2017-06-14 64 views
-2

image for the result我试图从DATABSE获得价值并将其设置为下拉框下拉列表但它没有得到任何价值 这里是我的代码如何从DATABSE获得价值并将其设置在PHP

<?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>'; 
    ?> 
+0

检查的print_r(mysqli_fetch_array($ SQL))有数据或者没有 –

+1

首先哪里是你的数据库连接变量 – JYoThI

+1

mysqli_query至少需要两个参数 –

回答

0

mysqli_query函数占用至少两个参数。您在调用mysqli_query()函数时没有提供连接变量。

mysqli_query (mysqli $link , string $query) 

作为每PHP参考:

链路:只有程序式:)由mysqli_connect()或mysqli_init(返回的链路标识符

查询:查询字符串。

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

0
<?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>'; 
} 
?>