Rabu, 11 September 2013

PHP Indonesia - Facebook: [ask] disini saya memiliki sebuah contoh ajax seba...

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[ask] disini saya memiliki sebuah contoh ajax seba...
Sep 11th 2013, 15:17, by Fauzi Fauzy

Fauzi Fauzy 3:17pm Sep 11
[ask]
disini saya memiliki sebuah contoh ajax sebagai berikut :

"datakaryawan.html"

<html>
<head>
<script>
var ajaxku;
function ambildata(nip){
ajaxku = buatajax();
var url="ambildata.php?b="+nip;
ajaxku.onreadystatechange=stateChanged;
ajaxku.open("GET",url,true);
ajaxku.send(null);
}
function buatajax(){
if (window.XMLHttpRequest){
return new XMLHttpRequest();
}
if (window.ActiveXObject){
return new ActiveXObject("Microsoft.XMLHTTP");
}
return null;
}
function stateChanged(){
var data;
if (ajaxku.readyState==4){
data=ajaxku.responseText;
if(data.length>0){
document.getElementById("alamat").value = data
}else{
document.getElementById("alamat").value = "";
}
}
}
</script>
</head>
<body>
Nama :
<select size="1" name="karyawan" id="karyawan"
onchange=ambildata(this.value)>
<option selected>--Pilih Karyawan--</option>
<option value="889456">Desrizal</option>
<option value="889457">Nurmi Yulita</option>
<option value="896543">Budi Hartono</option>
</select>
</p>
<p>
Alamat : <textarea rows="4" id="alamat" name="alamat"
cols="42"></textarea>
</p>
Pada contoh berikut anda akan memilih nama dan mengambil data alamat di
database<br>
berdasarkan nama tersebut dan menampilkannya di halaman ini tanpa harus
me-reload<br>
keseluruhan halaman
</body>
</html>

"ambildata.php"

<?php
mysql_connect("localhost","root","");
mysql_select_db("datakaryawan");
$nip = $_GET['b'];
if($nip){
$query = mysql_query("select alamat from tabelDataKaryawan where
nip=$nip");
while($d = mysql_fetch_array($query)){
echo $d['alamat'];
}
}
?>

Pertanyaanya ,minta penjelasan mengenai " $nip = $_GET['b'] " pada file ambildata.php ...? dan " var url="ambildata.php?b="+nip " pada file datakaryawan.html... minta pencerahannya y...?

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