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mysql - 如何编写 mysql 查询以从一个表中获取记录列表，其中列与多个其他表连接

问题描述

我的问题与此类似：MySQL concatenate values from a table into a record of another

但它不一样，我想是因为我试图利用来自其他几个表的多个连接列。

这是我的表：

CREATE TABLE `Albums` (
  `id` bigint(20) NOT NULL AUTO_INCREMENT,
  `userSettingsId` bigint(20) NOT NULL,
  `name` varchar(100) NOT NULL,
  `description` text DEFAULT NULL,
  `isFavorite` tinyint(1) NOT NULL DEFAULT 0,
  `isPublic` tinyint(1) NOT NULL DEFAULT 0,
  `created` datetime NOT NULL DEFAULT current_timestamp(),
  `lastEdited` datetime NOT NULL DEFAULT current_timestamp(),
  PRIMARY KEY (`id`),
  UNIQUE KEY `Albums_UN` (`userSettingsId`,`name`),
  CONSTRAINT `Albums_FK` FOREIGN KEY (`userSettingsId`) REFERENCES `UserSettings` (`id`) ON DELETE CASCADE ON UPDATE CASCADE
) ENGINE=InnoDB AUTO_INCREMENT=40 DEFAULT CHARSET=utf8mb4;

CREATE TABLE `AlbumsImages` (
  `albumsId` bigint(20) NOT NULL,
  `imagesId` bigint(20) NOT NULL,
  `isCoverImage` tinyint(1) NOT NULL DEFAULT 0,
  PRIMARY KEY (`albumsId`,`imagesId`),
  KEY `AlbumsImages_FK` (`imagesId`),
  CONSTRAINT `AlbumsImages_FK` FOREIGN KEY (`imagesId`) REFERENCES `Images` (`id`) ON DELETE CASCADE ON UPDATE CASCADE,
  CONSTRAINT `AlbumsImages_FK_1` FOREIGN KEY (`albumsId`) REFERENCES `Albums` (`id`) ON DELETE CASCADE ON UPDATE CASCADE
) ENGINE=InnoDB DEFAULT CHARSET=utf8mb4;

CREATE TABLE `Collaborators` (
  `albumsId` bigint(20) NOT NULL,
  `access` enum('view','put','edit') NOT NULL DEFAULT 'view',
  `email` varchar(100) NOT NULL,
  PRIMARY KEY (`albumsId`,`email`),
  CONSTRAINT `Collaborators_FK_1` FOREIGN KEY (`albumsId`) REFERENCES `Albums` (`id`) ON DELETE CASCADE ON UPDATE CASCADE
) ENGINE=InnoDB DEFAULT CHARSET=utf8mb4;

CREATE TABLE `Images` (
  `id` bigint(20) NOT NULL AUTO_INCREMENT,
  `src` longtext NOT NULL,
  `fileName` varchar(100) NOT NULL,
  `alt` varchar(100) NOT NULL,
  `userSettingsId` bigint(20) NOT NULL,
  PRIMARY KEY (`id`),
  UNIQUE KEY `Images_UN_FileName_UserSettingsId` (`fileName`,`userSettingsId`),
  KEY `Images_FK` (`userSettingsId`),
  CONSTRAINT `Images_FK` FOREIGN KEY (`userSettingsId`) REFERENCES `UserSettings` (`id`) ON DELETE CASCADE ON UPDATE CASCADE
) ENGINE=InnoDB AUTO_INCREMENT=91 DEFAULT CHARSET=utf8 COMMENT='All images in the application';

CREATE TABLE `UserSettings` (
  `id` bigint(20) NOT NULL AUTO_INCREMENT,
  `email` varchar(100) NOT NULL,
  `firstName` varchar(100) NOT NULL,
  `lastName` varchar(100) NOT NULL,
  `password` text NOT NULL,
  `isDarkThemeEnabled` tinyint(1) NOT NULL DEFAULT 1,
  `sessionId` varchar(255) DEFAULT NULL,
  `profilePicture` bigint(20) DEFAULT NULL,
  PRIMARY KEY (`id`),
  UNIQUE KEY `UserSettings_Email_UN` (`email`),
  UNIQUE KEY `UserSettings_SessionId_UN` (`sessionId`)
) ENGINE=InnoDB AUTO_INCREMENT=14 DEFAULT CHARSET=utf8 COMMENT='User Settings and Info'

注意：该Collaborators表不使用外键，UserSettings.id因为协作者不一定要有帐户。

因此，我想获取特定用户的所有专辑和有关它们的辅助信息。像这样的东西：

[
  {
    id: 1,
    name: 'Album Name',
    description: 'Description',
    isFavorite: 0,
    created: '2021-04-26 23:14:05',
    lastEdited: '2021-04-27 19:12:02',
    images: [
      {
        fileName: 'image.jpg',
        alt: 'Image Title',
        src: 'www.image.com/image.jpg',
        isCoverImage: 0,
      }, //...etc
    ],
    collaborators: [
      {
        email: 'someone@example.com',
        firstName: null,
        lastName: null,
        id: null,
        access: 'put',
      },
      {
        email: 'someoneelse@example.com',
        firstName: 'someone',
        lastName: 'else',
        id: 14,
        access: 'view',
      }, //...etc
    ],
  }, //...etc
]

这是我目前正在使用的查询。

SELECT 
    a.id,
    a.name,
    a.description,
    a.isFavorite,
    a.created,
    a.lastEdited,
    concat('[', group_concat(json_object(
        'fileName', i.fileName,
        'alt', i.alt,
        'src', i.src,
        'isCoverImage', ai.isCoverImage
    )), ']') as images,
    concat('[', group_concat(json_object(
        'email', c.email,
        'firstName', u.firstName,
        'lastName', u.lastName,
        'id', u.id,
        'access', c.access
    )), ']') as collaborators
from Albums a 
    left join AlbumsImages ai
        on a.id=ai.albumsId 
    left join Images i 
        on i.id=ai.imagesId 
    left join Collaborators c 
        on c.albumsId = a.id 
    left join UserSettings u
        on c.email = u.email 
where a.userSettingsId=?
group by id;

Aaand 它确实有效......有点。我得到了所有的专辑和他们的所有信息，但是合作者被图像数量所复制，反之亦然。作为目前的创可贴，我有一些在查询后立即运行的重复数据删除代码，但这显然是 hacky，而不是我想要长期使用的东西。

有没有办法解决这个问题来做我想做的事，或者我是一个白痴，想首先在一个查询中获取所有这些信息？

谢谢！

标签： mysqlsqlgroup-concat

解决方案

所以你有一个协作者和图像之间的笛卡尔积。因此，两者都乘以另一个结果的数量。

您可以运行多个查询，然后编写应用程序代码以将结果附加到更大的 JSON 文档中。

或者您可以使用相关子查询：

SELECT 
    a.id,
    a.name,
    a.description,
    a.isFavorite,
    a.created,
    a.lastEdited,
    concat('[', (
      select group_concat(json_object(
        'fileName', i.fileName,
        'alt', i.alt,
        'src', i.src,
        'isCoverImage', ai.isCoverImage))
      from Images i where i.id=ai.imagesId
    ), ']') as images,
    concat('[', (
      select group_concat(json_object(
        'email', c.email,
        'firstName', u.firstName,
        'lastName', u.lastName,
        'id', u.id,
        'access', c.access))
      from Collaborators c
      left join UserSettings u 
        on c.email = u.email 
      where c.albumsId=a.id
    ), ']') as collaborators
from Albums a 
    left join AlbumsImages ai
        on a.id=ai.albumsId 
where a.userSettingsId=?
group by id;

（未测试）

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