下面是我拥有的两个表:两个表中的所有列都是"text“类型,表名和列名都是粗体字体。
名字
--------------------------------
Name | DoB | Team |
--------------------------------
Harry | 3/12/85 | England
Kevin | 8/07/86 | England
James | 5/05/89 | England 分值
------------------------
ScoreName | Score
------------------------
James-1 | 120
Harry-1 | 30
Harry-2 | 40
James-2 | 56 我需要的最终结果是一个具有以下内容的表
NameScores
---------------------------------------------
Name | DoB | Team | ScoreData
---------------------------------------------
Harry | 3/12/85 | England | "{"ScoreName":"Harry-1", "Score":"30"}, {"ScoreName":"Harry-2", "Score":"40"}"
Kevin | 8/07/86 | England | null
James | 5/05/89 | England | "{"ScoreName":"James-1", "Score":"120"}, {"ScoreName":"James-2", "Score":"56"}"我需要使用一个SQL命令来完成这个任务,我将使用这个命令来创建一个物化视图。
我已经意识到它将涉及string_agg、JOIN和JSON的组合,但是还没有完全破解它。(请帮助:)
发布于 2016-01-17 18:31:25
我不认为join是棘手的。复杂的是构建JSON对象:
select n.name, n.dob, n.team,
json_agg(json_build_object('ScoreName', s.name,
'Score', s.score)) as ScoreData
from names n left join
scores s
ons.name like concat(s.name, '-', '%')
group by n.name, n.dob, n.team;注:json_build_object()是在Postgres 9.4中引入的。
编辑:
我认为您可以添加一个case语句来获得简单的NULL
(case when s.name is null then NULL
else json_agg(json_build_object('ScoreName', s.name,
'Score', s.score))
end) as ScoreData发布于 2016-01-17 18:34:18
使用json_agg()和row_to_json()将分数数据聚合为json值:
select n.*, json_agg(row_to_json(s)) "ScoreData"
from "Names" n
left join "Scores" s
on n."Name" = regexp_replace(s."ScoreName", '(.*)-.*', '\1')
group by 1, 2, 3;
Name | DoB | Team | ScoreData
-------+---------+---------+---------------------------------------------------------------------------
Harry | 3/12/85 | England | [{"ScoreName":"Harry-1","Score":30}, {"ScoreName":"Harry-2","Score":40}]
James | 5/05/89 | England | [{"ScoreName":"James-1","Score":120}, {"ScoreName":"James-2","Score":56}]
Kevin | 8/07/86 | England | [null]
(3 rows)https://stackoverflow.com/questions/34841885
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