2

鉴于此表：

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"

自 2004 年起，铜被淘汰，汞被引入。
我怎样才能得到(array_agg(value order by date desc) ) [1] as NULLfor的价值COPPER？

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}"
4

2 回答 2

0

要在结果数组中用 NULL 值“填充”缺失的行，请在完整的行网格和网格的LEFT JOIN实际值上构建查询。
鉴于此表定义：

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

我generate_series()用来获取代表年份的时间戳CROSS JOIN列表和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 ('2002-01-01 00:00:00'::timestamp
                      , '2004-01-01 00:00:00'::timestamp
                      , '1y') t(ts)
CROSS  JOIN (SELECT DISTINCT commodity FROM price) c(commodity)
LEFT   JOIN price p USING (ts, commodity)
GROUP  BY commodity;

结果：

COPPER  {NULL,99.45,96.45}
GOLD    {150.45,140.45,130.45}
MERCURY {60,NULL,NULL}
SILVER  {80.45,70.45,60.45}

SQL小提琴。
我将数组转换为小提琴中的文本，因为显示器很糟糕，否则会吞下 NULL 值。

于 2014-03-18T13:28:32.910 回答
0

SQL小提琴

select
    commodity,
    array_agg(
        case when commodity = 'COPPER' then null else price end
        order by date desc
    )
from CommodityPricing
group by commodity
;
于 2014-03-17T10:26:13.803 回答