聚合来自 JSONB 数组的不同值并结合 SQL 分组依据

Question

我正在尝试在 SQL GROUP BY 语句中聚合来自 JSONB 数组的不同值：

一个dataset有很多cfiles而一个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", "SampB", "SampD"]}
  3 |          1 | {"Sample Names": ["SampE"]}
  4 |          2 | {"Sample Names": ["SampA", "SampF"]}
  5 |          2 | {"Sample Names": ["SampG"]}

此查询有效，returns 是我想要的正确结果，但一团糟。

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

有没有更好的方法来编写这个查询而不需要所有的字符串操作？

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"

SQL 用于创建演示

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

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

Answer 1

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不必要地复杂化：

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

聚合来自 JSONB 数组的不同值并结合 SQL 分组依据

Aggregating distinct values from JSONB arrays combined with SQL group by

postgresql

query-optimization

jsonb