Question

我在桌子上有各种各样的地段。这些批次通常只出现一次，但可以重复。我希望获得每个产品代码（列）的最新信息，其中最早创建的日期（列）会带回与该批次相关的其他字段（到期日期，仓库等）。

到目前为止，我有以下查询：

SELECT DISTINCT lot_number, prod_code, date_received, expiry_date, quantity 
FROM scheme.stquem  
WHERE prod_code = '001'  
AND lot_number != ''     //removes blanks   
ORDER BY lot_number

如何通过选择收到最早日期的地段来删除重复项？

Answer 1

使用group by和join：

SELECT s.lot_number, s.prod_code, s.date_received, s.expiry_date, s.quantity 
FROM scheme.stquem s JOIN
     (SELECT lot_number, min(date_received) as mindr
      FROM scheme.stquem
      WHERE prod_code = '001' AND lot_number <> ''     //removes blanks   
      GROUP BY lot_number
     ) sl
     ON sl.lot_number = s.lot_number and sl.mindr = s.date_received 
WHERE prod_code = '001'
ORDER BY s.lot_number;

Answer 2

仅当没有其他行具有相同的prod_code但更早的日期时，才使用NOT EXISTS返回行：

SELECT lot_number, prod_code, date_received, expiry_date, quantity 
FROM scheme.stquem st1
WHERE prod_code = '001'  
AND lot_number != ''     //removes blanks   
AND NOT EXISTS (select 1 from scheme.stquem st2
                where st1.prod_code = st2.prod_code
                  and st2.date_received < st1.date_received)
ORDER BY lot_number

SQL Query返回列值的实例但返回其他字段

2 个答案: