Question

如何从单个Postgres查询创建多维数组。

我有3个表格，其中包含以下列：

tb_school（id，school_name）例如：

[ {id:"1", school_name:"School1"}, {id:"2", school_name:"School2"} ]
tb_profile（id，profile_name，school_id）例如：

[ {id:"1", profile_name:"John", school_id:"1"}, {id:"2", profile_name:"Peter", school_id:"1"}, {id:"3", profile_name:"Sam", school_id:"1"}, {id:"4", profile_name:"Susan", school_id:"2"}, {id:"5", profile_name:"Jude", school_id:"2"}, {id:"6", profile_name:"Kim", school_id:"2"} ]
tb_article（id，article_name，profile_id）例如：

[ {id:"1", article_name:"Headline News", profile_id:"1"}, {id:"2", article_name:"Sports Recap", profile_id:"2"}, {id:"3", article_name:"Weather", profile_id:"3"}, {id:"4", article_name:"Arts", profile_id:"4"}, {id:"5", article_name:"Other", profile_id:"5"}, {id:"6", article_name:"Example", profile_id:"6"} ]

我希望查询返回结果lat看起来像这样：

[
    {school_name:"School1", people:[
        {profile_name:"John", articles:[
            article_name:"Headline News"
        ]},
        {profile_name:"Peter", articles:[
            article_name:"Sports Recap"
        ]},
        {profile_name:"Sam", articles:[
            article_name:"Weather"
        ]},
    ]},
    {school_name:"School2", people:[
        {profile_name:"Susan", articles:[
            article_name:"Arts"
        ]},
        {profile_name:"Jude", articles:[
            article_name:"Other"
        ]},
        {profile_name:"Kim", articles:[
            article_name:"Example"
        ]},
    ]}
]

我知道可以通过嵌套for循环中的3个单独的选择查询来实现这一点。我想知道的是，这可以通过一个postgres查询来实现吗？

Answer 1

注意：我对PostgreSQL不太满意。可能做出错误的假设和代码可能不适合生产。

我几乎可以肯定你不能只得到一个嵌套对象。但是你可以收到JSON。所以

SELECT 
	json_build_object(
	  'school_name', school_name,
	  'people', (
		  SELECT 
		  	json_agg(json_build_object(
		  		'profile_name', profile_name,
		  		'articles', (
		  			SELECT 
		  				json_agg(json_build_object(
		  					'article_name', article_name
		  				))
		  			FROM tb_article
		  			WHERE profile_id = tb_profile.facid
		  		) 
		  	))
		  FROM tb_profile
		  WHERE school_id = tb_school.id
		 )
	) AS result
FROM tb_school

它会为您提供JSON行，您可以JSON.parse。

如果您想测试，可以使用this learning service。我用这段代码。

SELECT 
	json_build_object(
	  'name', firstname,
	  'slots', (
		  SELECT 
		  	json_agg(json_build_object(
		  		'number', slots,
		  		'facility', (
		  			SELECT 
		  				json_agg(json_build_object(
		  					'name', name
		  				))
		  			FROM cd.facilities
		  			WHERE facid = cd.bookings.facid
		  		) 
		  	))
		  FROM cd.bookings
		  WHERE memid = cd.members.memid
		 )
	) AS result
FROM cd.members

它可能会有所帮助或提供更多想法。

Postgres结果从查询多维

1 个答案: