获取 ID 和最大值 ORACLE SQL

Question

我目前有以下：

TABLE "QUARTO":

CREATE TABLE Quarto (
  Id                        number(2) NOT NULL, 
  LotacaoMaxima             number(1) NOT NULL, 
  TipoQuartoId              number(1) NOT NULL, 
  NumeroQuartoNumSequencial number(3) NOT NULL, 
  NumeroQuartoAndarId       varchar2(1) NOT NULL, 
  PRIMARY KEY (Id));

TABLE RESERVA:

CREATE TABLE Reserva (
  Id               number(3) NOT NULL, 
  ClienteNif       number(9) NOT NULL, 
  QuartoId         number(2) NOT NULL, 
  DataInicio       date NOT NULL, 
  DataFim          date NOT NULL, 
  NumPessoas       number(1) NOT NULL, 
  Estado           varchar2(15) NOT NULL, 
  DataCancelamento date, 
  PRIMARY KEY (Id));

我在两者中的一些数据是：

QUARTO:

| ID | LOTACAOMAXIMA | TIPOQUARTOID | NUMEROQUARTONUMSEQUENCIAL | NUMEROQUARTOANDARID |
| :--- | :--- | :--- | :--- | :--- |
| 1 | 1 | 1 | 1 | 1 |
| 2 | 1 | 1 | 2 | 1 |
| 3 | 1 | 1 | 3 | 1 |
| 4 | 1 | 1 | 4 | 1 |
| 5 | 1 | 1 | 5 | 1 |
| 6 | 1 | 1 | 6 | 1 |
| 7 | 1 | 1 | 7 | 1 |
| 8 | 1 | 1 | 8 | 1 |
| 9 | 1 | 1 | 9 | 1 |
| 10 | 1 | 1 | 10 | 1 |
| 11 | 2 | 2 | 11 | 1 |
| 12 | 2 | 2 | 12 | 1 |
| 13 | 2 | 2 | 13 | 1 |
| 14 | 2 | 2 | 14 | 1 |
| 15 | 2 | 2 | 15 | 1 |
| 16 | 2 | 2 | 16 | 1 |
| 17 | 2 | 2 | 17 | 1 |
| 18 | 2 | 2 | 18 | 1 |
| 19 | 2 | 2 | 19 | 1 |
| 20 | 2 | 2 | 20 | 1 |


RESERVA:

| ID | CLIENTENIF | QUARTOID | DATAINICIO | DATAFIM | NUMPESSOAS | ESTADO | DATACANCELAMENTO |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| 1 | 296837970 | 11 | 2020-06-01 00:00:00 | 2020-06-12 00:00:00 | 1 | Finalizada | NULL |
| 2 | 275784703 | 17 | 2020-06-13 00:00:00 | 2020-06-21 00:00:00 | 1 | Finalizada | NULL |
| 3 | 220347654 | 11 | 2020-07-07 00:00:00 | 2020-07-15 00:00:00 | 2 | Finalizada | NULL |
| 4 | 294772545 | 12 | 2020-08-01 00:00:00 | 2020-08-15 00:00:00 | 2 | Finalizada | NULL |
| 5 | 220347654 | 3 | 2020-01-01 00:00:00 | 2020-01-16 00:00:00 | 1 | Finalizada | NULL |

WITH CONTAGEM_QUARTO_POR_ID AS (SELECT q.ID, COUNT(r.QUARTOID) AS NUM_RESERVAS, q.TIPOQUARTOID, q.NUMEROQUARTOANDARID
                                FROM RESERVA r
                                         INNER JOIN QUARTO q on q.ID = r.QUARTOID
                                GROUP BY q.ID, q.TIPOQUARTOID, q.NUMEROQUARTOANDARID)
SELECT t.TIPOQUARTOID, t.NUMEROQUARTOANDARID, MAX(t.NUM_RESERVAS) AS MAX
FROM CONTAGEM_QUARTO_POR_ID t
GROUP BY t.TIPOQUARTOID, t.NUMEROQUARTOANDARID

输出如下：

| TIPOQUARTOID | NUMEROQUARTOANDARID | MAX |
| :----------- | :------------------ | :-- |
| 1            |                   2 |   2 |
| 2            |                   1 |   8 |
| 1            |                   1 |   1 |

除了我目前拥有的数据之外，我还想显示每一行的脚趾 ID，但是当我将 t.ID 添加到 SELECT 时，它会强制我将其添加到 GROUP BY 并且输出是这样的：

| TIPOQUARTOID | NUMEROQUARTOANDARID | MAX | ID |
| :----------- | :------------------ | :-- | :- |
| 2            | 1                   | 2   | 11 |
| 1            | 1                   | 1   | 1  |
| 1            | 1                   | 1   | 3  |
| 1            | 2                   | 2   | 21 |
| 2            | 1                   | 1   | 17 |
| 2            | 1                   | 1   | 12 |
| 2            | 1                   | 8   | 16 |

我只想获取最大值和与该 MAX 关联的 ID。

Answer 1

我不建议使用两个级别的聚合。只需使用窗口功能：

WITH CONTAGEM_QUARTO_POR_ID AS (
      SELECT q.ID, COUNT(*) AS NUM_RESERVAS, q.TIPOQUARTOID, q.NUMEROQUARTOANDARID,
             ROW_NUMBER() OVER (PARTITION BY q.TIPOQUARTOID, q.NUMEROQUARTOANDARID ORDER BY COUNT(*) DESC) as seqnum
      FROM RESERVA r INNER JOIN
           QUARTO q 
           ON q.ID = r.QUARTOID
      GROUP BY q.ID, q.TIPOQUARTOID, q.NUMEROQUARTOANDARID
     )
SELECT cq.*
FROM CONTAGEM_QUARTO_POR_ID cq
WHERE seqnum = 1;

我认为这比两个聚合的性能略好（但值得检查）。

这种方法的一个优点是它更灵活。如果您想要平局，只需将子查询中的ROW_NUMBER() 更改为RANK()。

也许更重要的是，ROW_NUMBER() 是 SQL 中的一个“惯用语”，用于每组返回一行（或特定数量的行）。学习如何使用它非常有价值。

Answer 2

您需要NUM_RESERVAS 和PARTITION BY TIPOQUARTOID, NUMEROQUARTOANDARID 列的MAX() OVER () 分析函数，以便按列表为分区内的这些列提供分组，例如

WITH CONTAGEM_QUARTO_POR_ID AS
(
  SELECT q.ID,
         COUNT(r.QUARTOID) AS NUM_RESERVAS,
         q.TIPOQUARTOID,
         q.NUMEROQUARTOANDARID
    FROM RESERVA r
    JOIN QUARTO q
      on q.ID = r.QUARTOID
   GROUP BY q.ID, q.TIPOQUARTOID, q.NUMEROQUARTOANDARID
), t AS
(
 SELECT t.TIPOQUARTOID, t.NUMEROQUARTOANDARID, t.NUM_RESERVAS, 
        MAX(t.NUM_RESERVAS) 
        OVER (PARTITION BY t.TIPOQUARTOID, t.NUMEROQUARTOANDARID) AS MAX, 
        t.ID
   FROM CONTAGEM_QUARTO_POR_ID t
)
SELECT TIPOQUARTOID, NUMEROQUARTOANDARID, NUM_RESERVAS, ID
  FROM t
 WHERE NUM_RESERVAS = MAX

或者更直接地使用HAVING子句

SELECT q.TIPOQUARTOID,
       q.NUMEROQUARTOANDARID,
       COUNT(r.QUARTOID) AS NUM_RESERVAS,
       q.ID
  FROM RESERVA r
  JOIN QUARTO q
    on q.ID = r.QUARTOID
 GROUP BY q.ID, q.TIPOQUARTOID, q.NUMEROQUARTOANDARID
 HAVING COUNT(r.QUARTOID) = q.TIPOQUARTOID

Answer 3

您可以在查询中使用KEEP 子句，而无需进行太多更改，如下所示：

WITH CONTAGEM_QUARTO_POR_ID AS 
  (SELECT q.ID, COUNT(r.QUARTOID) AS NUM_RESERVAS, q.TIPOQUARTOID, q.NUMEROQUARTOANDARID
     FROM RESERVA r
     INNER JOIN QUARTO q on q.ID = r.QUARTOID
   GROUP BY q.ID, q.TIPOQUARTOID, q.NUMEROQUARTOANDARID)
SELECT t.TIPOQUARTOID, t.NUMEROQUARTOANDARID, MAX(t.NUM_RESERVAS) AS MAX, 
       max(t.ID) keep(dense_rank first order by t.NUM_RESERVAS desc nulls last) as ID -- this
FROM CONTAGEM_QUARTO_POR_ID t
GROUP BY t.TIPOQUARTOID, t.NUMEROQUARTOANDARID