格式化Join Query的输出

Question

我有一个连接查询，它给出了以下输出。

SELECT DISTINCT LOC.STATE, ORD.* FROM 
( 
    SELECT COUNT (*) ORDERS, 
    ORDV.PERSON_ID PERSON_ID,
    PERSON.FIRST_NAME FN,
    PERSON.LAST_NAME LN,
    SOURCE 
    FROM MY_ORD_V ORDV, EIM_PERSON PERSON
    WHERE ORD_STATUS_ID NOT IN (A, C, D, F)
         AND PERSON.PERSON_ID = ORDV.PERSON_ID
         GROUP BY ORDV.PERSON_ID,
         SOURCE,
         PERSON.FIRST_NAME,
         PERSON.LAST_NAME
    ORDER BY 1 ASC
) ORD , MY_LOCATION_V LOC
    WHERE ORD.PERSON_ID = LOC.OBJ_ID ORDER BY LOC.STATE, ORD.FN, ORD.LN

当前输出和预期输出：

Answer 1

从Oracle 11g开始，您现在拥有LISTAGG analytic function。这可能是您正在寻找的（我用于测试的演示SQL）：

SELECT DISTINCT state, LISTAGG(TO_CHAR(orders)||src, '|') WITHIN GROUP (ORDER BY fn, ln) OVER (PARTITION BY state, fn, ln) as orders
     , fn
     , ln
FROM (
SELECT 'NY' as state, 2 as orders, 'GREG' as fn, 'BOOTINE' as ln, 'DBS' as src from dual
union all
SELECT 'NY' as state, 3 as orders, 'GREG' as fn, 'BOOTINE' as ln, 'PST' as src from dual
union all
SELECT 'MA' as state, 2 as orders, 'ANN' as fn, 'SILVEST' as ln, 'DBS' as src from dual
union all
SELECT 'MA' as state, 2 as orders, 'ANN' as fn, 'SILVEST' as ln, 'PST' as src from dual
)

使用你的SQL会像：

SELECT DISTINCT state
     , LISTAGG(TO_CHAR(orders)||src, '|') WITHIN GROUP (ORDER BY fn, ln) OVER (PARTITION BY state, fn, ln) as orders
     , fn
     , ln
FROM ( 
    SELECT COUNT (*) ORDERS, 
    ORDV.PERSON_ID PERSON_ID,
    PERSON.FIRST_NAME FN,
    PERSON.LAST_NAME LN,
    SOURCE 
    FROM MY_ORD_V ORDV, EIM_PERSON PERSON
    WHERE ORD_STATUS_ID NOT IN (A, C, D, F)
         AND PERSON.PERSON_ID = ORDV.PERSON_ID
         GROUP BY ORDV.PERSON_ID,
         SOURCE,
         PERSON.FIRST_NAME,
         PERSON.LAST_NAME
    ORDER BY 1 ASC
) ORD , MY_LOCATION_V LOC
    WHERE ORD.PERSON_ID = LOC.OBJ_ID 
  ORDER BY LOC.STATE, ORD.FN, ORD.LN

Answer 2

这是一个替代版本，只是一个显示想法的伪查询。您可以在oracle中与||连接。像这样你将始终拥有两个来源，当计数为0时。如果源需要是动态的，你仍然可以通过获取子查询中的所有不同来源（然后使用group和{{}来实现这一点。 1}}正如@Patrick Marchand所说。）

listagg