Question

我有一个名为LEDGER的表。它有三列名为PATID，CODE和TRANDATE。

示例数据：

PATID  CODE   TRANDATE
1      Z1110  07-15-1970 
1      D3330  07-15-1970
1      Z0330  07-15-1970
2      Z1298  07-20-1970
2      Z0987  07-20-1970
2      Z0330  07-20-1970
2      D1092  07-20-1970
3      Z0330  08-20-1980
3      Z1110  02-11-1982

我需要一个返回PATID的查询，如果他们有CODE Z0330但Z1110 并且两个代码都出现在同一个TRANDATE。如果存在这种情况，我只想要一个PATID返回。在上面的示例中，输出应为：

PATID  CODE   TRANDATE
2      Z0330  07-20-1970
3      Z0330  07-20-1980

如何编写查询来执行此操作？

Answer 1

您可以使用having子句聚合获得所需内容：

select patid, 'Z0330' as code, transdate
from example
group by patid, transdate
having sum(case when code = 'Z0330' then 1 else 0 end) > 0 and
       sum(case when code = 'Z1110' then 1 else 0 end) = 0;

Answer 2

试试这个，先选择代码为'z1110'的trandates，然后选择prev查询中没有transadate的行。

SELECT * FROM LEDGER WHERE TRANDATE NOT IN(
    SELECT TRANDATE FROM LEDGER WHERE CODE = 'Z1110'
)
AND CODE = 'Z3330'

Answer 3

这个怎么样？

SELECT DISTINCT PATID, TRANDATE
FROM LEDGER L1
where not exists (select 1 from LEDGER L2
              where L1.TRANDATE = L2.TRANDATE
                and L2.CODE = 'Z1110' );

SQL FIDDLE DEMO

Answer 4

这对你有用。如果您想要每个月的旧日，请将订单更改为ASC。目前，查询检索每个月的最后一天。

此处提供了示例SQL FIDDLE DEMO

;WITH cte_Result AS 
(
    SELECT   ROW_NUMBER() OVER(
                 PARTITION BY CODE,
                 CONVERT(VARCHAR(7), TRANDATE, 120) ORDER BY TRANDATE DESC
             ) AS ID,
             PATID,
             CODE,
             TRANDATE
    FROM     LEDGER
    WHERE    CODE = 'Z0330'
)
SELECT   PATID,
         CODE,
         TRANDATE
FROM     cte_Result
WHERE    ID = 1

SQL - 单个表中多行的结果

4 个答案: