<?php
include 'ASEngine/AS.php';
if(!$login->isLoggedIn())
header("Location: login.php");
$user = new ASUser(ASSession::get("user_id"));
$userInfo = $user->getInfo();
//basic include files
require_once("../db.php");
$nav = 'hotels';
$hotel_id = '1';
//Messages
include 'inc/messages.php';
$sql1 = mysqli_query($conn,"SELECT room_type_name FROM hotel_room_type WHERE hotel_id = '$hotel_id'");
?>
HTML 代码:
<div class="form-row row-fluid">
<div class="span12">
<div class="row-fluid">
<label class="form-label span3">Room Name</label>
<?php echo "<select>";
echo "<option value=''>Select One</option>";
$results = $conn->query($sqll);
foreach ($results as $data) {
echo "<option>$data[room_type_name]</option>";
}
echo "</select>";
?>
</div>
</div>
</div>
我正在从数据库的表中获取值,这些值应显示在下拉列表中。我已经使用了上面的代码,但它没有显示任何内容。有人可以帮忙解决这个问题吗?
最佳答案
你有一个错别字
$results = $conn->query($sqll);
应该是
$results = $conn->query($sql1);
这是一个 1,而不是 L
关于php - 将数据库表中的值显示到下拉列表中,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/34262671/