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Asrianor · Jan 11, 2017 · 0 Suka · 0 Tidak Suka
Mungkin ini bisa jadi referensi gan

1. Struktur databasenya nama database "dbsaya"
-- phpMyAdmin SQL Dump
-- version 4.5.1
-- http://www.phpmyadmin.net
--
-- Host: 127.0.0.1
-- Generation Time: Jan 11, 2017 at 03:58 AM
-- Server version: 10.1.19-MariaDB
-- PHP Version: 5.5.38

SET SQL_MODE = "NO_AUTO_VALUE_ON_ZERO";
SET time_zone = "+00:00";

CREATE TABLE `tabel1` (
  `id_tbl` varchar(5) NOT NULL,
  `na_tbl` varchar(10) NOT NULL,
  `al_tbl` varchar(10) NOT NULL,
  `no_tbl` varchar(10) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=latin1;


INSERT INTO `tabel1` (`id_tbl`, `na_tbl`, `al_tbl`, `no_tbl`) VALUES
('001', 'asri', 'bjb', '0811');


CREATE TABLE `tabel2` (
  `id_tbl` varchar(5) NOT NULL,
  `na_tbl` varchar(10) NOT NULL,
  `al_tbl` varchar(10) NOT NULL,
  `no_tbl` varchar(10) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=latin1;


INSERT INTO `tabel2` (`id_tbl`, `na_tbl`, `al_tbl`, `no_tbl`) VALUES
('001', 'rian', 'bjm', '0812');

ALTER TABLE `tabel1`
  ADD PRIMARY KEY (`id_tbl`);

--
-- Indexes for table `tabel2`
--
ALTER TABLE `tabel2`
  ADD PRIMARY KEY (`id_tbl`);


Buat file dengan nama proses-ajax.php
<?php
include 'koneksi.php';
$idterima = $_GET['idkirim'];
$tabelterima = $_GET['tabelkirim'];
$query = mysqli_query($koneksi, "select * from $tabelterima where id_tbl='$idterima'");
$ambil = mysqli_fetch_array($query);
$data = array(
            'data1'      =>  $ambil['na_tbl'],//na_tbl adalah nama field dari tabel di database
            'data2'   =>  $ambil['al_tbl'], //al_tbl adalah nama field dari tabel di database
            'data3'    =>  $ambil['no_tbl'],); //no_tbl adalah nama field dari tabel di database
 echo json_encode($data);
?>

File koneksi.php
<?php
// konfigurasi database
$host       =   "localhost";
$user       =   "root";
$password   =   "";
$database   =   "dbsaya";
// perintah php untuk akses ke database
$koneksi = mysqli_connect($host, $user, $password, $database);
?>

Untuk menampilkan nya yaitu file form-ajax.php
<!DOCTYPE html>
<html>
    <head>
        <title>Ajax Jquery - Belajarphp.net</title>
    </head>
    <body>
        <form action="">
            <table>
                <tr><td>ID</td><td><input type="text" id="id"></td></tr>
                <tr><td>TABEL</td><td><select type="text" onchange="isi_otomatis()" id="tabel">
                <option value="">Pilih tabel</option>
                            <option value="tabel1">tabel1</option>
                            <option value="tabel2">tabel2</option>
                        </select></td></tr>
                <tr><td>NAMA</td><td><input type="text" id="nama"></td></tr>
                <tr><td>ALAMAT</td><td><input type="text" id="alamat"></td></tr>
                <tr><td>NOHP</td><td><input type="text" id="nohp"></td></tr>
            </table>
        </form>
        <script src="https://ajax.googleapis.com/ajax/libs/jquery/1.12.4/jquery.min.js"></script>
        <script type="text/javascript">
            function isi_otomatis(){
                var idambil = $("#id").val();
                var tabelambil = $("#tabel").val();
                $.ajax({
                    url: 'proses-ajax.php',
                    data:"idkirim="+idambil+"&tabelkirim="+tabelambil,
                }).success(function (data) {
                    var json = data,
                    obj = JSON.parse(json);
                    $('#nama').val(obj.data1);
                    $('#alamat').val(obj.data2);
                    $('#nohp').val(obj.data3);
                });
            }
        </script>
    </body>
</html>

Saat di jalankan form-ajax.php masukkan dulu id baru pilih tabel