Postgresql中的json类型列的GROUP BY内容

问题描述 投票:1回答:2

我有一个Postgresql表,其中一个名为“food”的json列。

以下是一些行的示例:

food
["cheese", "salmon", "eggs"]
["salmon", "cheese", "eggs"]
["broccoli", "ham", "milk"]
["salmon", "cheese", "eggs", "pizza"]

目前的结果:

food                                       count
["cheese", "salmon", "eggs"]              | 1
["salmon", "cheese", "eggs"]              | 1
["broccoli", "ham", "milk"]               | 1
["salmon", "cheese", "eggs", "pizza"]     | 1

期望的结果:

food                                       count
["cheese", "salmon", "eggs"]              | 2
["broccoli", "ham", "milk"]               | 1
["salmon", "cheese", "eggs", "pizza"]     | 1

有没有办法在不考虑元素顺序的情况下对一个json字段的内容进行GROUP BY?如果两行具有相同的确切内容,那么我希望它们组合在一起。

我的计划是GROUP BY json_array_elements(食物),但由于某种原因,这只返回每行的第一个元素。

json postgresql
2个回答
1
投票

实际上类似于@Scoots的答案,但没有排序,窗口,aso:

SELECT (
    SELECT jsonb_agg(items order by items)
    FROM jsonb_array_elements(food) AS items
    ) AS food,
    count(*)
FROM test_json_grouping
GROUP BY 1;

...解释:

                                              QUERY PLAN                                              
------------------------------------------------------------------------------------------------------
 HashAggregate  (cost=1635.60..1890.60 rows=200 width=40)
   Group Key: (SubPlan 1)
   ->  Seq Scan on test_json_grouping  (cost=0.00..1629.25 rows=1270 width=32)
         SubPlan 1
           ->  Aggregate  (cost=1.25..1.26 rows=1 width=32)
                 ->  Function Scan on jsonb_array_elements items  (cost=0.00..1.00 rows=100 width=32)
(6 rows)

结果:

                 food                  | count 
---------------------------------------+-------
 ["cheese", "eggs", "salmon"]          |     2
 ["broccoli", "ham", "milk"]           |     1
 ["cheese", "eggs", "pizza", "salmon"] |     1
(3 rows)

0
投票

不直接 - 对于Postgres他们是不同的字符串。

然而,我们可以做的是通过json_array_elements解压缩这些json字符串,然后使用json_agg应用我们自己的排序重新打包它们。然后通过工作将它们同化为你的小组。

这是一个查询,说明我的意思:

select
    __food.food::text,
    count(1)
from (
    select
        json_agg(_unpack.food order by _unpack.food::text asc) as food
    from (
        select
            row_number() over(),
            json_array_elements(food) as food
        from
            YOUR_SCHEMA.YOUR_FOOD_TABLE
    ) as _unpack
    group by
        _unpack.row_number
) as __food
group by
    __food.food::text
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