PostgreSQL的 - 如何获得一个行最小值

Question

我已经表（t_image）与此列

datacd | imagecode | indexdate 
---------------------------------- 
    A  | 1  | 20170213 
    A  | 2  | 20170213 
    A  | 3  | 20170214 
    B  | 4  | 20170201 
    B  | 5  | 20170202 

期望的结果是这个

datacd | imagecode | indexdate 
    ---------------------------------- 
     A | 1  | 20170213 
     B | 4  | 20170201 

在上表中，我想为每个具有最小索引日期的datacd检索1行

这是我的查询，但结果返回2行，用于datacd A

select * 
from (
    select datacd, min(indexdate) as indexdate 
    from t_image 
    group by datacd 
) as t1 inner join t_image as t2 on t2.datacd = t1.datacd and t2.indexdate = t1.indexdate; 

Answer 1

Postgres的专有distinct on()运营商通常为greatest-n-per-group查询最快的解决方案：

select distinct on (datacd) * 
from t_image 
order by datacd, indexdate; 

Answer 2

可以选择使用ROW_NUMBER()：

SELECT t.datacd, 
     t.imagecode, 
     t.indexdate 
FROM 
(
    SELECT datacd, imagecode, indexdate, 
      ROW_NUMBER() OVER (PARTITION BY datacd ORDER BY indexdate) rn 
    FROM t_image 
) t 
WHERE t.rn = 1