选择1到n的关系“错误的方式”

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英文:

Select 1 to n relationship "the wrong way arround"

问题

你想要的结果似乎是从多个表中组合数据的嵌套结构。在SQLite中,你可以使用子查询和连接来实现这一点。以下是可能的SQL查询示例,以获取你所需的结果:

SELECT o.orderId AS id, o.address,
       json_group_array(
           json_object(
               'productid', p.productid,
               'price', p.price,
               'amount', op.amount
           )
       ) AS items
FROM orders AS o
JOIN orderedProducts AS op ON o.orderId = op.orderId
JOIN products AS p ON op.productid = p.productid
GROUP BY o.orderId, o.address;

这个查询将从 ordersorderedProductsproducts 表中联接数据,并按订单分组。每个订单将包含一个 idaddress 和一个包含订购的产品的嵌套JSON数组 items。这应该与你所描述的所需结果非常接近。

英文:

products:

productid price
0 10
1 20

orders:

orderId address
0 lala
1 lala

orderedProducts:

productid orderId amount
0 0 1
1 0 2
1 1 2

I want to get:

orders[
   {id, address, items[{productid , price, amount}, {productid , price, amount}]},
   {id, address, items[{productid , price, amount}]}
]

Is that possible with a single query? I am using SQLite with Python.

答案1

得分: 0

你可以使用 GROUP_CONCAT,类似这样:

SELECT   
  o.orderId,   
  o.address,  
  ''['' || GROUP_CONCAT('{"id": ' || p.productid || ', "price": ' || p.price || ', "amount": ' || op.amount || '}', ', ') || '']'' as items  
FROM orders o  
JOIN orderedProducts op ON o.orderId = op.orderId  
JOIN products p ON op.productid = p.productid  
GROUP BY o.orderId, o.address;  

这里有一个示例 SQL Fiddle

英文:

You can use GROUP_CONCAT, something like this:

SELECT   
  o.orderId,   
  o.address,  
  '[' || GROUP_CONCAT('{"id": ' || p.productid || ', "price": ' || p.price || ', "amount": ' || op.amount || '}', ', ') || ']' as items  
FROM orders o  
JOIN orderedProducts op ON o.orderId = op.orderId  
JOIN products p ON op.productid = p.productid  
GROUP BY o.orderId, o.address;  

Here's an example SQL Fiddle

答案2

得分: 0

您可以使用 json_group_arrayjson_object。有关用法的文档在 此处

您的查询可以如下所示:

SELECT
  o.id AS order_id,
  o.address,
  json_group_array(
    json_object(
      'productid', p.id,
      'price', p.price,
      'amount', op.amount
    )
  ) AS items
FROM
  orders o
LEFT JOIN
  orderedProducts op ON o.id = op.orderId
LEFT JOIN
  products p ON op.productid = p.id
GROUP BY
  o.id, o.address;

由于使用了 json_group_array,您需要使用 GROUP BY o.id, o.address

要创建一些演示查询,可以使用这个 网站

模拟数据创建:

CREATE TABLE products (
  id SERIAL PRIMARY KEY,
  price INTEGER
);

CREATE TABLE orders (
  id SERIAL PRIMARY KEY,
  address VARCHAR(100)
);

CREATE TABLE orderedProducts (
  productid INTEGER REFERENCES products(id),
  orderId INTEGER REFERENCES orders(id),
  amount INTEGER
);

INSERT INTO products (id, price) VALUES (0, 10), (1, 20);
INSERT INTO orders (id, address) VALUES (0, 'lala'), (1, 'lala');
INSERT INTO orderedProducts (productid, orderId, amount) VALUES (0, 0, 1), (1, 0, 2), (1, 1, 2);
英文:

You can use json_group_array and json_object. Documentation for usage is here

Your query could be this:

SELECT
  o.id AS order_id,
  o.address,
  json_group_array(
    json_object(
      'productid', p.id,
      'price', p.price,
      'amount', op.amount
    )
  ) AS items
FROM
  orders o
LEFT JOIN
  orderedProducts op ON o.id = op.orderId
LEFT JOIN
  products p ON op.productid = p.id
GROUP BY
  o.id, o.address;

You have to use GROUP BY o.id, o.address because of the json_group_array

To make some demo queries feel free to use this site.

Mock data creation:

CREATE TABLE products (
  id SERIAL PRIMARY KEY,
  price INTEGER
);

CREATE TABLE orders (
  id SERIAL PRIMARY KEY,
  address VARCHAR(100)
);

CREATE TABLE orderedProducts (
  productid INTEGER REFERENCES products(id),
  orderId INTEGER REFERENCES orders(id),
  amount INTEGER
);

INSERT INTO products (id, price) VALUES (0, 10), (1, 20);
INSERT INTO orders (id, address) VALUES (0, 'lala'), (1, 'lala');
INSERT INTO orderedProducts (productid, orderId, amount) VALUES (0, 0, 1), (1, 0, 2), (1, 1, 2);

huangapple
  • 本文由 发表于 2023年6月29日 18:45:27
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