这是我的表结构:
___语言:
|--------|------------|
| LAN_Id | LAN_En |
|--------|------------|
| DI | Direct |
| WE | Web |
| OT | Other |
|--------|------------|
___分割:
|--------|------------|
| SEG_Id | SEG_Code |
|--------|------------|
| 1 | DI |
| 2 | WE |
| 3 | OT |
|--------|------------|
___预订:
|--------|------------------|
| BOO_Id | BOO_Segmentation |
|--------|------------------|
| 1 | 1 |
| 2 | 1 |
| 3 | 2 |
|--------|------------------|
___计费数据:
|--------|---------------|------------|------------|
| BIL_Id | BIL_BookingId | BIL_Date | BIL_Item |
|--------|---------------|------------|------------|
| 1 | 1 | 2017-02-21 | Night |
| 2 | 1 | 2017-02-22 | Night |
| 3 | 1 | 2017-02-23 | Night |
| 4 | 1 | 2017-02-24 | Night |
| 5 | 2 | 2017-02-25 | Night |
| 6 | 2 | 2017-02-26 | Night |
| 7 | 3 | 2017-02-28 | Night |
| 8 | 3 | 2017-03-01 | Night |
| 9 | 3 | 2017-03-02 | Night |
| 10 | 3 | 2017-03-03 | Night |
|--------|---------------|------------|------------|
我想知道某个日期范围内最受欢迎的分割。
所需的结果应该是以下日期范围内的结果:
表格 2017-02-01 至 2017-02-28(含)
|------------|------------|------------|--------------|------------|
| ROO_Name | Night_Nb | Percentage | Booking_Nb | Percentage |
|------------|------------|------------|--------------|------------|
| Direct | 6 | 85.71 | 2 | 66.66 |
| Website | 1 | 14.28 | 1 | 33.33 |
| Other | 0 | 0 | 0 | 0 |
|------------|------------|------------|--------------|------------|
我已经尝试过的:
SELECT r.SEG_Id
, Sum(CASE WHEN BOO_Id IS NULL THEN 0 ELSE 1 END) Night_Nb
, Concat(
Format(
Sum(CASE WHEN BOO_Id IS NULL THEN 0 ELSE 1 END)
/ TotalBookings
* 100
, 0) ) AS PercentageTotal
FROM ( ___Segmentations r LEFT JOIN ___Bookings b ON r.SEG_Id = b.BOO_Segmentation
) INNER JOIN (SELECT BOO_HotelId
, Count(*) AS TotalBookings
FROM ___Bookings
GROUP BY BOO_HotelId
) AS TotalHotelBookings
ON r.SEG_HotelId = TotalHotelBookings.BOO_HotelId
WHERE r.SEG_HotelId = :hotel_id
GROUP BY r.SEG_Id
ORDER BY NumBookings DESC
但实际上并没有用。
谁能帮我解决这个问题？
您可以使用 SQL Fiddle: http://sqlfiddle.com/#!9/1aa10a
最佳答案
我建议我们逐步构建查询。验证查询结果是否如我们在每一步所期望的那样。当某事“不起作用”时，备份一个步骤。
我们想要返回三行，一个用于 ___Segmentations 中的每一行，用于特定的 hotelid
SELECT r.seg_id
, r.seg_text
FROM ___Segmentations r
WHERE r.seg_hotelid = :hotel_id
ORDER BY r.seg_id
将外部联接添加到 __Bookings
SELECT r.seg_id
, r.seg_text
, b.boo_id
FROM ___Segmentations r
LEFT
JOIN ___Bookings b
ON b.boo_segmentation = r.seg_id
WHERE r.seg_hotelid = :hotel_id
ORDER
BY r.seg_id
, b.boo_id
将外部连接添加到 ___BillableDatas
SELECT r.seg_id
, r.seg_text
, b.boo_id
, d.bil_id
FROM ___Segmentations r
LEFT
JOIN ___Bookings b
ON b.boo_segmentation = r.seg_id
LEFT
JOIN `___BillableDatas` d
ON d.bil_bookingid = b.boo_id
WHERE r.seg_hotelid = :hotel_id
ORDER
BY r.seg_id
, b.boo_id
, d.bil_id
如果这是我们感兴趣的行，我们可以进行聚合。
SELECT r.seg_id
, r.seg_text
, COUNT(DISTINCT b.boo_id) AS cnt_bookings
, COUNT(DISTINCT d.bil_id) AS cnt_billable
FROM ___Segmentations r
LEFT
JOIN ___Bookings b
ON b.boo_segmentation = r.seg_id
LEFT
JOIN `___BillableDatas` d
ON d.bil_bookingid = b.boo_id
WHERE r.seg_hotelid = :hotel_id
GROUP
BY r.seg_id
, r.seg_text
ORDER
BY r.seg_text
现在获取带有“总数”的聚合。
我将采用的方法是使用 CROSS JOIN 操作制作行的“副本”。我们可以对我们编写的第一个查询返回的行进行连接，引用为内联 View 。 (在下面别名为 q。)
如果我们有一组完整的行，并为每个 seg_id/seg_text(我们编写的第一个查询)重复，我们可以使用条件聚合。
我们写的最后一个查询(就在上面)是下面查询中的一个内联 View ，别名为 c。
所有行的 cnt_bookings 的 SUM 是总数。
对于单个计数，我们可以仅包含具有匹配 seg_id 的行，即该子集的总数。
SELECT q.seg_id
, q.seg_text
, SUM(IF(c.seg_id=q.seg_id,c.cnt_bookings,0)) AS cnt_bookings
, SUM(c.cnt_bookings) AS tot_bookings
, SUM(IF(c.seg_id=q.seg_id,c.cnt_billable,0)) AS cnt_billable
, SUM(c.cnt_billable) AS tot_billable
FROM ( SELECT t.seg_id
, t.seg_text
FROM ___Segmentations t
WHERE t.seg_hotelid = :hotel_id_1
ORDER BY t.seg_id
) q
CROSS
JOIN ( SELECT r.seg_id
, COUNT(DISTINCT b.boo_id) AS cnt_bookings
, COUNT(DISTINCT d.bil_id) AS cnt_billable
FROM ___Segmentations r
LEFT
JOIN ___Bookings b
ON b.boo_segmentation = r.seg_id
LEFT
JOIN `___BillableDatas` d
ON d.bil_bookingid = b.boo_id
WHERE r.seg_hotelid = :hotel_id
GROUP
BY r.seg_id
) c
GROUP
BY q.seg_id
, q.seg_text
ORDER
BY q.seg_text
在SELECT列表中，我们可以做除法得到百分比:cnt_bookings * 100.0/tot_bookings
例如
SELECT q.seg_id
, q.seg_text
, SUM(IF(c.seg_id=q.seg_id,c.cnt_bookings,0)) AS cnt_bookings
, SUM(c.cnt_bookings) AS tot_bookings
, SUM(IF(c.seg_id=q.seg_id,c.cnt_bookings,0))
* 100.0 / SUM(c.cnt_bookings) AS pct_bookings
, SUM(IF(c.seg_id=q.seg_id,c.cnt_billable,0)) AS cnt_billable
, SUM(c.cnt_billable) AS tot_billable
, SUM(IF(c.seg_id=q.seg_id,c.cnt_billable,0))
* 100.0 / SUM(c.cnt_billable) AS pct_billable
修改 ORDER BY 子句以按您想要的顺序返回行
从 SELECT 列表中删除返回 tot_bookings 和 tot_billable 的表达式。
编辑
我想我错过了日期标准。我们可以把外连接变成内连接，把 CROSS JOIN 换成 LEFT JOIN。我们有可能为 cnt_bookings 和 cnt_billable 返回 NULL 值，我们可以将它们包装在 IFNULL() 或 COALESCE() 函数中以用零替换 NULL。
SELECT q.seg_id
, q.seg_text
, SUM(IF(c.seg_id=q.seg_id,c.cnt_bookings,0)) AS cnt_bookings
, SUM(c.cnt_bookings) AS tot_bookings
, SUM(IF(c.seg_id=q.seg_id,c.cnt_bookings,0))
* 100.0 / SUM(c.cnt_bookings) AS pct_bookings
, SUM(IF(c.seg_id=q.seg_id,c.cnt_billable,0)) AS cnt_billable
, SUM(c.cnt_billable) AS tot_billable
, SUM(IF(c.seg_id=q.seg_id,c.cnt_billable,0))
* 100.0 / SUM(c.cnt_billable) AS pct_billable
FROM ( SELECT t.seg_id
, t.seg_text
FROM ___Segmentations t
WHERE t.seg_hotelid = :hotel_id_1
ORDER BY t.seg_id
) q
LEFT
JOIN ( SELECT r.seg_id
, COUNT(DISTINCT b.boo_id) AS cnt_bookings
, COUNT(DISTINCT d.bil_id) AS cnt_billable
FROM ___Segmentations r
JOIN ___Bookings b
ON b.boo_segmentation = r.seg_id
JOIN `___BillableDatas` d
ON d.bil_bookingid = b.boo_id
AND d.bil_date BETWEEN '2017-02-21' AND '2017-02-28'
WHERE r.seg_hotelid = :hotel_id
GROUP
BY r.seg_id
) c
ON 1=1
GROUP
BY q.seg_id
, q.seg_text
ORDER
BY q.seg_text
关于mysql - 使用 mySQL 从多个表返回排名，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/42726428/