Question

我有房间清单

豪华
标准
常规

我根据用户每天添加的内容进行了显示记录。

如果用户添加新事件并使用其中一个房间。

示例：今天是2018年8月26日。预订了一个事件，它将显示ff。从数据库中检索。

豪华 -已预订1张

我要显示的是这个

豪华 -已预订1张

标准 -无

常规 -无

我想添加所有房间列表，即使某些房间在数据库中没有价值或数据库中不存在。

但是我该怎么办呢？请帮忙！

我的示例代码：

    $connect = mysqli_connect("localhost", "root", "", "bookings");
    $sql = "SELECT * FROM events WHERE (start BETWEEN '$start' AND '$end' OR 
    end BETWEEN '$start' AND '$end');  
    $result = mysqli_query($connect, $sql);
    while($rows = mysqli_fetch_array($result))
    {
    echo $rows['room'];
    }

Answer 1

基于OP的评论，您应该使用GROUP BY子句获取每个类别的预订房间数。如果未找到特定房间类别的条目，则应让您的应用程序代码来处理。这是示例代码：

$connect = mysqli_connect("localhost", "root", "", "bookings");
$sql = "SELECT room, COUNT(id) as rooms_booked 
         FROM events 
         WHERE ((start BETWEEN '$start' AND '$end') OR 
                (end BETWEEN '$start' AND '$end')
               ) 
         GROUP BY room";  
$result = mysqli_query($connect, $sql);

// all possible room categories
$all_room_categories = array('Deluxe', 'Standard', 'Regular');
// room categories for which there is some booking done
$booked_room_categories = array();

while($row = mysqli_fetch_array($result, MYSQLI_ASSOC))
{
    echo $row['room'] . " - " . $row['rooms_booked'] . " booked" . "<\br>";
    $booked_room_categories[] = $row['room'];
}

// Now find all the room categories for which no booking was found
$unbooked_room_categories = array_diff($all_room_categories, $booked_room_categories);

 // Loop over unbooked rooms to print their status as well
foreach ($unbooked_room_categories as $cat)
{
    echo $cat . " - " . " none" . "<\br>";
}

如果数据库中不存在所有列表记录，该如何显示？ MySQL的PHP

1 个答案: