按日期最后注册进行分组

Question

各位下午好...

我有以下查询，返回以下结果：

SELECT 
    r.rec_num as Column_A
    ,c.amount as Column_B
    ,c.status as Column_C
    ,c.currency_code_to as Column_D
    ,c.due_date as Column_E
    ,c.id_recurrency as Column_F
FROM purchase c
JOIN recurrency r ON c.id_recurrency = r.id
JOIN item_product ip ON c.item_estoque = ip.id
JOIN signature a ON a.id = ip.signature
WHERE a.id = :id
    AND r.plano_signature IS NOT NULL
    AND r.rec_num IN (1,2,3,4)
    AND r.rec_num NOT IN (
        SELECT rn.recurrence
        FROM negotiate n
        JOIN negotiate_recurrence rn ON rn.negotiate = n.id
        JOIN purchase purchase ON purchase.id = n.purchase
        WHERE n.subscription = a.id
            AND rn.recurrence = r.rec_num
            AND purchase.status IN (
                'VALID'
                ,'FILED'
                )
        )
    AND r.rec_num NOT IN (
        SELECT r2.rec_num
        FROM purchase c2
        JOIN recurrency r2 ON c2.id_recurrency = r2.id
        JOIN item_product ip2 ON c2.item_estoque = ip2.id
        WHERE ip2.signature = a.id
            AND c2.status IN ('VALID','FILED')
        )
    AND c.id NOT IN (
        SELECT n1.purchase
        FROM negotiate n1
        WHERE n1.subscription = a.id
        )
    AND c.status = 'DELAYED'

A 栏	B 栏	C 栏	D 栏	E 栏	F 栏
2	1000	延迟	巴西利亚	2023-10-04T19:48:17.000Z	11102501
2	250	延迟	巴西利亚	2023-10-04T19:54:16.000Z	11102502
2	250	延迟	巴西利亚	2023-10-09T23:30:20.000Z	11102503
2	250	延迟	巴西利亚	2023-10-14T18:14:20.000Z	12042504
3	250	延迟	巴西利亚	2023-11-04T12:51:33.000Z	13842790

我需要它按日期和重复次数返回最后的结果：

A 栏	B 栏	C 栏	D 栏	E 栏	F 栏
2	250	延迟	巴西利亚	2023-10-14T18:14:20.000Z	12042504
3	250	延迟	巴西利亚	2023-11-04T12:51:33.000Z	13842790

我尝试在 DUE_DATE 上设置 MAX 并按 rec_num 分组，但结果如下：

SELECT 
    r.rec_num as Column_A
    ,c.amount as Column_B
    ,c.status as Column_C
    ,c.currency_code_to as Column_D
    ,max(c.due_date) as Column_E
    ,c.id_recurrency as Column_F
FROM purchase c
JOIN recurrency r ON c.id_recurrency = r.id
JOIN item_product ip ON c.item_estoque = ip.id
JOIN signature a ON a.id = ip.signature
WHERE a.id = :id
    AND r.plano_signature IS NOT NULL
    AND r.rec_num IN (1,2,3,4)
    AND r.rec_num NOT IN (
        SELECT rn.recurrence
        FROM negotiate n
        JOIN negotiate_recurrence rn ON rn.negotiate = n.id
        JOIN purchase purchase ON purchase.id = n.purchase
        WHERE n.subscription = a.id
            AND rn.recurrence = r.rec_num
            AND purchase.status IN (
                'VALID'
                ,'FILED'
                )
        )
    AND r.rec_num NOT IN (
        SELECT r2.rec_num
        FROM purchase c2
        JOIN recurrency r2 ON c2.id_recurrency = r2.id
        JOIN item_product ip2 ON c2.item_estoque = ip2.id
        WHERE ip2.signature = a.id
            AND c2.status IN ('VALID','FILED')
        )
    AND c.id NOT IN (
        SELECT n1.purchase
        FROM negotiate n1
        WHERE n1.subscription = a.id
        )
GROUP BY r.rec_num
    

A 栏	B 栏	C 栏	D 栏	E 栏	F 栏
2	1000	延迟	巴西利亚	2023-10-14T18:14:20.000Z	11102501
3	250	延迟	巴西利亚	2023-11-04T12:51:33.000Z	13842790

有人可以帮助我吗？

Answer 1

无需更改当前查询即可执行的一种快速解决方案是将每个

rec_num

及其

MAX(c.due_date)

分别提取为派生表，并将

JOIN

与当前查询一起提取。例如：

SELECT 
    r.rec_num as Column_A
    ,c.amount as Column_B
    ,c.status as Column_C
    ,c.currency_code_to as Column_D
    ,c.due_date as Column_E
    ,c.id_recurrency as Column_F
FROM purchase c
JOIN recurrency r ON c.id_recurrency = r.id
JOIN item_product ip ON c.item_estoque = ip.id
JOIN signature a ON a.id = ip.signature
/*add this*/
JOIN (SELECT rec_num, MAX(due_date) AS dd FROM purchase
        JOIN recurrency ON id_recurrency = id GROUP BY rec_num) dt
  ON r.rec_num = dt.rec_num AND c.due_date = dt.dd
... /*the rest of the query*/

并从查询中删除

GROUP BY r.rec_num

。