将来自 JSONB 数组的不同值与 SQL group 组合在一起

如何解决将来自 JSONB 数组的不同值与 SQL group 组合在一起

我试图在 SQL GROUP BY 语句中聚合来自 JSONB 数组的不同值：

一个dataset有多个cfiles，一个cfile只有一个dataset

SELECT * FROM cfiles;
 id | dataset_id |                property_values (jsonb)                
----+------------+-----------------------------------------------
  1 |          1 | {"Sample Names": ["SampA","SampB","SampC"]}
  2 |          1 | {"Sample Names": ["SampA","SampD"]}
  3 |          1 | {"Sample Names": ["SampE"]}
  4 |          2 | {"Sample Names": ["SampA","SampF"]}
  5 |          2 | {"Sample Names": ["SampG"]}

此查询有效并返回正确的我想要的结果，但它很混乱。

SELECT distinct(datasets.id) as dataset_id,ARRAY_TO_STRING(
  ARRAY(
    SELECT DISTINCT * FROM unnest(
      STRING_TO_ARRAY(
        STRING_AGG(
          DISTINCT REPLACE(
            REPLACE(
              REPLACE(
                REPLACE(
                  cfiles.property_values ->> 'Sample Names','",' || chr(32) || '"',';'
                ),'[',''
              ),'"',''
            ),']',''
          ),';'
        ),';'
      )
    ) ORDER BY 1 ASC
  ),'; '
) as sample_names
FROM datasets
JOIN cfiles ON cfiles.dataset_id=datasets.id
GROUP BY datasets.id

 dataset_id |           sample_names            
------------+-----------------------------------
          1 | SampA; SampB; SampC; SampD; SampE
          2 | SampA; SampF; SampG

是否有更好的方法来编写此查询而不进行所有字符串操作？

我累了 jsonb_array_elements 但它给了我错误子查询使用来自外部查询的未分组列“cfiles.property_values”。然后我将 cfiles.property_values 添加到 GROUP BY 但它不再仅按 dataset_id 分组

不是我想要的结果：

SELECT DISTINCT datasets.id as dataset_id,ARRAY_TO_STRING(
  ARRAY(
    SELECT DISTINCT * FROM jsonb_array_elements(
      cfiles.property_values -> 'Sample Names'
    ) ORDER BY 1 ASC
  ),'; '
) as sample_names
FROM datasets
JOIN cfiles ON cfiles.dataset_id=datasets.id
GROUP BY datasets.id,cfiles.property_values

 dataset_id |       sample_names        
------------+---------------------------
          1 | "SampA"; "SampB"; "SampC"
          1 | "SampA"; "SampB"; "SampD"
          1 | "SampE"
          2 | "SampA"; "SampF"
          2 | "SampG"

用于创建演示的 SQL

CREATE TABLE datasets (
  id INT PRIMARY KEY
);

CREATE TABLE cfiles (
  id INT PRIMARY KEY,dataset_id INT,property_values JSONB,FOREIGN KEY (dataset_id) REFERENCES datasets(id)
);

INSERT INTO datasets values (1),(2);

INSERT INTO cfiles values
  (1,1,'{"Sample Names":["SampA","SampC"]}'),(2,"SampD"]}'),(3,'{"Sample Names":["SampE"]}');

INSERT INTO cfiles values 
  (4,2,"SampF"]}'),(5,'{"Sample Names":["SampG"]}');

解决方法

jsonb_array_elements 是一个集合返回函数，应该在 FROM 子句中使用。在 SELECT 列表中使用它会使事情变得不必要地复杂：

select c.dataset_id,string_agg(distinct n.name,'; ' order by n.name)
from cfiles c
  cross join jsonb_array_elements_text(c.property_values -> 'Sample Names') as n(name)
group by c.dataset_id
order by c.dataset_id;  

Online example