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mysql - 使用数据透视表时如何选择？

问题描述

考虑以下SQLFiddle。

DROP TABLE IF EXISTS appointments;
CREATE TABLE appointments (
    id INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY,
    name VARCHAR(255) NOT NULL
);

DROP TABLE IF EXISTS services;
CREATE TABLE services (
    id INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY,
    name VARCHAR(255) NOT NULL
);

DROP TABLE IF EXISTS join_services_appointment;
CREATE TABLE join_services_appointment (
    service_id INT(6) UNSIGNED,
    appointment_id INT(6) UNSIGNED,
    FOREIGN KEY(service_id) references services(id),
    FOREIGN KEY(appointment_id) references appointments(id)
);

INSERT INTO appointments (name) VALUES ('appointment1');
INSERT INTO appointments (name) VALUES ('appointment2');
INSERT INTO appointments (name) VALUES ('appointment3');

INSERT INTO services (name) VALUES ('services1');
INSERT INTO services (name) VALUES ('services2');
INSERT INTO services (name) VALUES ('services3');
INSERT INTO services (name) VALUES ('services4');
INSERT INTO services (name) VALUES ('services5');

INSERT INTO join_services_appointment (service_id, appointment_id) VALUES (1, 1);
INSERT INTO join_services_appointment (service_id, appointment_id) VALUES (2, 1);
INSERT INTO join_services_appointment (service_id, appointment_id) VALUES (2, 2);
INSERT INTO join_services_appointment (service_id, appointment_id) VALUES (3, 2);
INSERT INTO join_services_appointment (service_id, appointment_id) VALUES (3, 3);
INSERT INTO join_services_appointment (service_id, appointment_id) VALUES (4, 3);

我将使用什么查询来获取以下结果？

appointment_name   services
appointment1       services1,services2
appointment2       services2,services3
appointment3       services3,services4

标签： mysqlsql

解决方案

如评论中所述，您可以使用GROUP_CONCAT()来实现所需的输出。

SELECT 
  a.name AS appointment_name,
  GROUP_CONCAT(DISTINCT s.name ORDER BY s.name SEPARATOR ', ') AS services
FROM appointments a
INNER JOIN join_services_appointment sa ON a.id = sa.appointment_id
INNER JOIN services s ON s.id = sa.service_id
GROUP BY a.id;

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