这是我的表“AuctionDetails”
以下选择:
select string_agg("AuctionNO",',' ) as "AuctionNO"
,sum("QuntityInAuction" ) as "QuntityInAuction"
,"AmmanatPattiID"
,"EntryPassDetailsId"
,"BrokerID"
,"TraderID"
,"IsSold"
,"IsActive"
,"IsExit"
,"IsNew"
,"CreationDate"
from "AuctionDetails"
group by "AmmanatPattiID"
,"EntryPassDetailsId"
,"TraderID"
,"IsSold"
,"IsActive"
,"IsExit"
,"IsNew"
,"BrokerID"
,"CreationDate"
给我这个结果:
但我需要这样的记录
AuctionNo QunatityInAuction AmmanatpattiID EntryPassDetailID BrokerID Trader ID IsSold ISActive ISExit IsNew CreationDate
AU8797897,AU8797886,AU596220196F37379 1050 -1 228,229 42 42 f t f t 2013-10-10
最后我需要交易者和经纪人的最新条目,在我们的例子中是“42”,数量总和,以及拍卖编号的串联......
最佳答案
Postgres wiki 描述了如何定义您自己的 FIRST 和 LAST 聚合函数。例如:
-- Create a function that always returns the last non-NULL item
CREATE OR REPLACE FUNCTION public.last_agg ( anyelement, anyelement )
RETURNS anyelement LANGUAGE SQL IMMUTABLE STRICT AS $$
SELECT $2;
$$;
-- And then wrap an aggregate around it
CREATE AGGREGATE public.LAST (
sfunc = public.last_agg,
basetype = anyelement,
stype = anyelement
);
页面在这里:https://wiki.postgresql.org/wiki/First/last_(aggregate)
关于sql - 在 PostgreSQL 中使用 group by 时如何获取最后一条记录,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/19289189/