MySQL:带有LEFT JOIN的GROUP_CONCAT

问题描述 投票:48回答:3

我遇到了MySQL的“GROUP_CONCAT”功能问题。我将使用一个简单的帮助台数据库来说明我的问题:

CREATE TABLE Tickets (
 id INTEGER NOT NULL PRIMARY KEY,
 requester_name VARCHAR(255) NOT NULL,
 description TEXT NOT NULL);

CREATE TABLE Solutions (
 id INTEGER NOT NULL PRIMARY KEY,
 ticket_id INTEGER NOT NULL,
 technician_name VARCHAR(255) NOT NULL,
 solution TEXT NOT NULL,
 FOREIGN KEY (ticket_id) REFERENCES Tickets.id);

INSERT INTO Tickets VALUES(1, 'John Doe', 'My computer is not booting.');
INSERT INTO Tickets VALUES(2, 'Jane Doe', 'My browser keeps crashing.');
INSERT INTO Solutions VALUES(1, 1, 'Technician A', 'I tried to solve this but was unable to. I will pass this on to Technician B since he is more experienced than I am.');
INSERT INTO Solutions VALUES(2, 1, 'Technician B', 'I reseated the RAM and that fixed the problem.');
INSERT INTO Solutions VALUES(3, 2, 'Technician A', 'I was unable to figure this out. I will again pass this on to Technician B.');
INSERT INTO Solutions VALUES(4, 2, 'Technician B', 'I re-installed the browser and that fixed the problem.');

请注意,此帮助台数据库有两个票证,每个票证都有两个解决方案条目。我的目标是使用SELECT语句创建数据库中所有票证的列表及其相应的解决方案条目。这是我正在使用的SELECT语句:

SELECT Tickets.*, GROUP_CONCAT(Solutions.solution) AS CombinedSolutions
FROM Tickets
LEFT JOIN Solutions ON Tickets.id = Solutions.ticket_id
ORDER BY Tickets.id;

上面的SELECT语句的问题是它只返回一行:

id: 1
requester_name: John Doe
description: My computer is not booting.
CombinedSolutions: I tried to solve this but was unable to. I will pass this on to Technician B since he is more experienced than I am.,I reseated the RAM and that fixed the problem.,I was unable to figure this out. I will again pass this on to Technician B.,I re-installed the browser and that fixed the problem.

请注意,它将使用故障单1和故障单2的解决方案条目返回故障单1的信息。

我究竟做错了什么?谢谢!

sql mysql join aggregate-functions group-concat
3个回答
85
投票

使用:

   SELECT t.*,
          x.combinedsolutions
     FROM TICKETS t
LEFT JOIN (SELECT s.ticket_id,
                  GROUP_CONCAT(s.soution) AS combinedsolutions
             FROM SOLUTIONS s 
         GROUP BY s.ticket_id) x ON x.ticket_id = t.ticket_id

备用:

   SELECT t.*,
          (SELECT GROUP_CONCAT(s.soution)
             FROM SOLUTIONS s 
            WHERE s.ticket_id = t.ticket_id) AS combinedsolutions
     FROM TICKETS t

2
投票

我认为@Dylan Valade的评论是最简单的答案,所以我将其作为另一个答案发布:只需在OP的SELECT中添加GROUP BY Tickets.id即可解决问题。它解决了我自己的问题。

但是,对于不小的数据库,接受的答案,特别是如果Tickets.id上有任何谓词似乎不涉及总表扫描,所以虽然前一段返回正确的结果,但在我的情况下似乎效率低得多。


1
投票

您只需要添加GROUP_BY:

SELECT Tickets.*, GROUP_CONCAT(Solutions.solution) AS CombinedSolutions FROM Tickets 
LEFT JOIN Solutions ON Tickets.id = Solutions.ticket_id 
GROUP_BY Tickets.id 
ORDER BY Tickets.id;
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