Question

注意：按combination，我会参考svc_id＆amp; loc_id

如果combination将status设置为零，我有兴趣找到combination设置为零的最新唯一status，但稍后会在再次设置为1，然后应忽略combination！

MySQL 5.6架构设置：

CREATE TABLE `activity_log` (
  `id` int(11) NOT NULL,
  `prj_id` bigint(20) NOT NULL,
  `svc_id` bigint(20) NOT NULL,
  `loc_id` bigint(20) NOT NULL,
  `status` tinyint(1) NOT NULL,
  `created_by` bigint(20) NOT NULL,
  `created_at` datetime NOT NULL DEFAULT CURRENT_TIMESTAMP,
  `deleted_by` bigint(20) DEFAULT NULL,
  `deleted_at` datetime DEFAULT NULL ON UPDATE CURRENT_TIMESTAMP
) ENGINE=MyISAM DEFAULT CHARSET=utf8mb4 COLLATE=utf8mb4_unicode_520_ci;

INSERT INTO `activity_log` (`id`, `prj_id`, `svc_id`, `loc_id`, `status`, `created_by`, `created_at`, `deleted_by`, `deleted_at`) VALUES
(1, 7751, 7825, 5321, 1, 1, '2018-02-26 06:16:52', NULL, NULL),
(2, 7751, 7821, 5321, 0, 1, '2018-02-26 06:17:05', 1, '2018-02-26 06:17:33'),
(3, 7751, 7821, 5323, 1, 1, '2018-02-26 06:17:05', NULL, NULL),
(4, 7751, 7825, 5324, 0, 1, '2018-02-26 06:16:52', 1, '2018-02-26 06:17:33'),
(5, 7751, 7825, 5325, 1, 1, '2018-02-26 06:16:52', NULL, NULL),
(6, 7751, 7825, 5326, 0, 1, '2018-02-26 06:16:52', 1, '2018-02-26 06:17:33'),
(7, 7751, 7821, 5327, 0, 1, '2018-02-26 06:17:05', 1, '2018-02-26 06:17:33'),
(8, 7751, 7825, 5326, 1, 1, '2018-02-26 06:16:52', NULL, NULL);

查询1 ：显示所有结果

SELECT * FROM `activity_log`

Results ：

| id | prj_id | svc_id | loc_id | status | created_by |           created_at | deleted_by |           deleted_at |
|----|--------|--------|--------|--------|------------|----------------------|------------|----------------------|
|  1 |   7751 |   7825 |   5321 |   true |          1 | 2018-02-26T06:16:52Z |     (null) |               (null) |
|  2 |   7751 |   7821 |   5321 |  false |          1 | 2018-02-26T06:17:05Z |          1 | 2018-02-26T06:17:33Z |
|  3 |   7751 |   7821 |   5323 |   true |          1 | 2018-02-26T06:17:05Z |     (null) |               (null) |
|  4 |   7751 |   7825 |   5324 |  false |          1 | 2018-02-26T06:16:52Z |          1 | 2018-02-26T06:17:33Z |
|  5 |   7751 |   7825 |   5325 |   true |          1 | 2018-02-26T06:16:52Z |     (null) |               (null) |
|  6 |   7751 |   7825 |   5326 |  false |          1 | 2018-02-26T06:16:52Z |          1 | 2018-02-26T06:17:33Z |
|  7 |   7751 |   7821 |   5327 |  false |          1 | 2018-02-26T06:17:05Z |          1 | 2018-02-26T06:17:33Z |
|  8 |   7751 |   7825 |   5326 |   true |          1 | 2018-02-26T06:16:52Z |     (null) |               (null) |

查询2 ：显示离线组合

SELECT * FROM `activity_log` WHERE `status` = 0 GROUP BY `svc_id`, `loc_id` ORDER BY `id` DESC;

Results ：

| id | prj_id | svc_id | loc_id | status | created_by |           created_at | deleted_by |           deleted_at |
|----|--------|--------|--------|--------|------------|----------------------|------------|----------------------|
|  7 |   7751 |   7821 |   5327 |  false |          1 | 2018-02-26T06:17:05Z |          1 | 2018-02-26T06:17:33Z |
|  6 |   7751 |   7825 |   5326 |  false |          1 | 2018-02-26T06:16:52Z |          1 | 2018-02-26T06:17:33Z |
|  4 |   7751 |   7825 |   5324 |  false |          1 | 2018-02-26T06:16:52Z |          1 | 2018-02-26T06:17:33Z |
|  2 |   7751 |   7821 |   5321 |  false |          1 | 2018-02-26T06:17:05Z |          1 | 2018-02-26T06:17:33Z |

预期结果： ID 6应排除，因为{8}已在ID 8中激活

combination

SQL Fiddle Link

Answer 1

CREATE TABLE `activity_log` (
  `id` int(11) NOT NULL,
  `svc_id` bigint(20) NOT NULL,
  `loc_id` bigint(20) NOT NULL,
  `status` tinyint(1) NOT NULL
);

INSERT INTO `activity_log` (`id`, `svc_id`, `loc_id`, `status`) VALUES
(1,  7825, 5321, 1),
(2,  7821, 5321, 0),
(3,  7821, 5323, 1),
(4,  7825, 5324, 0),
(5,  7825, 5325, 1),
(6,  7825, 5326, 0),
(7,  7821, 5327, 0),
(8,  7825, 5326, 1);

SELECT a.* 
  FROM activity_log a 
  JOIN 
     ( SELECT MAX(id) id FROM activity_log GROUP BY svc_id,loc_id) b 
    ON b.id = a.id 
 WHERE a.status = 0;
+----+--------+--------+--------+
| id | svc_id | loc_id | status |
+----+--------+--------+--------+
|  2 |   7821 |   5321 |      0 |
|  4 |   7825 |   5324 |      0 |
|  7 |   7821 |   5327 |      0 |
+----+--------+--------+--------+

编辑：为了证明cdaiga解决方案的谬误，请考虑这个简化的例子......

DROP TABLE IF EXISTS activity_log;

CREATE TABLE `activity_log` 
(`id` int(11) NOT NULL
,`svc_id` bigint(20) NOT NULL
,`loc_id` bigint(20) NOT NULL
,`status` tinyint(1) NOT NULL
, INDEX(loc_id,svc_id)
);

INSERT INTO `activity_log` (`id`, `svc_id`, `loc_id`, `status`) VALUES
( 1, 5, 1, 1),
( 2, 1, 1, 0),
( 3, 1, 3, 1),
( 4, 5, 4, 0),
( 5, 5, 5, 1),
( 6, 5, 6, 0),
( 7, 1, 7, 0),
( 8, 5, 6, 1),
( 9, 5, 1, 0),
(10, 5, 1, 1),
(11, 1, 1, 0);

正确答案如下：

SELECT a.* 
  FROM activity_log a 
  JOIN 
     ( SELECT MAX(id) id FROM activity_log GROUP BY svc_id,loc_id) b 
    ON b.id = a.id 
 WHERE a.status = 0;

+----+--------+--------+--------+
| id | svc_id | loc_id | status |
+----+--------+--------+--------+
|  4 |      5 |      4 |      0 |
|  7 |      1 |      7 |      0 |
| 11 |      1 |      1 |      0 |
+----+--------+--------+--------+

然而，cdaiga的回答可能会给出错误的结果，如下所示：

SELECT A.*
FROM `activity_log` A
WHERE A.`status` = 0 AND
NOT EXISTS (SELECT null
            FROM `activity_log` B
            WHERE A.`svc_id`=B.`svc_id` AND  A.`loc_id`=B.`loc_id`
            AND B.`status`=1)
GROUP BY `svc_id`, `loc_id`
ORDER BY `id` DESC;
+----+--------+--------+--------+
| id | svc_id | loc_id | status |
+----+--------+--------+--------+
|  7 |      1 |      7 |      0 |
|  4 |      5 |      4 |      0 |
|  2 |      1 |      1 |      0 |
+----+--------+--------+--------+

用于识别最新非活动组合的SQL查询，同时保留活动组合

1 个答案: