Как использовать MySQL и считать и ранжировать

Я разработчик MS-SQL, теперь я использую этот запрос (MySQL) ↓

SELECT A.place_idx,A.place_id,B.TODAY_CNT,C.TOTAL_CNT FROM CUSTOM_LIST 

AS A

INNER JOIN
(SELECT place_id,COUNT(place_id) AS TODAY_CNT from COUNT_TABLE where DATE(place_date) = DATE(NOW()) GROUP BY place_id)
AS B ON B.place_id=A.place_id

INNER JOIN
(SELECT place_id,COUNT(place_id) AS TOTAL_CNT from COUNT_TABLE GROUP BY place_id)
AS C ON C.place_id=A.place_id

Результат:

Я хочу это:

Ответ 1

Попробуйте что-нибудь вроде этого:

SELECT ..., C.TOTAL_CNT, (@r := @r + 1) AS rank FROM CUSTOM_LIST, (SELECT  @r := 0) t
...
ORDER BY C.TOTAL_CNT DESC

Весь запрос:

SELECT A.place_idx,A.place_id,B.TODAY_CNT,C.TOTAL_CNT, (@r := @r + 1) AS rank 
FROM CUSTOM_LIST AS A, (SELECT  @r := 0) t

INNER JOIN
(SELECT place_id,COUNT(place_id) AS TODAY_CNT from COUNT_TABLE where DATE(place_date) = DATE(NOW()) GROUP BY place_id)
AS B ON B.place_id=A.place_id

INNER JOIN
(SELECT place_id,COUNT(place_id) AS TOTAL_CNT from COUNT_TABLE GROUP BY place_id)
AS C ON C.place_id=A.place_id

ORDER BY C.TOTAL_CNT DESC

Что делать, если мы получили два одинаковых значения в Total_CNT?

Возможно, что-то вроде этого:

SELECT ..., (@last := C.TOTAL_CNT) AS TOTAL_CNT, 
  IF(@last = C.TOTAL_CNT, @r, @r := @r + 1) AS rank
FROM CUSTOM_LIST, (SELECT  @r := 0, @last := -1) t
...

Ответ 2

Обновление

RANK() OVER (ORDER BY TOTAL_CNT DESC DESC) AS Rank

Здесь я получил еще одно очень хорошее решение.

SELECT A.place_idx,A.place_id,B.TODAY_CNT,C.TOTAL_CNT, RANK() OVER (ORDER BY TOTAL_CNT DESC) AS Rank FROM CUSTOM_LIST 

AS A

INNER JOIN
(SELECT place_id,COUNT(place_id) AS TODAY_CNT from COUNT_TABLE where DATE(place_date) = DATE(NOW()) GROUP BY place_id)
AS B ON B.place_id=A.place_id

INNER JOIN
(SELECT place_id,COUNT(place_id) AS TOTAL_CNT from COUNT_TABLE GROUP BY place_id)
AS C ON C.place_id=A.place_id