Postgres - 将行数据作为 JSON 数组或数组返回

Question

在 PG v9.4 及更高版本上，我想将一些数据（基于 SELECT 语句）导出为 JSON 数组数组。

标准 json_agg 返回我想要的，但它返回对象数组（其中对象键是列名） 例如这个查询：

SELECT json_agg(data_rows)
FROM (
        -- in memory table sample
    SELECT * FROM 
    (VALUES
        ('John',now(),1940,'Winston','Lennon'),
        ('Paul',now(),1942,'','McCartney'),
        ('George',now(),1943,NULL,'Harrison'),
        ('Ringo',now(),1940,'my passions are ring,drum and shades','Starr')
    ) AS x("FirstName", "CurrentDt", "BirthYear", "MiddleName", "LastName")
        ORDER BY "BirthYear" DESC, "FirstName" DESC
) AS data_rows

返回以下内容：

[
 {"FirstName":"George","CurrentDt":"2016-09-12T13:13:07.862485-04:00","BirthYear":1943,"MiddleName":null,"LastName":"Harrison"}, 
 {"FirstName":"Paul","CurrentDt":"2016-09-12T13:13:07.862485-04:00","BirthYear":1942,"MiddleName":"","LastName":"McCartney"}, 
 {"FirstName":"Ringo","CurrentDt":"2016-09-12T13:13:07.862485-04:00","BirthYear":1940,"MiddleName":"my passions are ring,drum and shades","LastName":"Starr"}, 
 {"FirstName":"John","CurrentDt":"2016-09-12T13:13:07.862485-04:00","BirthYear":1940,"MiddleName":"Winston","LastName":"Lennon"}
]

但我想要的是：

[
 ["George","2016-09-12T13:13:07.862485-04:00",1943,null,"Harrison"}, 
 ["Paul","2016-09-12T13:13:07.862485-04:00",1942,"","McCartney"}, 
 ["Ringo","2016-09-12T13:13:07.862485-04:00",1940,"my passions are ring,drum and shades","Starr"}, 
 ["John","2016-09-12T13:13:07.862485-04:00",1940,"Winston","Lennon"}
]

我曾尝试使用技巧mentioned here 首先将行转换为 hstore，但这样做的问题是未保留列顺序.... 所以这个查询：

SELECT json_agg(avals(hstore(data_rows)))
FROM (
        -- in memory table sample
    SELECT * FROM 
    (VALUES
        ('John',  now(),1940,'Winston','Lennon'),
        ('Paul',  now(),1942,'','McCartney'),
        ('George',now(),1943,NULL,'Harrison'),
        ('Ringo', now(),1940,'my passions are ring,drum and shades','Starr')
    ) AS x("FirstName", "CurrentDt", "BirthYear", "MiddleName", "LastName")
        ORDER BY "BirthYear" DESC, "FirstName" DESC
) AS data_rows
-- placing order by 'outside' did not make any difference
--ORDER BY "BirthYear" DESC, "FirstName" DESC

返回以下（列顺序错误）

[
 ["Harrison","1943","2016-09-12 14:07:06.772227-04","George",null], 
 ["McCartney","1942","2016-09-12 14:07:06.772227-04","Paul",""], 
 ["Starr","1940","2016-09-12 14:07:06.772227-04","Ringo","my passions are ring,drum and shades"], 
 ["Lennon","1940","2016-09-12 14:07:06.772227-04","John","Winston"]
]

有谁知道如何以 JSON 数组形式获取结果？

Answer 1

你可能想要这样的东西：

SELECT json_agg(info)
FROM (
    SELECT json_build_array("LastName","BirthYear","CurrentDt","FirstName","MiddleName") AS info
    FROM 
        (VALUES
            ('John',now(),1940,'Winston','Lennon'),
            ('Paul',now(),1942,'','McCartney'),
            ('George',now(),1943,NULL,'Harrison'),
            ('Ringo',now(),1940,'my passions are ring,drum and shades','Starr')
        ) AS x("FirstName", "CurrentDt", "BirthYear", "MiddleName", "LastName")
    ORDER BY "BirthYear" DESC, "FirstName" DESC
) as t;

我使用json_build_array 将每个人的所有值放入一个json 数组中，然后在外部查询中，我使用json_agg 将所有这些数组收集到一个数组中。

您还可以将ORDER BY 子句移动到聚合函数内以获得以下内容：

SELECT json_agg(json_build_array("LastName","BirthYear","CurrentDt","FirstName","MiddleName") ORDER BY "BirthYear" DESC, "FirstName" DESC)
    FROM 
        (VALUES
            ('John',now(),1940,'Winston','Lennon'),
            ('Paul',now(),1942,'','McCartney'),
            ('George',now(),1943,NULL,'Harrison'),
            ('Ringo',now(),1940,'my passions are ring,drum and shades','Starr')
        ) AS x("FirstName", "CurrentDt", "BirthYear", "MiddleName", "LastName");