Mysql 如何连接多个外键？

我正在使用PHP和SQL创建一个博客，我正在尝试在1 SQL查询中查询所有内容，这样我就可以带出作者的所有评论和作者的所有博客。它们中的每一个都有

user.uuid

作为外键

我的博客表如下所示：

CREATE TABLE `blogs` (
  `uuid` int(255) NOT NULL,
  `title` varchar(1024) NOT NULL,
  `detail` varchar(1024) NOT NULL,
  `user_id` int(255) NOT NULL,
  `created` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP
) ENGINE=InnoDB DEFAULT CHARSET=utf8;

CREATE TABLE `blogs_comments` (
  `uuid` int(255) NOT NULL,
  `blog_uuid` int(255) NOT NULL,
  `user_uuid` int(255) NOT NULL,
  `comment` varchar(250) NOT NULL,
  `created` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP
) ENGINE=InnoDB DEFAULT CHARSET=utf8;

CREATE TABLE `users` (
  `uuid` int(255) NOT NULL,
  `username` varchar(15) COLLATE utf8_unicode_ci NOT NULL,
  `password` varchar(64) COLLATE utf8_unicode_ci NOT NULL,
  `foreName` varchar(30) COLLATE utf8_unicode_ci NOT NULL,
  `surName` varchar(30) COLLATE utf8_unicode_ci NOT NULL,
  `email` text COLLATE utf8_unicode_ci NOT NULL,
  `number` int(11) NOT NULL,
  `created` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP,
  `is_admin` tinyint(1) NOT NULL DEFAULT '0'
) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci;

我的博客评论表如下所示：

CREATE TABLE `blogs` (
  `uuid` int(255) NOT NULL,
  `title` varchar(1024) NOT NULL,
  `detail` varchar(1024) NOT NULL,
  `user_id` int(255) NOT NULL,
  `created` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP
) ENGINE=InnoDB DEFAULT CHARSET=utf8;

CREATE TABLE `blogs_comments` (
  `uuid` int(255) NOT NULL,
  `blog_uuid` int(255) NOT NULL,
  `user_uuid` int(255) NOT NULL,
  `comment` varchar(250) NOT NULL,
  `created` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP
) ENGINE=InnoDB DEFAULT CHARSET=utf8;

CREATE TABLE `users` (
  `uuid` int(255) NOT NULL,
  `username` varchar(15) COLLATE utf8_unicode_ci NOT NULL,
  `password` varchar(64) COLLATE utf8_unicode_ci NOT NULL,
  `foreName` varchar(30) COLLATE utf8_unicode_ci NOT NULL,
  `surName` varchar(30) COLLATE utf8_unicode_ci NOT NULL,
  `email` text COLLATE utf8_unicode_ci NOT NULL,
  `number` int(11) NOT NULL,
  `created` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP,
  `is_admin` tinyint(1) NOT NULL DEFAULT '0'
) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci;

我的用户表如下所示：

CREATE TABLE `blogs` (
  `uuid` int(255) NOT NULL,
  `title` varchar(1024) NOT NULL,
  `detail` varchar(1024) NOT NULL,
  `user_id` int(255) NOT NULL,
  `created` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP
) ENGINE=InnoDB DEFAULT CHARSET=utf8;

CREATE TABLE `blogs_comments` (
  `uuid` int(255) NOT NULL,
  `blog_uuid` int(255) NOT NULL,
  `user_uuid` int(255) NOT NULL,
  `comment` varchar(250) NOT NULL,
  `created` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP
) ENGINE=InnoDB DEFAULT CHARSET=utf8;

CREATE TABLE `users` (
  `uuid` int(255) NOT NULL,
  `username` varchar(15) COLLATE utf8_unicode_ci NOT NULL,
  `password` varchar(64) COLLATE utf8_unicode_ci NOT NULL,
  `foreName` varchar(30) COLLATE utf8_unicode_ci NOT NULL,
  `surName` varchar(30) COLLATE utf8_unicode_ci NOT NULL,
  `email` text COLLATE utf8_unicode_ci NOT NULL,
  `number` int(11) NOT NULL,
  `created` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP,
  `is_admin` tinyint(1) NOT NULL DEFAULT '0'
) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci;

以下是我当前的SQL：

SELECT
b.uuid AS 'Blog ID', b.title AS 'Blog Title', b.detail AS 'Blog Body', u.uuid AS 'Blog Author ID', u.username AS 'Blog Author Username',
(SELECT bc.comment WHERE u.uuid = bc.user_uuid) AS 'Comment',
(SELECT u.uuid WHERE u.uuid = bc.user_uuid) AS 'Comment Author ID',
(SELECT u.username WHERE u.uuid = bc.user_uuid) AS 'Comment Author Username' 
FROM blogs b
INNER JOIN blogs_comments bc ON b.uuid = bc.blog_uuid
INNER JOIN users u ON b.user_id = u.uuid

查询工作正常，但有意外输出，如下所示：

CREATE TABLE `blogs` (
  `uuid` int(255) NOT NULL,
  `title` varchar(1024) NOT NULL,
  `detail` varchar(1024) NOT NULL,
  `user_id` int(255) NOT NULL,
  `created` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP
) ENGINE=InnoDB DEFAULT CHARSET=utf8;

CREATE TABLE `blogs_comments` (
  `uuid` int(255) NOT NULL,
  `blog_uuid` int(255) NOT NULL,
  `user_uuid` int(255) NOT NULL,
  `comment` varchar(250) NOT NULL,
  `created` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP
) ENGINE=InnoDB DEFAULT CHARSET=utf8;

CREATE TABLE `users` (
  `uuid` int(255) NOT NULL,
  `username` varchar(15) COLLATE utf8_unicode_ci NOT NULL,
  `password` varchar(64) COLLATE utf8_unicode_ci NOT NULL,
  `foreName` varchar(30) COLLATE utf8_unicode_ci NOT NULL,
  `surName` varchar(30) COLLATE utf8_unicode_ci NOT NULL,
  `email` text COLLATE utf8_unicode_ci NOT NULL,
  `number` int(11) NOT NULL,
  `created` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP,
  `is_admin` tinyint(1) NOT NULL DEFAULT '0'
) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci;

我希望输出将注释作者的所有注释放在一行中，然后每一行都特定于注释作者。即：

Comment Author    Comment        Comment
Jaquarh           TEST           Lorem Ipsum

---

你需要知道每次你会有多少评论，但如果你这样做，一个轴心可能会给你你所需要的。我对您的查询结构有点困惑，但我认为它可能看起来像这样，带有一个轴心：

SELECT
blog_id AS 'Blog ID',
title AS 'Blog Title',
detail AS 'Blog Body',
blog_author_id AS 'Blog Author ID',
username AS 'Blog Author Username',
Comment_Author_ID AS 'Comment Author ID',
Comment_Author_Username AS 'Comment Author Username',
[0] AS 'Comment 1', 
[1] AS 'Comment 2', 
[2] AS 'Comment 3', 
[3] AS 'Comment 4', 
[4] AS 'Comment 5', 
[5] AS 'Comment 6'

FROM
  (SELECT --A1
  b.uuid as blog_id, b.title, b.detail, u.uuid as blog_author_id, u.username, 
  (SELECT bc.comment WHERE u.uuid = bc.user_uuid) AS Comment,
  (SELECT u.uuid WHERE u.uuid = bc.user_uuid) AS Comment_Author_ID,
  (SELECT u.username WHERE u.uuid = bc.user_uuid) AS Comment_Author_Username,
  row_number() over(partition by b.uuid order by bc.comment) AS row_num
  FROM blogs b
  INNER JOIN blogs_comments bc ON b.uuid = bc.blog_uuid 
  INNER JOIN users u ON b.user_id = u.uuid AND u.uuid = bc.user_uuid
  ) as A1
PIVOT
(max([comment]) FOR row_num in ([0], [1], [2], [3], [4], [5])) AS pvt

---

这篇文章可能也会对你有所帮助：

你需要知道每次你会有多少评论，但如果你这样做，一个轴心可能会给你你所需要的。我对您的查询结构有点困惑，但我认为它可能看起来像这样，带有一个轴心：

SELECT
blog_id AS 'Blog ID',
title AS 'Blog Title',
detail AS 'Blog Body',
blog_author_id AS 'Blog Author ID',
username AS 'Blog Author Username',
Comment_Author_ID AS 'Comment Author ID',
Comment_Author_Username AS 'Comment Author Username',
[0] AS 'Comment 1', 
[1] AS 'Comment 2', 
[2] AS 'Comment 3', 
[3] AS 'Comment 4', 
[4] AS 'Comment 5', 
[5] AS 'Comment 6'

FROM
  (SELECT --A1
  b.uuid as blog_id, b.title, b.detail, u.uuid as blog_author_id, u.username, 
  (SELECT bc.comment WHERE u.uuid = bc.user_uuid) AS Comment,
  (SELECT u.uuid WHERE u.uuid = bc.user_uuid) AS Comment_Author_ID,
  (SELECT u.username WHERE u.uuid = bc.user_uuid) AS Comment_Author_Username,
  row_number() over(partition by b.uuid order by bc.comment) AS row_num
  FROM blogs b
  INNER JOIN blogs_comments bc ON b.uuid = bc.blog_uuid 
  INNER JOIN users u ON b.user_id = u.uuid AND u.uuid = bc.user_uuid
  ) as A1
PIVOT
(max([comment]) FOR row_num in ([0], [1], [2], [3], [4], [5])) AS pvt

这篇文章也可能对您有所帮助：