MySQL：带有LEFT JOIN的GROUP_CONCAT

Posted 2021-03-30

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！

答案

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

另一答案

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。

另一答案

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;