SQL合并行而不是UNION ALL 查询结果

编程问答 2022-06-11

问题描述

我正在做UNION ALL来获得结果，如下表所示。这种方法导致不必要的行。 DESK，SEGMENT和SUPERVISOR这三列是独立的，没有关系。

代码 

SELECT ID,DESK,'' as SEGMENT,'' as SUPERVISOR FROM myTable1 
UNION ALL 
SELECT ID,'' AS DESK,SEGMENT,'' as SUPERVISOR FROM myTable2 
UNION ALL 
SELECT ID,SUPERVISOR FROM myTable3

结果： 

+------+------------+---------+------------+
| ID   | DESK       | SEGMENT | SUPERVISOR | TOTAL ENTRIES
+------+------------+---------+------------+
| 4782 | OIL & GAS  |         |            |  23
+------+------------+---------+------------+
| 4782 | AUTOMOTIVE |         |            |  23
+------+------------+---------+------------+
| 4782 |            | GLOBAL  |            |  23
+------+------------+---------+------------+
| 4782 |            |         | DANIEL     |  23
+------+------------+---------+------------+
| 4782 |            |         | JAMES      |  23
+------+------------+---------+------------+

如何查询以获得以下结果？

预期结果： 

+------+------------+---------+------------+
| ID   | DESK       | SEGMENT | SUPERVISOR | TOTAL ENTRIES
+------+------------+---------+------------+
| 4782 | OIL & GAS  | GLOBAL  | DANIEL     |  23
+------+------------+---------+------------+
| 4782 | AUTOMOTIVE |         | JAMES      |  23
+------+------------+---------+------------+

解决方法

对于这三个表，可以将ROW_NUMBER()分析函数与ID一起按FULL OUTER JOIN列进行分区：

SELECT NVL(NVL(t2.ID,t3.ID),t1.ID) AS ID,desk,segment,supervisor 
  FROM ( SELECT t1.*,ROW_NUMBER() OVER (PARTITION BY ID ORDER BY 0) AS rn FROM myTable1 t1 ) t1
  FULL JOIN ( SELECT t2.*,ROW_NUMBER() OVER (PARTITION BY ID ORDER BY 0) AS rn FROM myTable2 t2 ) t2
    ON t2.ID = t1.ID AND t2.rn = t1.rn
  FULL JOIN ( SELECT t3.*,ROW_NUMBER() OVER (PARTITION BY ID ORDER BY 0) AS rn FROM myTable3 t3 ) t3
    ON t3.ID = t1.ID AND t3.rn = t1.rn;


ID   DESK        SEGMENT  SUPERVISOR
---- ----------  -------  ----------
4782 AUTOMOTIVE  GLOBAL   JAMES
4782 OIL & GAS            DANIEL

Demo

P.S：我离开了ORDER BY 0，因为ORDER BY必须使用ROW_NUMBER()选项，您可以用合适的列或标识符替换零。

,

您可以尝试以下方法：

SELECT table1.ID,table1.DESK,table2.SEGMENT,(select SUPERVISOR from (select SUPERVISOR,ROWNUM AS RN FROM table3) WHERE RN = 1) SUPERVISOR
FROM table1 JOIN table2 on table1.ID = table2.ID
WHERE table1.DESK = 'OIL & GAS'
UNION ALL
SELECT table1.ID,null SEGMENT,ROWNUM AS RN FROM table3) WHERE RN = 2) SUPERVISOR
FROM table1 JOIN table2 on table1.ID = table2.ID
WHERE table1.DESK = 'AUTOMOTIVE'

enter image description here

https://dbfiddle.uk/?rdbms=oracle_11.2&fiddle=c5594bb1d99579611d2669f6bab675a2

,

您可以尝试以下查询。我不知道23的来源，因此我没有将其计入查询中，但是如果它是三个表之一中的一列，则可以使用类似的逻辑将其添加到结果中。

查询

WITH
    table1 (id,desk)
    AS
        (SELECT 4782,'OIL & GAS' FROM DUAL
         UNION ALL
         SELECT 4782,'AUTOMOTIVE' FROM DUAL),table2 (id,segment) AS (SELECT 4782,'GLOBAL' FROM DUAL),table3 (id,supervisor)
    AS
        (SELECT 4782,'DANIEL' FROM DUAL
         UNION ALL
         SELECT 4782,'JAMES' FROM DUAL)
SELECT *
  FROM (SELECT t1.id,CASE WHEN t1.desk = LAG (t1.desk) OVER (ORDER BY t1.desk) THEN NULL ELSE t1.desk END
                   AS desk,CASE
                   WHEN t2.segment = LAG (t2.segment) OVER (ORDER BY t2.segment) THEN NULL
                   ELSE t2.segment
               END
                   AS segment,CASE
                   WHEN t3.supervisor = LAG (t3.supervisor) OVER (ORDER BY t3.supervisor) THEN NULL
                   ELSE t3.supervisor
               END
                   AS supervisor
          FROM table1 t1,table2 t2,table3 t3
         WHERE t1.id = t2.id AND t1.id = t3.id)
 WHERE desk IS NOT NULL OR segment IS NOT NULL OR supervisor IS NOT NULL;

结果

     ID          DESK    SEGMENT    SUPERVISOR
_______ _____________ __________ _____________
   4782 AUTOMOTIVE    GLOBAL     DANIEL
   4782 OIL & GAS                JAMES
oracleoraclesqlsqlunion-all