将表中的值从数据库显示到下拉列表中

问题描述：

<?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 code:

<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>

I am getting values from a table from a database and the values should be displayed inside a dropdown list. I have used the above code but it is not displaying anything. Can anyone help in this issue?

 ＆lt;？php 
 include'ONEngine / AS.php'; 
 
 if（  ！$ login-＆gt; isLoggedIn（））
 header（“Location：login.php”）; 
 $ user = new ASUser（ASSession :: get（“user_id”））; 
 $ userInfo = $ user-  ＆gt; getInfo（）; 
 
 //基本包含文件
 require_once（“../ db.php”）; 
 
 $ nav ='hotels'; 
 
 $ hotel_id ='1'  ; 
 
 //消息
包含'inc / messages.php'; 
 
 $ sql1 = mysqli_query（$ conn，“SELECT room_type_name FROM hotel_room_type WHERE hotel_id ='$ hotel_id'”）; 
 
  ？＆gt; 
  code>  pre> 
 
  HTML代码： p> 
 
 
 ＆lt; div class =“form-row row-fluid”  ＆gt; 
＆lt; div class =“span12”＆gt; 
＆lt; div class =“row-fluid”＆gt; 
＆lt; label class =“form-label span3”＆gt;房间名称＆lt; / label＆gt; \  n＆lt;？php echo“＆lt; select＆gt;”; 
 echo“＆lt; option value =''＆gt;选择一个＆lt; / option＆gt;”;  
 $ results = $ conn-＆gt; query（$ sqll）; 
 foreach（$ results as $ data）{
 echo“＆lt; option＆gt; $ data [room_type_name]＆lt; / option＆gt;”;  
} 
 echo“＆lt; / select＆gt;”; 
？＆gt; 
 
＆lt; / div＆gt; 
＆lt; / div＆gt; 
＆lt; / div＆gt; 
  code>  pre  > 
 
 我从数据库中的表中获取值，并且值应显示在下拉列表中。 我使用了上面的代码，但它没有显示任何内容。 任何人都可以帮忙解决这个问题吗？ p> 
  div>

答

You have a typo

 $results =  $conn->query($sqll);

should be

 $results =  $conn->query($sql1);

it is a one, not an L

将表中的值从数据库显示到下拉列表中

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