PostgreSQL不同不起作用 [英] postgresql distinct not working

问题描述

我正在使用以下代码从数据库中获取价值:

I am using the following code for getting value from the database:

但是当我编写这段代码(测试以查看问题出在哪里..)时，我注意到查询没有从数据库中获取不同的值:这是查询

选择"distinct(ca.id)"作为"id"，"acc.name"作为"accName"，"pIsu.name"作为"pareentIsu"，"sIsu.name"作为"subIsu"，dv.domain_value作为contractType，acc.id作为accId，dvct.domain_value作为contractstatus
来自contracts_account的ca大约在ca.parent_isu_fk上加入business_unit pIsu = pIsu.id
在ca.sub_isu_fk上左加入business_unit sIsu = sIsu.id
在ca.account_fk上左加入business_unit acc = acc.id
con.contracts_account_fk = ca.id
上的左加盟合约condv.id = con.contract_type_fk上的左连接domain_values dv
dvct.id = con.contract_status_fk上的左联接domain_values dvct
其中ca.id不为null，并且按(ca.id)以(4466079)顺序con.contract_type_fk

But when i wrote this code (testing as to see where the problem is..) i noticed the query is not fetching the distinct value out from the database: Here is the query

select distinct(ca.id)as id, acc.name as accName,pIsu.name as pareentIsu, sIsu.name as subIsu , dv.domain_value as contractType,acc.id as accId,dvct.domain_value as contractstatus
from contracts_account ca left join business_unit pIsu on ca.parent_isu_fk = pIsu.id
left join business_unit sIsu on ca.sub_isu_fk = sIsu.id
left join business_unit acc on ca.account_fk = acc.id
left join contracts con on con.contracts_account_fk = ca.id
left join domain_values dv on dv.id = con.contract_type_fk
left join domain_values dvct on dvct.id = con.contract_status_fk
where ca.id is not null and con.contract_type_fk in ( 4466079 ) order by ca.id

此查询只是不返回我'不同的ID '
我究竟做错了什么?我正在使用postgres 8.2

This query is simply not returning me 'Distinct Id'
What am i doing wrong? I am using postgres 8.2

推荐答案

您的查询被解释为:

select distinct (ca.id) as id, acc.name as accName, . . .

这是标准的 distinct 语句.如果您只希望每个 ca.id 包含第一行的值(基于 order by )，则可以使用 distinct on :

This is a standard distinct statement. If you want just one row per ca.id with values from the first row (based on the order by), then use distinct on:

select distinct on (ca.id) ca.id, acc.name as accName, . . .

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