array_agg group by and null

Question

给定此表:

SELECT * FROM CommodityPricing order by dateField
"SILVER";60.45;"2002-01-01"
"GOLD";130.45;"2002-01-01"
"COPPER";96.45;"2002-01-01"
"SILVER";70.45;"2003-01-01"
"GOLD";140.45;"2003-01-01"
"COPPER";99.45;"2003-01-01"
"GOLD";150.45;"2004-01-01"
"MERCURY";60;"2004-01-01"
"SILVER";80.45;"2004-01-01"

2004年，铜被丢弃，汞被引入。
如何将(array_agg(value order by date desc) ) [1]的值作为COPPER的NULL ?

select commodity,(array_agg(value order by date desc) ) --[1]
from CommodityPricing
group by commodity
"COPPER";"{99.45,96.45}"
"GOLD";"{150.45,140.45,130.45}"
"MERCURY";"{60}"
"SILVER";"{80.45,70.45,60.45}"

Answer 1

SQL Fiddle

select
    commodity,
    array_agg(
        case when commodity = 'COPPER' then null else price end
        order by date desc
    )
from CommodityPricing
group by commodity
;

Answer 2

To "pad"如果在结果数组中缺少具有空值的行，则在完整的行和 LEFT JOIN 上构建查询，以获取网格的实际值。

给定这个表定义:

CREATE TABLE price (
  commodity text
, value     numeric
, ts        timestamp  -- using ts instead of the inappropriate name date 
);

使用generate_series()获取代表年份的时间戳列表，使用CROSS JOIN获取所有商品的唯一列表(SELECT DISTINCT ...)。

SELECT commodity, array_agg(value ORDER BY ts DESC) AS years
FROM   generate_series (timestamp '2002-01-01'
                      , timestamp '2004-01-01'
                      , '1 year') t(ts)
CROSS  JOIN (SELECT DISTINCT commodity FROM price) c(commodity)
LEFT   JOIN price p USING (ts, commodity)
GROUP  BY commodity;

结果:

<表类> 商品金道明> {150.45,140.45,130.45}{60,零,零} 银 {80.45,70.45,60.45}tbody>