mysql - MySQLの使用方法とカウントおよびランク付け

Question

私はMS-SQL開発者ですが、このクエリ（MySQL）を使用しています↓</ p>

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

結果は次のとおりです。

これ欲しい：

Answer 1

次のように考えてみてください：

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

クエリ全体:

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

Total_CNT に同じ値が 2 つある場合はどうなるでしょうか。

多分このようなもの：

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
...

Answer 2

更新しました

RANK() OVER (ORDER BY TOTAL_CNT DESC DESC) AS ランク

ここで私は別の非常に良い解決策を得ました.:

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