Mysql 如果我想要'；t显示前10个记录列表

我有一个房间预订系统，用户可以选择一个房间，然后为给定的房间预订时间段

我不想再扩展我的系统了，我可以选择预订最多的前十名房间

您可以在此处查看我的数据库及其外键：

我知道我可以计算所有房间id为2 fx的预订：

SELECT COUNT(*) FROM `bookings` WHERE `room_id` = 2

但我想实现的是，有这样一个列表：

Room1 = 84 bookings
Room2 = 70 bookings
Room3 = 54 bookings
etc.

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有人能给我指出正确的方向吗？

我认为你应该使用GROUPBY子句

SELECT COUNT(*) as bCount, room_id 
FROM `bookings`
GROUP BY room_id
ORDER BY bCount
LIMIT 10

SELECT Count(*) as Count , room_id
FROM bookings
GROUP BY room_id
ORDER BY Count

---

我认为你应该使用GROUPBY子句

SELECT Count(*) as Count , room_id
FROM bookings
GROUP BY room_id
ORDER BY Count

---

我认为你应该使用GROUPBY子句

SELECT Count(*) as Count , room_id
FROM bookings
GROUP BY room_id
ORDER BY Count

---

我认为你应该使用GROUPBY子句

SELECT Count(*) as Count , room_id
FROM bookings
GROUP BY room_id
ORDER BY Count

试试这个查询

   select count(*) as booking_count , room_id as room from `bookings`
   group by room order by booking_count 

试试这个查询

   select count(*) as booking_count , room_id as room from `bookings`
   group by room order by booking_count 

试试这个查询

   select count(*) as booking_count , room_id as room from `bookings`
   group by room order by booking_count 

试试这个查询

   select count(*) as booking_count , room_id as room from `bookings`
   group by room order by booking_count 

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令人惊叹的。谢谢这不仅仅是一个指针。然而，现在预订的房间中，预订最少的在顶部，预订最多的在底部。我该如何扭转这种局面？@Adnaves你对排序数组那么不熟悉吗？太棒了。谢谢这不仅仅是一个指针。然而，现在预订的房间中，预订最少的在顶部，预订最多的在底部。我该如何扭转这种局面？@Adnaves你对排序数组那么不熟悉吗？太棒了。谢谢这不仅仅是一个指针。然而，现在预订的房间中，预订最少的在顶部，预订最多的在底部。我该如何扭转这种局面？@Adnaves你对排序数组那么不熟悉吗？太棒了。谢谢这不仅仅是一个指针。然而，现在预订的房间中，预订最少的在顶部，预订最多的在底部。我该如何扭转这种局面？@Adnaves你对排序数组不熟悉吗？