Question

我有两张桌子：

categories =＆gt; Category_ID，Title，Description，Default_Points，Groups

transactions =＆gt; Transaction_ID，Datetime，Giver_ID，Recipient_ID，Points，Category_ID，Reason

教师奖励积分，选择一个类别（如“积极态度和行为”）和一个理由（如“今日优秀作品”），将一个条目放入transactions表。

典型的categories行可能是：

INSERT INTO `categories` (`Category_ID`, `Title`, `Description`, `Default_Points`, `Groups`) VALUES
(17, 'Olympic Values', 'Please clearly state the correct Olympic Value that''s being used currently in the REASON box.', 5, '');

典型的transactions行可能是：

INSERT INTO `transactions` (`Transaction_ID`, `Datetime`, `Giver_ID`, `Recipient_ID`, `Points`, `Category_ID`, `Reason`) VALUES
(50, '2011-09-07', 35023, 90236, 5, 17, 'Excellent work during PE');

我想尝试使用MySQL做的是生成一个总点数列表（即每个类别SUM(transactions.Points)，还有一些样本Reasons。

我想这必须使用CONCAT？

我需要：

SUM(transactions.Points)
categories.title
每个类别5个唯一transactions.reason

这可能看起来像......

Points      Title                   Sample
14252       Olympic Values          Excellent work in PE!|Great display of friendship|Well done!
15532       Outstanding Effort      Amazing work!|Worked so hard|Great piece!

这可能吗？

提前致谢，

Answer 1

这是你想要的GROUP_CONCAT。

你需要做这样的事情：

SELECT SUM(t.Points), 
    c.title, 
    SUBSTRING_INDEX(GROUP_CONCAT(transactions.reasons SEPERATOR '|'), '|', 5)
FROM transactions t JOIN categories c ON (t.Category_ID=c.Category_ID)
GROUP BY c.Category_ID

Answer 2

感谢@ChrisPatrick的回答，这是我使用的代码：

SELECT
 SUM(t.Points) AS Total_Points, 
    c.Title, 
    SUBSTRING_INDEX(GROUP_CONCAT(DISTINCT t.reason SEPARATOR '|'), '|', 5) AS Sample_Reasons
FROM
 transactions t JOIN categories c ON (t.Category_ID=c.Category_ID)
GROUP BY c.Category_ID
ORDER BY c.Title ASC

使用CONCAT进行MySQL查询

2 个答案: