加入的設置返回函數（json_array_elements）結果與表列

Question

比方說，我有兩個表：

User_Combination

+--------+----------------+ 
| id | combination | 
+--------+----------------+ 
| 6 |  [1, 2]  | 
| 9 |  [2, 3]  | 
+--------+----------------+ 

顏色

+--------+----------------+ 
| id | color  | 
+--------+----------------+ 
| 1 |  Blue  | 
| 2 |  Yellow | 
| 3 |  Green | 
+--------+----------------+ 

我想加盟結果json_array_elements(color)與元素的id。例如，

select json_array_elements(color) as CombinationID 
from User_Combination where id = 6; 

結果是

+-------------------+ 
| CombinationID | 
+-------------------+ 
| 1    | 
| 2    | 
+-------------------+ 

我無法加入CombinationID與Colors.id。當我嘗試SQL命令，例如：

select json_array_elements(article_data) AS articlesInOutfits (color) as CombinationID 
from User_Combination uc JOIN Colors co ON co.id = articlesInOutfits; 

或

select json_array_elements(article_data) AS articlesInOutfits (color) as CombinationID 
from User_Combination uc JOIN Colors co ON co.id = uc.articlesInOutfits; 

它說articlesInOutfits不存在。
有什麼建議嗎？

Answer 1

使用unnest()得到解壓組合：

select id, unnest(combination) cid 
from user_combination; 

id | cid 
----+----- 
    6 | 1 
    6 | 2 
    9 | 2 
    9 | 3 
(4 rows)  

使用解壓cids加入與colors：

select u.id, color 
from (
    select id, unnest(combination) cid 
    from user_combination 
    ) u 
join colors c 
on cid = c.id; 

id | color 
----+-------- 
    6 | Blue 
    6 | Yellow 
    9 | Yellow 
    9 | Green 
(4 rows) 

使用聚集函數（如json_agg()）即可加入顏色彙總的對於用戶：

​​

---

如果combination的類型是json你應該使用在橫向json_array_elements()的加入：

select u.id, json_agg(color) 
from (
    select id, cid 
    from user_combination, 
    lateral json_array_elements(combination) cid 
    ) u 
join colors c 
on cid::text::int = c.id 
group by 1;