有没有办法在 SQL 中加入两个查询，每个查询都有一个 order by？

Posted 2023-03-21

【中文标题】有没有办法在 SQL 中加入两个查询，每个查询都有一个 order by？

【英文标题】：Is there a way to join two queries in SQL each with an order by?

【发布时间】：2021-08-19 05:44:44

【问题描述】：

我有两个查询从两个表中返回数据：

SELECT TOP 3 
    AE.id, AE.name, COUNT(R.id) 'number of reserves'
FROM 
    AIRPORT AE
INNER JOIN 
    FLY V ON V.id_destiny = AE.id
INNER JOIN 
    RESERVE R ON R.id_fly = V.id
GROUP BY 
    AE.id, AE.name
ORDER BY 
    COUNT(R.id) DESC;

返回数据示例：

id	name	number of reserves
6	name1	27
4	name2	18
14	name3	14

和

SELECT TOP 3 
    AE.id, AE.name, COUNT(R.id) 'number of reserves'
FROM 
    AEROPUERTO AE
LEFT JOIN 
    FLY V ON V.id_destiny = AE.id
LEFT JOIN 
    RESERVE R ON R.id_fly = V.id
GROUP BY 
    AE.id, AE.name
ORDER BY 
    COUNT(R.id) ASC;

第二个查询返回的数据示例：

id	name	number of reserves
7	name4	0
11	name5	0
12	name6	0

我需要将它们组合成一个输出，首先是第一个查询（以相同的顺序），然​​后是第二个查询，顺序相同，如下所示：

id	name	number of reserves
6	name1	27
4	name2	18
14	name3	14
7	name4	0
11	name5	0
12	name6	0

有办法吗？

编辑：我已经尝试过union all 选项，但我不能在每个查询中使用group by，因此返回的表与我需要的不同

(SELECT TOP 3 AE.id, AE.name, COUNT(R.id) 'number of reserves'
FROM AIRPORT AE
    INNER JOIN FLY V
ON V.id_destiny = AE.id
    INNER JOIN RESERVE R
ON R.id_fly = V.id
GROUP BY AE.id, AE.name)
UNION ALL
(SELECT TOP 3 AE.id, AE.name, COUNT(R.id) 'number of reserves'
FROM AEROPUERTO AE
    LEFT JOIN FLY V
ON V.id_destiny= AE.id
    LEFT JOIN RESERVE R
ON R.id_fly = V.id
GROUP BY AE.id, AE.name)
ORDER BY COUNT(R.id) ASC;

【问题讨论】：

使用UNION ALL

您需要：1) 将每个查询括在括号中，2) 使用 UNION ALL 将它们合并为一个，3) 添加 ORDER BY 以指定合并结果集的顺序。

已经尝试过了，但没有成功，因为如果 group by 存在，它会发送错误。

union all 使用人工列标识源，然后按该列排序并计算计数

【参考方案1】：

当您将每个查询括在括号中时，它就像一个派生表。您将需要一个SELECT 子句来从派生表中进行选择。

SELECT *
FROM
(
    -- Your first query here
) AS Q1

UNION ALL

SELECT *
FROM
(
    -- Your second query here
) AS Q2

你也可以使用CTE来做

WITH 
Q1 AS
(
    -- Your first query here
),
Q2 AS
(
    -- Your second query here
)
SELECT *
FROM   Q1
UNION ALL
SELECT *
FROm   Q2

---

编辑：如果您还希望两个查询中的最终结果具有相同的顺序，请为最终查询添加另一列ORDER BY

WITH 
Q1 AS
(
    SELECT TOP 3 
           AE.id, AE.name, COUNT(R.id) 'number of reserves',
           Q  = 1,
           RN = ROW_NUMBER() OVER (ORDER BY COUNT(R.id) DESC)
    FROM 
           AIRPORT AE
    INNER JOIN 
           FLY V ON V.id_destiny = AE.id
    INNER JOIN 
           RESERVE R ON R.id_fly = V.id
    GROUP BY 
           AE.id, AE.name
    ORDER BY 
           COUNT(R.id) DESC
),
Q2 AS
(
    SELECT TOP 3 
           AE.id, AE.name, COUNT(R.id) 'number of reserves',
           Q  = 2,
           RN = ROW_NUMBER() OVER (ORDER BY COUNT(R.id) DESC)
    FROM 
           AEROPUERTO AE
    LEFT JOIN 
           FLY V ON V.id_destiny = AE.id
    LEFT JOIN 
           RESERVE R ON R.id_fly = V.id
    GROUP BY 
           AE.id, AE.name
    ORDER BY 
           COUNT(R.id) ASC
)
SELECT *
FROM   Q1
UNION ALL
SELECT *
FROM   Q2
ORDER BY Q, RN

【讨论】：

这并没有解释如何明确按照 OP 的规定对结果进行排序。无法保证union all 会产生预期的结果。

我认为 OP 只需要每个查询中的 ORDER BY。无论如何，我已经编辑了答案并使用另一个查询进行了更新。

【参考方案2】：

我打破了查询并制作了两个临时表。您也可以在 CTE（Commn Table 表达式）中执行此操作。

 SELECT TOP 3 AE.id, AE.name, COUNT(R.id) 'number of reserves'
 INTO #Temp_1
 FROM AIRPORT AE
    INNER JOIN FLY V
       ON V.id_destiny = AE.id
    INNER JOIN RESERVE R
      ON R.id_fly = V.id
  GROUP BY AE.id, AE.name


 SELECT TOP 3 AE.id, AE.name, COUNT(R.id) 'number of reserves'
 into #Temp_2
 FROM AEROPUERTO AE
    LEFT JOIN FLY V
      ON V.id_destiny= AE.id
    LEFT JOIN RESERVE R
      ON R.id_fly = V.id
  GROUP BY AE.id, AE.nam

  SELECT *
  FROM #Temp_1
  UNION ALL
 SELECT * 
  FROM #Temp_2

 drop table #Temp_1
  drop table #Temp_2