Question

我希望在SQL Server 2014中运行SQL，这将使我获得表中多于一行的项目（INV_ITEM_ID），和具有相同的生效日期（ EFFDT），如下例所示。

SETID   INV_ITEM_ID VENDOR_SETID    VENDOR_ID   UNIT_OF_MEASURE   EFFDT
-----------------------------------------------------------------------------------------
SHARE   1           SHARE           ERSUT       BX                2018-07-16 00:00:00.000
SHARE   1           SHARE           ERSUT       BX                2018-09-01 00:00:00.000
SHARE   10          SHARE           ERSUT       RL                1901-01-01 00:00:00.000
SHARE   100         SHARE           JJHS        CS                1901-01-01 00:00:00.000
SHARE   100         SHARE           PSMED       CS                1901-01-01 00:00:00.000

上述输出中的项目100将是一个很好的例子，该项目多次出现并且两行具有相同的EFFDT（1901-01-01）。

这是我尝试过的SQL，但是没有给我预期的输出：

SELECT 
    SETID, INV_ITEM_ID, VENDOR_SETID, VENDOR_ID, UNIT_OF_MEASURE, EFFDT
FROM 
    PS_ITM_VNDR_UOM_PR A
WHERE 
    EXISTS (SELECT 1 FROM PS_ITM_VNDR_UOM_PR B
            WHERE B.INV_ITEM_ID = A.INV_ITEM_ID AND B.EFFDT = A.EFFDT)
GROUP BY 
    SETID, INV_ITEM_ID, VENDOR_SETID, VENDOR_ID, UNIT_OF_MEASURE, EFFDT
HAVING 
    COUNT(INV_ITEM_ID) > 1 AND COUNT(EFFDT) > 1
ORDER BY 
    INV_ITEM_ID

Answer 1

SELECT
    * 
FROM 
    TABLENAME a
WHERE EXISTS
(SELECT
    COUNT(1) cnt
    ,inv_item_id
    ,effdt
FROM
    TABLENAME b
WHERE b.inv_item_id = a.inv_item_id
GROUP BY inv_item_id,effdt
HAVING COUNT(1) > 1)

Answer 2

以下查询应会得到您想要的：

with x as (
select inv_item_id, effdt
  from ps_itm_vndr_uom_pr
  group by inv_item_id, effdt
  having count(*) > 1 
)
select inv_item_id
  from x
  group by inv_item_id
  having count(*) = 1

Answer 3

您要选择INV_ITEM_ID和EFFDT> 1的行数的所有行。使用COUNT OVER可以轻松解决：

select setid, inv_item_id, vendor_setid, vendor_id, unit_of_measure, effdt
from
(
  select *, count(*) over (partition by inv_item_id, effdt) as cnt
  from ps_itm_vndr_uom_pr
) t
where cnt > 1
order by inv_item_id, effdt;

获取具有多于1行且具有相同生效日期的项目

3 个答案: