PHP + MySQL-數據庫自動完成功能無法從表中獲取數據

Question

我正在做一個關於討論室服務的小型大學項目。 現在，我被要求實施自動完成功能的名稱命令。 我已經用谷歌搜索了一些教程。 我不確定出了什么問題，當我嘗試鍵入名稱時，前面沒有數據。 這是我的表單代碼：

<?php
$host = "localhost";
$user = "root";
$pass = "";
$name = "pinjamruang";
$koneksi = mysqli_connect($host, $user, $pass, $name);

//Periksa apakah koneksi berhasil
if(mysqli_connect_errno()){
    echo "Error: ";
    echo mysqli_connect_error();
    echo "<br   /> Error Code: ";
    echo mysqli_connect_errno();
    die();
}
$sql = "SELECT * FROM ruangan
        WHERE id = $_GET[id]";
$hasil = mysqli_query($koneksi,$sql);
$row = mysqli_fetch_assoc($hasil);
$sql2 = "SELECT * FROM shift
        WHERE id = $_GET[shift]";
$hasil2 = mysqli_query($koneksi,$sql2);
$row2 = mysqli_fetch_assoc($hasil2);
?>
<link rel="stylesheet" href="//code.jquery.com/ui/1.11.4/themes/smoothness/jquery-ui.css">
<script src="//code.jquery.com/jquery-1.10.2.js"></script>
<script src="//code.jquery.com/ui/1.11.4/jquery-ui.js"></script>
<script>
  $(function() {
    $( "#typeahead" ).autocomplete({
      source: 'typeaheads.php';
    });
  });
</script>
<h1> Konfirmasi Pemesanan Ruang <?php echo $row['kode']; ?></h1><br>
<form class="form-horizontal" action="process/process-order-ruang.php" method="post">
    <div class="form-group">
        <label for="inputNamaPemesan" class="col-sm-2 control-label">Nama</label>
        <div class="col-sm-10">
            <input type="text" name="nama_pemesan" class="form-control" id="typeahead" placeholder="Nama Pemesan">
        </div>
    </div>
    <div class="form-group">
        <label for="inputKeperluan" class="col-sm-2 control-label">Keperluan</label>
        <div class="col-sm-10">
            <select name="keperluan" class="form-control" id="inputKeperluan">
                <option value="Diskusi Belajar">Diskusi Belajar</option>
                <option value="Diskusi Tugas">Diskusi Tugas</option>
                <option value="Dokumentasi">Dokumentasi</option>
                <option value="Lain-lain">Lain-lain</option>
            </select>
        </div>
    </div>
    <div class="form-group">
        <label for="inputWaktu" class="col-sm-2 control-label">Waktu</label>
        <div class="col-sm-10">
        <input type="text" class="col-sm-5" name="waktu" value="<?php $row2['shift'];
                        $timestamp = strtotime($row2['shift']);
                        $waktuk = date('H.i A', $timestamp);
                        $int = (int)$waktuk;
                        echo $int; ?>:00" disabled> - <input type="text" class="col-sm-5"value="<?php $row2['shift'];
                        $timestamp = strtotime($row2['shift']);
                        $waktuk = date('H.i A', $timestamp);
                        $int = (int)$waktuk;
                        echo $int+2; ?>:00" disabled>
        </div>
    </div>
    <?php $shift = $_GET['shift'];
          $ruangan = $_GET['id'];?> 
    <input type="hidden" value="<?php $int2 = (int)$shift;?>" name="shift">
    <input type="hidden" value="<?php $int3 = (int)$ruangan;?>" name="ruangan">
    <div class="form-group">
        <div class="col-sm-offset-2 col-sm-10">
            <button type="submit" class="btn btn-primary">Pesan</button>
        </div>
    </div>
</form>

這是我的代碼，應該從表中返回json數據

<?php
$host = "localhost";
$user = "root";
$pass = "";
$name = "pinjamruang";
$koneksi = mysqli_connect($host, $user, $pass, $name);
//connect with the database

//get search term
$searchTerm = $_GET['term'];
//get matched data from table
$query = $koneksi->query("SELECT * FROM user 
                          WHERE nama LIKE '%".$searchTerm."%' ORDER BY nama ASC");
while ($row = $query->fetch_assoc()) {
    $data[] = $row['nama'];
}
//return json data
echo json_encode($data);
?>

任何幫助將非常感激。 非常感謝！

Answer 1

在腳本中使用以下代碼。 從源代碼中刪除分號。 您可以將冒號用於其他參數。

<script>
    $(function() {
        $( "#typeahead" ).autocomplete({
            source: 'typeaheads.php'
        });
    });
</script>