Question

我有一个mysql表，其中包含与其订单相关的项目。

CREATE DATABASE IF NOT EXISTS `sqltest`;
USE `sqltest`;

DROP TABLE IF EXISTS `testdata`;
CREATE TABLE `testdata` (
  `orderID` varchar(10) DEFAULT NULL,
  `itemID` varchar(10) DEFAULT NULL,
  `qtyOrdered` int(10) DEFAULT NULL,
  `sellingPrice` decimal(10,2) DEFAULT NULL
)

INSERT INTO `testdata`(`orderID`,`itemID`,`qtyOrdered`,`sellingPrice`) 
values ('1','a',1,'7.00'),('1','b',2,'8.00'),('1','c',3,'3.00'),('2','a',1,'7.00'),('2','c',4,'3.00');

预期结果：

A =（1 + 1）2

B = 2

C =（2 + 4）6＆lt; - 最受欢迎

如何为每个项目添加所有数量并得出最高的数量？

它应该是相当紧张的前进，但我是SQL的新手，我无法解决这个问题：S

解决方案需要是mysql和/或php。

我想每个商品ID都需要某种临时的tally变量，但是看起来它可能会因太多物品而变得混乱。

解答：

（感谢nuqqsa）

SELECT itemID, SUM(qtyOrdered) AS total FROM testdata GROUP BY itemID ORDER BY total DESC LIMIT 1;

Answer 1

这个怎么样：

SELECT itemID, SUM(qtyOrdered) AS total FROM testdata GROUP BY itemID ORDER BY total DESC;

Answer 2

SELECT itemID, SUM(qtyOrdered) as blah FROM sqltest GROUP BY itemID ORDER BY blah DESC应该这样做

Answer 3

SELECT      *
FROM        testdata
ORDER BY    SUM(gtyOrdered) DESC
GROUP BY    itemID

Answer 4

SELECT SUM( qtyOrdered ) AS sum_ordered, itemID
FROM testdata
GROUP BY itemID
ORDER BY sum_ordered

Answer 5

select count(qtyOrdered), qtyOrdered from testdata group by qtyOrdered

SQL最受欢迎

如何为每个项目添加所有数量并得出最高的数量？

解答：

5 个答案: