Question

DROP TABLE IF EXISTS `items_stock`;
CREATE TABLE IF NOT EXISTS `items_stock` (
  `ItemStockID` int(10) unsigned NOT NULL AUTO_INCREMENT,
  `ItemID` mediumint(8) unsigned NOT NULL,
  `Date` date NOT NULL,
  `CurrentStock` decimal(14,2) NOT NULL DEFAULT '0.00',
  `CreatedBy` int(5) NOT NULL,
  `CreatedDate` datetime NOT NULL,
  `ModifiedBy` int(5) DEFAULT NULL,
  `ModifiedDate` timestamp NOT NULL DEFAULT '0000-00-00 00:00:00' ON UPDATE CURRENT_TIMESTAMP,
  `Active` enum('1','0') NOT NULL DEFAULT '1',
  PRIMARY KEY (`ItemStockID`)
) ENGINE=InnoDB AUTO_INCREMENT=5 DEFAULT CHARSET=utf8;

INSERT INTO `items_stock` (`ItemStockID`, `ItemID`, `Date`, `CurrentStock`, `CreatedBy`, `CreatedDate`, `ModifiedBy`, `ModifiedDate`, `Active`) VALUES 

(1, 1, '2010-11-03', 500.00, 1, '2010-11-03 11:12:12', NULL, '2010-11-04 09:39:29', '1'), 
(2, 1, '2010-11-04', 400.00, 1, '2010-11-04 11:12:12', NULL, '2010-11-04 10:10:09', '1'), 
(3, 1, '2010-11-04', 700.00, 1, '2010-11-04 11:14:12', NULL, '0000-00-00 00:00:00', '1'), 
(4, 1, '2010-11-03', 600.00, 1, '2010-11-04 11:19:12', NULL, '2010-11-04 10:11:26', '1'),
(5, 2, '2010-11-05', 800.00, 1, '2010-11-05 11:19:12', NULL, '2010-11-05 10:11:26', '1');

现在我想得到今天（2010-11-05）所有项目的当前库存，例如ID为1当前库存的物品为700（ItemStockID为3）

我尝试了以下查询

SELECT MAX(DATE) AS DATE, ItemStockID, ItemID, CurrentStock FROM items_stock WHERE DATE <= '2010-11-05' AND Active = 1 GROUP BY ItemID;

+------------+-------------+--------+--------------+
| DATE       | ItemStockID | ItemID | CurrentStock |
+------------+-------------+--------+--------------+
| 2010-11-04 |           1 |      1 |       500.00 |
| 2010-11-05 |           5 |      2 |       800.00 |
+------------+-------------+--------+--------------+

但如果同一日期超过一次，那么它没有提供适当的当前股票

实际结果应该是这样的

+------------+-------------+--------+--------------+
| DATE       | ItemStockID | ItemID | CurrentStock |
+------------+-------------+--------+--------------+
| 2010-11-04 |           3 |      1 |       700.00 |
| 2010-11-05 |           5 |      2 |       800.00 |
+------------+-------------+--------+--------------+

如果我使用以下查询，则会给出以上结果

SELECT ISTI.ItemStockID, ISTI.DATE, ISTI.ItemID, ISTI.CurrentStock FROM items_stock AS ISTI
JOIN (SELECT MAX(ItemStockID) AS ItemStockID 
        FROM items_stock AS IST
            JOIN (SELECT MAX(DATE) AS DATE, ItemID FROM items_stock WHERE DATE <= '2010-11-05' AND Active = 1 GROUP BY ItemID) AS SD ON SD.Date = IST.Date AND IST.ItemID = SD.ItemID
        GROUP BY IST.ItemID
    ) AS ISTO ON ISTO.ItemStockID = ISTI.ItemStockID;

您能告诉我如何以优化的方式获得上述结果

Answer 1

好吧，我认为mysql查询的行为应该如此。你需要一个不同的查询。原始查询将返回组中与您的组匹配的第一行。但是，你想要相反。我想你想要与你的小组相对应的最后一行（最新日期）。此查询应该为您提供所需的结果。

select A.ItemId,A.DATE,A.CurrentStock,A.ItemStockID from (select ItemId,MAX(DATE) DATE from items_stock where DATE <= '2010-11-05' AND Active = 1 group by ItemId) I, items_stock A where A.ItemStockID=( select MAX(ItemStockID) from items_stock where ItemID=I.ItemId and DATE=I.DATE and Active=1) ;

Answer 2

我想要这样的解决方案，但是要以优化的方式

SELECT ISTI.ItemStockID, ISTI.DATE, ISTI.ItemID, ISTI.CurrentStock FROM items_stock AS ISTI
JOIN (SELECT MAX(ItemStockID) AS ItemStockID 
        FROM items_stock AS IST
            JOIN (SELECT MAX(DATE) AS DATE, ItemID FROM items_stock WHERE DATE <= '2010-11-05' AND Active = 1 GROUP BY ItemID) AS SD ON SD.Date = IST.Date AND IST.ItemID = SD.ItemID
        GROUP BY IST.ItemID
    ) AS ISTO ON ISTO.ItemStockID = ISTI.ItemStockID;

Max（）GROUP BY（Aggregate）具有相同列值的函数问题

2 个答案: