merhabalar,
selectbox a databaseden bilgileri ajax ile yazdirabildim ama ayni bilgileri text area ya yazdiramadim.
yardimci olabilecek birisi varmi?
saygilar
<html>
<head>
<title>Roshan's Triple Ajax dropdown code</title>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">
<script language="javascript" type="text/javascript">
// Roshan's Ajax dropdown code with php
// This notice must stay intact for legal use
// Copyright reserved to Roshan Bhattarai - nepaliboy007@yahoo.com
// If you have any problem contact me at http://roshanbh.com.np
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) {
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" action="" name="form1">
<table width="600" border="0" align="center" cellpadding="0" cellspacing="0">
<tr>
<td width="150">Country</td>
<td width="150">
<? $countryId=intval($_GET['country']);
$stateId=intval($_GET['state']);
$link = mysql_connect('localhost', 'root', 'emre'); //changet the configuration in required
if (!$link) {
die('Could not connect: ' . mysql_error());
}
mysql_select_db('db_ajax');
$query="SELECT id, country FROM country ";
$result=mysql_query($query);
?>
<select name="country" onChange="getState(this.value)">
<option>Select City</option>
<? while($row=mysql_fetch_array($result)) { ?>
<option value="<?=$row['id']?>"><?=$row['country']?></option>
<? } ?>
</select> </td>
<td width="150" align="center">State</td>
<td width="150"><div id="statediv"><select name="state" >
<option>Select Country First</option>
</select></div></td>
<td width="150" align="center">City </td>
<td width="150"><div id="citydiv"><select name="city">
<option>Select State First</option>
</select></div></td>
</tr>
</table>
<table width="600" align="center">
<tr>
<td><label>
<textarea name="textarea" id="textarea" cols="45" rows="5"></textarea>
</label></td>
</tr>
</table>
</form>
</body>
</html>