如何从数据库中获取值并将其设置为php中的下拉列表.结果的

Question

图像我试图从数据库中获取值并将其设置为下拉框，但它没有得到任何值这是我的代码

<?php
$sql=mysqli_query("SELECT RoomTypeId from roomtypemaster");
if(mysqli_num_rows($sql)){
echo '<select name="select">';
while($rs=mysqli_fetch_array($sql)){
      echo '<option value="'.$rs['RoomTypeId'].'">'.$rs['RoomTypeId'].'</option>';
  }
}
echo '</select>';
     ?> 

Answer 1

mysqli_query函数至少接受两个参数。调用 mysqli_query(( 函数时未提供连接变量。

mysqli_query ( mysqli $link , string $query)

根据 PHP 参考：

link：仅过程样式：由 mysqli_connect(( 或 mysqli_init(( 返回的链接标识符

查询

：查询字符串。

http://php.net/manual/en/mysqli.query.php

Answer 2

<?php
// you should connect the database
$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "myDB";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} 
$sql=mysqli_query("SELECT RoomTypeId from roomtypemaster");
$result = $conn->query($sql);
//after this you will check  the number of rows from  your  database table
$count = mysqli_num_rows();
if($count>0){
   echo '<select>';
   while($data = $result->mysql_fetch_aasoc()){
    echo '<option value="'.$data['RoomTypeId'].'">'.$data['RoomTypeId'].'</option>';
   }
   echo '</select>';
}
?>