多个连接和每个父级的最后 N 行
Multiple joins and last N rows per each parent
我有 3 个 table。
companies
- id
- name
- user_id
departments
- id
- name
- user_id
- company_id
invoices
- id
- department_id
- price
- created_at
出于性能目的，我试图在 1 个 mysql 大查询中获取 "dashboard" 屏幕所需的所有数据。值得一提的是，发票 table 有大约 70 万条记录，并且只会不断增加。
所以我需要获取所有用户的公司、部门和每个部门的最后 2 张发票（每个 ID 的 2 个最高日期）。
现在我对前 2 个没有问题，我可以轻松做到，例如：
SELECT companies.id as company_id, companies.name as company_name, departments.id as department_id, departments.name as department_name
FROM companies
LEFT JOIN departments
ON companies.id = departments.company_id
WHERE companies.user_id = 1
我正在努力获取每个部门的最新 2 张发票。在同一个查询中执行此操作的最佳方法是什么？
这是请求的数据，SQL Fiddle相同。
CREATE TABLE `companies` (
`id` int(10) UNSIGNED NOT NULL,
`name` varchar(255) COLLATE utf8_unicode_ci NOT NULL,
`user_id` int(11) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci;
CREATE TABLE `departments` (
`id` int(10) UNSIGNED NOT NULL,
`name` varchar(255) COLLATE utf8_unicode_ci NOT NULL,
`user_id` int(11) NOT NULL,
`company_id` int(11) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci;
CREATE TABLE `invoices` (
`id` int(10) UNSIGNED NOT NULL,
`price` decimal(6,2) NOT NULL,
`created_at` timestamp NULL DEFAULT NULL,
`department_id` int(11) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci;
ALTER TABLE `companies`
ADD PRIMARY KEY (`id`);
ALTER TABLE `departments`
ADD PRIMARY KEY (`id`);
ALTER TABLE `invoices`
ADD PRIMARY KEY (`id`);
ALTER TABLE `companies`
MODIFY `id` int(10) UNSIGNED NOT NULL AUTO_INCREMENT, AUTO_INCREMENT=1;
ALTER TABLE `departments`
MODIFY `id` int(10) UNSIGNED NOT NULL AUTO_INCREMENT, AUTO_INCREMENT=1;
ALTER TABLE `invoices`
MODIFY `id` int(10) UNSIGNED NOT NULL AUTO_INCREMENT, AUTO_INCREMENT=1;
INSERT INTO companies
(`name`, `user_id`)
VALUES
('Google', 1),
('Apple', 1),
('IBM', 1)
;
INSERT INTO departments
(`name`, `user_id`, `company_id`)
VALUES
('Billing', 1, 1),
('Support', 1, 1),
('Tech', 1, 1),
('Billing Dept', 1, 2),
('Support Dept', 1, 2),
('Tech Dept', 1, 2),
('HR', 1, 3),
('IT', 1, 3),
('Executive', 1, 3)
;
INSERT INTO invoices
(`price`, `created_at`, `department_id`)
VALUES
(155.23, '2016-04-07 14:39:29', 1),
(123.23, '2016-04-07 14:40:26', 1),
(150.50, '2016-04-07 14:40:30', 1),
(123.23, '2016-04-07 14:41:38', 1),
(432.65, '2016-04-07 14:44:15', 1),
(323.23, '2016-04-07 14:44:22', 2),
(541.43, '2016-04-07 14:44:33', 2),
(1232.23, '2016-04-07 14:44:36', 2),
(433.42, '2016-04-07 14:44:37', 2),
(1232.43, '2016-04-07 14:44:39', 2),
(850.40, '2016-04-07 14:44:46', 3),
(133.32, '2016-04-07 14:45:11', 3),
(12.43, '2016-04-07 14:45:15', 3),
(154.23, '2016-04-07 14:45:25', 3),
(132.43, '2016-04-07 14:46:01', 3),
(859.55, '2016-04-07 14:53:11', 4),
(123.43, '2016-04-07 14:53:45', 4),
(433.33, '2016-04-07 14:54:14', 4),
(545.12, '2016-04-07 14:54:54', 4),
(949.99, '2016-04-07 14:55:10', 4),
(1112.32, '2016-04-07 14:53:40', 5),
(132.32, '2016-04-07 14:53:44', 5),
(42.43, '2016-04-07 14:53:48', 5),
(545.34, '2016-04-07 14:53:56', 5),
(2343.32, '2016-04-07 14:54:05', 5),
(3432.43, '2016-04-07 14:54:02', 6),
(231.32, '2016-04-07 14:54:22', 6),
(1242.33, '2016-04-07 14:54:54', 6),
(232.32, '2016-04-07 14:55:12', 6),
(43.12, '2016-04-07 14:55:23', 6),
(4343.23, '2016-04-07 14:55:24', 7),
(1123.32, '2016-04-07 14:55:31', 7),
(4343.32, '2016-04-07 14:55:56', 7),
(354.23, '2016-04-07 14:56:04', 7),
(867.76, '2016-04-07 14:56:12', 7),
(45.76, '2016-04-07 14:55:54', 8),
(756.65, '2016-04-07 14:56:08', 8),
(153.74, '2016-04-07 14:56:14', 8),
(534.86, '2016-04-07 14:56:23', 8),
(867.65, '2016-04-07 14:56:55', 8),
(433.56, '2016-04-07 14:56:32', 9),
(1423.43, '2016-04-07 14:56:54', 9),
(342.56, '2016-04-07 14:57:11', 9),
(343.75, '2016-04-07 14:57:23', 9),
(1232.43, '2016-04-07 14:57:34', 9)
;
这是您想要的结果。
company_id| company_name| department_id | department_name | invoice_price | invoice_created_at
1| Google | 1 | Billing | 123.23 | 2016-04-07 14:41:38 |
1| Google | 1 | Billing | 432.65 | 2016-04-07 14:44:15 |
1| Google | 2 | Support | 433.42 | 2016-04-07 14:44:37 |
1| Google | 2 | Support | 1232.43 | 2016-04-07 14:44:39 |
1| Google | 3 | Tech | 154.23 | 2016-04-07 14:45:25 |
1| Google | 3 | Tech | 132.43 | 2016-04-07 14:46:01 |
2| Apple | 4 | Billing Dept | 545.12 | 2016-04-07 14:54:54 |
2| Apple | 4 | Billing Dept | 949.99 | 2016-04-07 14:55:10 |
2| Apple | 5 | Support Dept | 545.34 | 2016-04-07 14:53:56 |
2| Apple | 5 | Support Dept | 2343.32 | 2016-04-07 14:54:05 |
2| Apple | 6 | Tech Dept | 232.32 | 2016-04-07 14:55:12 |
2| Apple | 6 | Tech Dept | 43.12 | 2016-04-07 14:55:23 |
3| IBM | 7 | HR | 354.23 | 2016-04-07 14:56:04 |
3| IBM | 7 | HR | 867.76 | 2016-04-07 14:56:12 |
3| IBM | 8 | IT | 534.86 | 2016-04-07 14:56:23 |
3| IBM | 8 | IT | 867.65 | 2016-04-07 14:56:55 |
3| IBM | 9 | Executive | 343.75 | 2016-04-07 14:57:23 |
3| IBM | 9 | Executive | 1232.43 | 2016-04-07 14:57:34 |
一个想法是在 invoices table
中再包含一个 JOIN
LEFT JOIN invoices i ON i.department_id = departments.id
这样您就可以获得每个部门的所有发票。但是您需要将它们限制为每个部门的最后两个。一种方法是使用 LIMIT 2
的相关子查询的附加 IN 条件
LEFT JOIN invoices i
ON i.department_id = departments.id
AND i.id IN (
SELECT i1.id
FROM invoices i1
WHERE i1.department_id = departments.id
ORDER BY i1.id DESC
LIMIT 2
)
但是由于某些奇怪的原因，MySQL 不允许在 IN 语句中使用 LIMIT。所以我们需要更巧妙地避免 IN 条件。相反，我们可以使用 >= 和 select 第二高的 ID 使用 LIMIT 1 OFFSET 1:
AND i.id >= (
SELECT i1.id
FROM invoices i1
WHERE i1.department_id = departments.id
ORDER BY i1.id DESC
LIMIT 1
OFFSET 1
)
现在最后一个问题：如果只有一张发票，我们就找不到第二张了。子查询将返回 NULL 并且条件将始终失败。在那种情况下，我们使用 COALESCE.
将 NULL 替换为 0
所以最终的查询看起来像：
SELECT companies.id as company_id,
companies.name as company_name,
departments.id as department_id,
departments.name as department_name,
i.id as invoice_id,
i.price as invoice_price
FROM companies
LEFT JOIN departments
ON companies.id = departments.company_id
LEFT JOIN invoices i
ON i.department_id = departments.id
AND i.id >= COALESCE((
SELECT i1.id
FROM invoices i1
WHERE i1.department_id = departments.id
ORDER BY i1.id DESC
LIMIT 1
OFFSET 1
), 0)
WHERE companies.user_id = 1
我不得不承认，我对您的结果集与您的描述和数据集的匹配方式有点费劲，但这里有一些东西可以玩...
SELECT x.price
, x.created_at
, x.department_id
, x.department
, x.department_user
, x.company_id
, x.company
, x.company_user
FROM
( SELECT i.id
, i.price
, i.created_at
, i.department_id
, d.name department
, d.user_id department_user
, d.company_id
, c.name company
, c.user_id company_user
, CASE WHEN @prev=department_id THEN @i:=@i+1 ELSE @i:=1 END i
, @prev := i.department_id
FROM invoices i
JOIN departments d
ON d.id = i.department_id
JOIN companies c
ON c.id = d.company_id
JOIN (SELECT @prev:=null, @i:=0) vars
ORDER
BY department_id
, created_at DESC
) x
WHERE i<=2;
这里是概念化相同想法的较慢方式（我省略了不太相关的部分）...
SELECT x.*
FROM invoices x
JOIN invoices y
ON y.department_id = x.department_id
AND y.created_at <= x.created_at
GROUP
BY x.department_id
, x.created_at
HAVING COUNT(*) <=2;
我有 3 个 table。
companies
- id
- name
- user_id
departments
- id
- name
- user_id
- company_id
invoices
- id
- department_id
- price
- created_at
出于性能目的，我试图在 1 个 mysql 大查询中获取 "dashboard" 屏幕所需的所有数据。值得一提的是，发票 table 有大约 70 万条记录，并且只会不断增加。
所以我需要获取所有用户的公司、部门和每个部门的最后 2 张发票（每个 ID 的 2 个最高日期）。
现在我对前 2 个没有问题，我可以轻松做到，例如：
SELECT companies.id as company_id, companies.name as company_name, departments.id as department_id, departments.name as department_name
FROM companies
LEFT JOIN departments
ON companies.id = departments.company_id
WHERE companies.user_id = 1
我正在努力获取每个部门的最新 2 张发票。在同一个查询中执行此操作的最佳方法是什么？
这是请求的数据，SQL Fiddle相同。
CREATE TABLE `companies` (
`id` int(10) UNSIGNED NOT NULL,
`name` varchar(255) COLLATE utf8_unicode_ci NOT NULL,
`user_id` int(11) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci;
CREATE TABLE `departments` (
`id` int(10) UNSIGNED NOT NULL,
`name` varchar(255) COLLATE utf8_unicode_ci NOT NULL,
`user_id` int(11) NOT NULL,
`company_id` int(11) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci;
CREATE TABLE `invoices` (
`id` int(10) UNSIGNED NOT NULL,
`price` decimal(6,2) NOT NULL,
`created_at` timestamp NULL DEFAULT NULL,
`department_id` int(11) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci;
ALTER TABLE `companies`
ADD PRIMARY KEY (`id`);
ALTER TABLE `departments`
ADD PRIMARY KEY (`id`);
ALTER TABLE `invoices`
ADD PRIMARY KEY (`id`);
ALTER TABLE `companies`
MODIFY `id` int(10) UNSIGNED NOT NULL AUTO_INCREMENT, AUTO_INCREMENT=1;
ALTER TABLE `departments`
MODIFY `id` int(10) UNSIGNED NOT NULL AUTO_INCREMENT, AUTO_INCREMENT=1;
ALTER TABLE `invoices`
MODIFY `id` int(10) UNSIGNED NOT NULL AUTO_INCREMENT, AUTO_INCREMENT=1;
INSERT INTO companies
(`name`, `user_id`)
VALUES
('Google', 1),
('Apple', 1),
('IBM', 1)
;
INSERT INTO departments
(`name`, `user_id`, `company_id`)
VALUES
('Billing', 1, 1),
('Support', 1, 1),
('Tech', 1, 1),
('Billing Dept', 1, 2),
('Support Dept', 1, 2),
('Tech Dept', 1, 2),
('HR', 1, 3),
('IT', 1, 3),
('Executive', 1, 3)
;
INSERT INTO invoices
(`price`, `created_at`, `department_id`)
VALUES
(155.23, '2016-04-07 14:39:29', 1),
(123.23, '2016-04-07 14:40:26', 1),
(150.50, '2016-04-07 14:40:30', 1),
(123.23, '2016-04-07 14:41:38', 1),
(432.65, '2016-04-07 14:44:15', 1),
(323.23, '2016-04-07 14:44:22', 2),
(541.43, '2016-04-07 14:44:33', 2),
(1232.23, '2016-04-07 14:44:36', 2),
(433.42, '2016-04-07 14:44:37', 2),
(1232.43, '2016-04-07 14:44:39', 2),
(850.40, '2016-04-07 14:44:46', 3),
(133.32, '2016-04-07 14:45:11', 3),
(12.43, '2016-04-07 14:45:15', 3),
(154.23, '2016-04-07 14:45:25', 3),
(132.43, '2016-04-07 14:46:01', 3),
(859.55, '2016-04-07 14:53:11', 4),
(123.43, '2016-04-07 14:53:45', 4),
(433.33, '2016-04-07 14:54:14', 4),
(545.12, '2016-04-07 14:54:54', 4),
(949.99, '2016-04-07 14:55:10', 4),
(1112.32, '2016-04-07 14:53:40', 5),
(132.32, '2016-04-07 14:53:44', 5),
(42.43, '2016-04-07 14:53:48', 5),
(545.34, '2016-04-07 14:53:56', 5),
(2343.32, '2016-04-07 14:54:05', 5),
(3432.43, '2016-04-07 14:54:02', 6),
(231.32, '2016-04-07 14:54:22', 6),
(1242.33, '2016-04-07 14:54:54', 6),
(232.32, '2016-04-07 14:55:12', 6),
(43.12, '2016-04-07 14:55:23', 6),
(4343.23, '2016-04-07 14:55:24', 7),
(1123.32, '2016-04-07 14:55:31', 7),
(4343.32, '2016-04-07 14:55:56', 7),
(354.23, '2016-04-07 14:56:04', 7),
(867.76, '2016-04-07 14:56:12', 7),
(45.76, '2016-04-07 14:55:54', 8),
(756.65, '2016-04-07 14:56:08', 8),
(153.74, '2016-04-07 14:56:14', 8),
(534.86, '2016-04-07 14:56:23', 8),
(867.65, '2016-04-07 14:56:55', 8),
(433.56, '2016-04-07 14:56:32', 9),
(1423.43, '2016-04-07 14:56:54', 9),
(342.56, '2016-04-07 14:57:11', 9),
(343.75, '2016-04-07 14:57:23', 9),
(1232.43, '2016-04-07 14:57:34', 9)
;
这是您想要的结果。
company_id| company_name| department_id | department_name | invoice_price | invoice_created_at
1| Google | 1 | Billing | 123.23 | 2016-04-07 14:41:38 |
1| Google | 1 | Billing | 432.65 | 2016-04-07 14:44:15 |
1| Google | 2 | Support | 433.42 | 2016-04-07 14:44:37 |
1| Google | 2 | Support | 1232.43 | 2016-04-07 14:44:39 |
1| Google | 3 | Tech | 154.23 | 2016-04-07 14:45:25 |
1| Google | 3 | Tech | 132.43 | 2016-04-07 14:46:01 |
2| Apple | 4 | Billing Dept | 545.12 | 2016-04-07 14:54:54 |
2| Apple | 4 | Billing Dept | 949.99 | 2016-04-07 14:55:10 |
2| Apple | 5 | Support Dept | 545.34 | 2016-04-07 14:53:56 |
2| Apple | 5 | Support Dept | 2343.32 | 2016-04-07 14:54:05 |
2| Apple | 6 | Tech Dept | 232.32 | 2016-04-07 14:55:12 |
2| Apple | 6 | Tech Dept | 43.12 | 2016-04-07 14:55:23 |
3| IBM | 7 | HR | 354.23 | 2016-04-07 14:56:04 |
3| IBM | 7 | HR | 867.76 | 2016-04-07 14:56:12 |
3| IBM | 8 | IT | 534.86 | 2016-04-07 14:56:23 |
3| IBM | 8 | IT | 867.65 | 2016-04-07 14:56:55 |
3| IBM | 9 | Executive | 343.75 | 2016-04-07 14:57:23 |
3| IBM | 9 | Executive | 1232.43 | 2016-04-07 14:57:34 |
一个想法是在 invoices table
LEFT JOIN invoices i ON i.department_id = departments.id
这样您就可以获得每个部门的所有发票。但是您需要将它们限制为每个部门的最后两个。一种方法是使用 LIMIT 2
的相关子查询的附加 IN 条件LEFT JOIN invoices i
ON i.department_id = departments.id
AND i.id IN (
SELECT i1.id
FROM invoices i1
WHERE i1.department_id = departments.id
ORDER BY i1.id DESC
LIMIT 2
)
但是由于某些奇怪的原因，MySQL 不允许在 IN 语句中使用 LIMIT。所以我们需要更巧妙地避免 IN 条件。相反，我们可以使用 >= 和 select 第二高的 ID 使用 LIMIT 1 OFFSET 1:
AND i.id >= (
SELECT i1.id
FROM invoices i1
WHERE i1.department_id = departments.id
ORDER BY i1.id DESC
LIMIT 1
OFFSET 1
)
现在最后一个问题：如果只有一张发票，我们就找不到第二张了。子查询将返回 NULL 并且条件将始终失败。在那种情况下，我们使用 COALESCE.
0
所以最终的查询看起来像：
SELECT companies.id as company_id,
companies.name as company_name,
departments.id as department_id,
departments.name as department_name,
i.id as invoice_id,
i.price as invoice_price
FROM companies
LEFT JOIN departments
ON companies.id = departments.company_id
LEFT JOIN invoices i
ON i.department_id = departments.id
AND i.id >= COALESCE((
SELECT i1.id
FROM invoices i1
WHERE i1.department_id = departments.id
ORDER BY i1.id DESC
LIMIT 1
OFFSET 1
), 0)
WHERE companies.user_id = 1
我不得不承认，我对您的结果集与您的描述和数据集的匹配方式有点费劲，但这里有一些东西可以玩...
SELECT x.price
, x.created_at
, x.department_id
, x.department
, x.department_user
, x.company_id
, x.company
, x.company_user
FROM
( SELECT i.id
, i.price
, i.created_at
, i.department_id
, d.name department
, d.user_id department_user
, d.company_id
, c.name company
, c.user_id company_user
, CASE WHEN @prev=department_id THEN @i:=@i+1 ELSE @i:=1 END i
, @prev := i.department_id
FROM invoices i
JOIN departments d
ON d.id = i.department_id
JOIN companies c
ON c.id = d.company_id
JOIN (SELECT @prev:=null, @i:=0) vars
ORDER
BY department_id
, created_at DESC
) x
WHERE i<=2;
这里是概念化相同想法的较慢方式（我省略了不太相关的部分）...
SELECT x.*
FROM invoices x
JOIN invoices y
ON y.department_id = x.department_id
AND y.created_at <= x.created_at
GROUP
BY x.department_id
, x.created_at
HAVING COUNT(*) <=2;