Oracle SQL 到 return 来自唯一行的列值

Question

我有一个 table 具有多个子批次的单独批次：

+--------+----------+
¦Batch   +Sub-batch ¦
¦--------¦----------¦
¦B_01    ¦SB_01     ¦
¦B_02    ¦SB_02     ¦
¦B_02    ¦SB_03     ¦
¦B_03    ¦SB_04     ¦
+--------+----------+

其中B_和SB_只是批次的名称（标识符），不需要聚合。

我愿意return:

+--------+-----------+-----------+
¦Batch   +Sub_Batch1 ¦Sub_Batch2 ¦
¦--------¦-----------¦-----------¦
¦B_01    ¦SB_01      ¦           ¦
¦B_02    ¦SB_02      ¦SB_03      ¦
¦B_03    ¦SB_04      ¦           ¦
+--------+-----------+-----------+

看起来 PARTITION BY 或 PIVOT 是可行的方法，但我似乎只能找到聚合函数。

感谢收到任何帮助！

Answer 1

可以先应用row_number()函数，再使用conditional aggregation :

with t as
(
select t0.*,
       row_number() over (partition by Batch order by Sub_batch) as rn
  from t0
)
select Batch,
       max(case when rn = 1 then Sub_batch end) as Sub_Batch1, 
       max(case when rn = 2 then Sub_batch end) as Sub_Batch2
  from t
 group by Batch
 order by Batch;

Demo

Answer 2

如果只有 2 个 sub_batches 那么您可以简单地使用 MIN 和 MAX 聚合函数，如下所示：

SQL> WITH YOUR_TABLE(Batch, Sub_batch) AS
  2  (SELECT 'B_01', 'SB_01' FROM DUAL UNION ALL
  3  SELECT 'B_02', 'SB_02' FROM DUAL UNION ALL
  4  SELECT 'B_02', 'SB_03' FROM DUAL UNION ALL
  5  SELECT 'B_03', 'SB_04' FROM DUAL)
  6  SELECT
  7      BATCH,
  8      MIN(SUB_BATCH) AS SUB_BATCH1,
  9      CASE
 10          WHEN MIN(SUB_BATCH) <> MAX(SUB_BATCH) THEN MAX(SUB_BATCH)
 11      END AS SUB_BATCH2
 12  FROM YOUR_TABLE
 13  GROUP BY BATCH
 14  ORDER BY BATCH;

BATCH      SUB_BATCH1 SUB_BATCH2
---------- ---------- ----------
B_01       SB_01
B_02       SB_02      SB_03
B_03       SB_04

SQL>

干杯！！

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