['id']?>><?=$row['statename']?></option> <? } ?> </select> </code> Please let me know where the problem is!! or does this script requires any resource.! as a newbie i dont understand the problem exactly by looking at the code. Thanks!! Quote Link to comment Share on other sites More sharing options...
Muddy_Funster Posted December 31, 2012 Share Posted December 31, 2012 could you please edit you post to use code tags around your script? That's rather hard to read like that. Quote Link to comment Share on other sites More sharing options...
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