单个字段的子查询创建多个记录答案

【问题标题】：Subqueries for individual fields create multiple records单个字段的子查询创建多个记录
【发布时间】：2023-03-21 01:32:01
【问题描述】：

我的场景：一个请求可以有多个（最多 3 个）拒绝代码。第一个是主要拒绝代码，次要（第二和第三）拒绝代码是可选的。要确定rejection_code 排名，有一个名为rejection_rowId 的字段。如果rejection_rowId = '1'，那么它是主要拒绝。如果rejection_rowId = '2'，则为第二次拒绝。如果是“3”，则为第三个。现在，我试图在一条记录中显示属于一个请求的所有拒绝代码值，而不是为每个拒绝代码存储 1 条记录。我遇到了麻烦，想知道是否有人可以解释为什么我的 SQL 不能正常工作。

我的SQL基本上是这样的；

SELECT DISTINCT rea.id
    ,(SELECT rejection_code 
      FROM rejection t1
      INNER JOIN reason t2 ON t1.rowId = t2.rowId
      WHERE t1.rejection_rowId = '1'
        AND t1.rowId = REA.rowId) AS rejection1

    ,(SELECT rejection_code 
      FROM rejection t1
      INNER JOIN reason t2 ON t1.rowId = t2.rowId
      WHERE t1.rejection_rowId = '2'
        AND t1.rowId = REA.rowId) AS rejection2

    ,(SELECT rejection_code 
      FROM rejection t1
      INNER JOIN reason t2 ON t1.rowId = t2.rowId
      WHERE t1.rejection_rowId = '3'
        AND t1.rowId = REA.rowId) AS rejection3

FROM reason rea
INNER JOIN rejection rej ON rea.rowId = rej.rowId
WHERE rea.id = '12345';

上述SQL生成的输出是……

id    | rejection1 | rejection2 | rejection3
12345 | 26         | NULL       | NULL
12345 | NULL       | 51         | NULL
12345 | NULL       | NULL       | 3

我正在尝试（但失败）生成的输出是...

id    | rejection1 | rejection2 | rejection3
12345 | 26         | 51         | 3

如果有人可以帮助我识别并解决问题，我将不胜感激。谢谢！

【问题讨论】：

标签： sql subquery teradata

【解决方案1】：

我认为你正在尝试做这样的事情：

SELECT
  rea.id,
  MAX(CASE WHEN rej.rejection_rowId = '1' THEN rej.rejection_code END) AS rejection1,
  MAX(CASE WHEN rej.rejection_rowId = '2' THEN rej.rejection_code END) AS rejection2,
  MAX(CASE WHEN rej.rejection_rowId = '3' THEN rej.rejection_code END) AS rejection3,
FROM
  reason rea
  INNER JOIN rejection rej ON rea.rowId = rej.rowId
WHERE
  rea.id = '12345'
GROUP BY rea.id

没有必要一次又一次地在同一个表之间连接，使用子查询来实现你所需要的根本没有必要，而且成本很高。

使用GROUP BY 子句可以将结果限制为一行。使用聚合函数中的条件，您仅在实际存在拒绝时才在列中显示拒绝，并且您可以将它们拆分为单独的列。

【讨论】：

太棒了，谢谢！这成功了。我之前尝试过使用case语句，但我没有想到使用聚合函数和group by语句。

【解决方案2】：

试试这个：

求和：

SELECT DISTINCT rea.id
    ,SUM(SELECT rejection_code 
      FROM rejection t1
      INNER JOIN reason t2 ON t1.rowId = t2.rowId
      WHERE t1.rejection_rowId = '1'
        AND t1.rowId = REA.rowId) AS rejection1

    ,SUM(SELECT rejection_code 
      FROM rejection t1
      INNER JOIN reason t2 ON t1.rowId = t2.rowId
      WHERE t1.rejection_rowId = '2'
        AND t1.rowId = REA.rowId) AS rejection2

    ,SUM(SELECT rejection_code 
      FROM rejection t1
      INNER JOIN reason t2 ON t1.rowId = t2.rowId
      WHERE t1.rejection_rowId = '3'
        AND t1.rowId = REA.rowId) AS rejection3

FROM reason rea
INNER JOIN rejection rej ON rea.rowId = rej.rowId
WHERE rea.id = '12345'
GROUP BY rea.id ;

最大：

SELECT DISTINCT rea.id
    ,MAX(SELECT rejection_code 
      FROM rejection t1
      INNER JOIN reason t2 ON t1.rowId = t2.rowId
      WHERE t1.rejection_rowId = '1'
        AND t1.rowId = REA.rowId) AS rejection1

    ,MAX(SELECT rejection_code 
      FROM rejection t1
      INNER JOIN reason t2 ON t1.rowId = t2.rowId
      WHERE t1.rejection_rowId = '2'
        AND t1.rowId = REA.rowId) AS rejection2

    ,MAX(SELECT rejection_code 
      FROM rejection t1
      INNER JOIN reason t2 ON t1.rowId = t2.rowId
      WHERE t1.rejection_rowId = '3'
        AND t1.rowId = REA.rowId) AS rejection3

FROM reason rea
INNER JOIN rejection rej ON rea.rowId = rej.rowId
WHERE rea.id = '12345'
GROUP BY rea.id ;

【讨论】：