5

在一个简单的库存管理数据库中，添加新库存的数量并发货，直到数量达到零。每个库存移动都分配了一个参考，仅使用最新的参考。

在提供的示例中，从不显示最新的引用，股票 ID 的 1,4 应该分别有引用 charlie、foxtrot，而是显示 alpha、delta。

如何将多个条件上的 GROUP BY 和 LEFT JOIN 关联起来以显示最新记录？

http://sqlfiddle.com/#!2/6bf37/107

CREATE TABLE stock (
  id tinyint PRIMARY KEY,
  quantity int,
  parent_id tinyint
);

CREATE TABLE stock_reference (
  id tinyint PRIMARY KEY,
  stock_id tinyint,
  stock_reference_type_id tinyint,
  reference varchar(50)
);

CREATE TABLE stock_reference_type (
  id tinyint PRIMARY KEY,
  name varchar(50)
);

INSERT INTO stock VALUES 
(1, 10, 1),
(2, -5, 1),
(3, -5, 1),
(4, 20, 4),
(5, -10, 4),
(6, -5, 4);

INSERT INTO stock_reference VALUES 
(1, 1, 1, 'Alpha'),
(2, 2, 1, 'Beta'),
(3, 3, 1, 'Charlie'),
(4, 4, 1, 'Delta'),
(5, 5, 1, 'Echo'),
(6, 6, 1, 'Foxtrot');

INSERT INTO stock_reference_type VALUES 
(1, 'Customer Reference');

SELECT stock.id, SUM(stock.quantity) as quantity, customer.reference
FROM stock
LEFT JOIN stock_reference AS customer ON stock.id = customer.stock_id AND stock_reference_type_id = 1
GROUP BY stock.parent_id
4

1 回答 1

1

您可以使用子查询来提取每个股票组的最新 ID：

SELECT g.parent_id, g.quantity, customer.reference
FROM (
    SELECT parent_id, SUM(stock.quantity) as quantity, MAX(id) as LatestID
    FROM stock
    GROUP BY parent_id
) g LEFT JOIN stock_reference AS custome
    ON g.LatestID = customer.stock_id AND stock_reference_type_id = 1
于 2012-09-10T16:02:41.287 回答