我正在使用Postgres 9.5,并且具有以下表格:
用户
图片
发布
帖子的内容如下:
[
{ "type": "text", "text": "learning pg" },
{ "type": "image", "image_id": "8f4422b4-3936-49f5-ab02-50aea5e6755f" },
{ "type": "image", "image_id": "57efc97c-b9b4-4cd5-b1e1-3539f5853835" },
{ "type": "text", "text": "pg is awesome" }
]
现在,我想加入内容的图像类型,并使用
image_id
填充它们,例如:{
"id": "cb1267ca-b1ac-4daa-8c7e-72d4c000e9fa",
"title": "Learning join jsonb in Postgres",
"author_id": "deba01b7-ec58-4cc2-b3ae-7dc42e582767",
"content": [
{ "type": "text", "text": "learning pg" },
{
"type": "image",
"image": {
"id": "8f4422b4-3936-49f5-ab02-50aea5e6755f",
"key": "/upload/test1.jpg",
"width": 800,
"height": 600
}
},
{
"type": "image",
"image": {
"id": "57efc97c-b9b4-4cd5-b1e1-3539f5853835",
"key": "/upload/test2.jpg",
"width": 1280,
"height": 720
}
},
{ "type": "text", "text": "pg is awesome" }
]
}
这是我的测试sql文件:
CREATE EXTENSION IF NOT EXISTS "uuid-ossp";
DROP TABLE IF EXISTS Users;
DROP TABLE IF EXISTS Images;
DROP TABLE IF EXISTS Posts;
CREATE TABLE Users (
id UUID PRIMARY KEY DEFAULT uuid_generate_v4(),
name text NOT NULL
);
CREATE TABLE Images (
id UUID PRIMARY KEY DEFAULT uuid_generate_v4(),
key TEXT,
width INTEGER,
height INTEGER,
creator_id UUID
);
CREATE TABLE Posts (
id UUID PRIMARY KEY DEFAULT uuid_generate_v4(),
title TEXT,
author_id UUID,
content JSONB
);
DO $$
DECLARE user_id UUID;
DECLARE image1_id UUID;
DECLARE image2_id UUID;
BEGIN
INSERT INTO Users (name) VALUES ('test user') RETURNING id INTO user_id;
INSERT INTO Images (key, width, height, creator_id) VALUES ('upload/test1.jpg', 800, 600, user_id) RETURNING id INTO image1_id;
INSERT INTO Images (key, width, height, creator_id) VALUES ('upload/test2.jpg', 600, 400, user_id) RETURNING id INTO image2_id;
INSERT INTO Posts (title, author_id, content) VALUES (
'test post',
user_id,
('[ { "type": "text", "text": "learning pg" }, { "type": "image", "image_id": "' || image1_id || '" }, { "type": "image", "image_id": "' || image2_id || '" }, { "type": "text", "text": "pg is awesome" } ]') :: JSONB
);
END $$;
有什么办法可以实现这一要求?
最佳答案
假设至少为Postgres 9.5,它将完成以下工作:
SELECT jsonb_pretty(to_jsonb(p)) AS post_row_as_json
FROM (
SELECT id, title, author_id, c.content
FROM posts p
LEFT JOIN LATERAL (
SELECT jsonb_agg(
CASE WHEN c.elem->>'type' = 'image' AND i.id IS NOT NULL
THEN elem - 'image_id' || jsonb_build_object('image', i)
ELSE c.elem END) AS content
FROM jsonb_array_elements(p.content) AS c(elem)
LEFT JOIN images i ON c.elem->>'type' = 'image'
AND i.id = (elem->>'image_id')::uuid
) c ON true
) p;
如何?
jsonb
数组的嵌套,每个数组元素产生1行:jsonb_array_elements(p.content) AS c(elem)
LEFT JOIN
转换为images
一种。键“类型”的值为“图像”:c.elem->>'type' = 'image'
b。 image_id
中的UUID匹配:i.id = (elem->>'image_id')::uuid
请注意,content
中的无效UUID会引发异常。 c.elem->>'type' = 'image' AND i.id IS NOT NULL
删除键'image_id'并将相关图像行添加为
jsonb
值:elem - 'image_id' || jsonb_build_object('image', i)
否则保留原始元素。
content
将修改后的元素重新聚合到新的jsonb_agg()
列中。也可以与普通的ARRAY constructor一起使用。
LEFT JOIN LATERAL
转换为posts
并选择所有列,仅将p.content
替换为生成的替换c.content
SELECT
中,使用简单的jsonb
将整个行转换为to_jsonb()
。jsonb_pretty()
对于人类可读的表示形式是完全可选的。 All
jsonb
functions are documented in the manual here.关于sql - 如何在Postgres中加入jsonb数组元素?,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/40413861/