我正在进行一个关于讨论室服务的小型大学项目。现在我的任务是实现订购名称的自动完成功能。我已经用谷歌搜索了一些教程。我不确定出了什么问题，当我尝试输入名称时，前面没有输入任何数据。
这是我的表单代码：
<?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>
这是我的代码，它应该从我的表返回 json 数据
<?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);
?>
任何帮助将非常感激。多谢！
在脚本中使用以下代码。从源中删除分号。您可以使用冒号作为其他参数。
<script>
$(function() {
$( "#typeahead" ).autocomplete({
source: 'typeaheads.php'
});
});
</script>
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