image for the result我试图从数据库获取价值并将其设置为下拉框,但它没有获得任何价值 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>';
?>
答案 0 :(得分:0)
mysqli_query 函数至少有两个参数。调用mysqli_query()函数时没有提供连接变量。
mysqli_query ( mysqli $link , string $query)
根据PHP参考:
link:仅限程序样式:mysqli_connect()或mysqli_init()返回的链接标识符
query:查询字符串。
答案 1 :(得分: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>';
}
?>