我有表A,其中的JSON列f
具有以下内容:
[{"name": "abc", "id": "1"}, {"name": "abcd", "id": "2"}, {"name": "abcde", "id": "3"} ]
我想在上述JSON对象内的id
上加入另一个表B,但还要从JSON对象中获取属性name
。
我设法创建了以下查询:
WITH sample_data_array(arr) AS (
SELECT f FROM A
), sample_data_elements(elem) AS (
SELECT json_array_elements(arr) FROM sample_data_array
)
SELECT CAST(elem->>'id' AS int) AS id, elem->'name' AS name FROM sample_data_elements
返回以下结果:
id, name
1, "abc"
2, "abcd"
3, "abcde"
表B中的样本数据:
id, title, slug
1, "title 1", "title-1"
2, "title 2", "title-2"
3, "title 3", "title-3"
如何将结果与表B合并,并从表中添加更多数据(列)?
预期结果:
id, name, title, slug
1, "abc", "title 1", "title-1"
2, "abcd", "title 2", "title-2"
3, "abcde", "title 3", "title-3"
SELECT
b.id,
elems ->> 'name' as name, -- 3
b.title
FROM
a,
json_array_elements(f) as elems -- 1
JOIN
b ON b.id = (elems ->> 'id')::int -- 2
json_array_elements()
从json数组中取出元素id
属性为text
,并将其转换为int
值)b
和数组元素(如name
)中获取所有相关值