关于mysql取出数据库中连续日期（值）的例子，实现Oracle的row_number()

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转自：http://www.tuicool.com/articles/uyEZJf

在网上看到一道有意思的题目，大意是如何在mysql查询连续的时间内登录的次数。原文链接：

http://www.oschina.net/question/573517_118821

首先建表，填充测试数据：

CREATE TABLE `tmysql_test_lianxu_3` (  `id` int(11) NOT NULL AUTO_INCREMENT,  `uid` int(11) DEFAULT NULL,  `sts` datetime DEFAULT NULL COMMENT '登录时间',  `ets` datetime DEFAULT NULL COMMENT '离线时间',  PRIMARY KEY (`id`)) ENGINE=InnoDB AUTO_INCREMENT=9 DEFAULT CHARSET=utf8 COLLATE=utf8_bin

测试数据为：

INSERT INTO `tmysql_test_lianxu_3` VALUES (1, 1, '2014-1-1 21:00:00', '2014-1-2 07:00:00');INSERT INTO `tmysql_test_lianxu_3` VALUES (2, 1, '2014-1-2 15:37:57', '2014-1-2 21:00:00');INSERT INTO `tmysql_test_lianxu_3` VALUES (3, 2, '2014-1-1 09:00:00', '2014-1-1 15:00:00');INSERT INTO `tmysql_test_lianxu_3` VALUES (4, 2, '2014-1-2 09:00:00', '2014-2-1 16:00:00');INSERT INTO `tmysql_test_lianxu_3` VALUES (5, 1, '2014-1-4 10:00:00', '2014-1-4 18:00:00');INSERT INTO `tmysql_test_lianxu_3` VALUES (6, 1, '2014-1-5 12:00:00', '2014-1-5 13:00:00');INSERT INTO `tmysql_test_lianxu_3` VALUES (7, 2, '2014-1-10 00:00:00', '2014-1-10 06:00:00');INSERT INTO `tmysql_test_lianxu_3` VALUES (8, 2, '2014-1-11 13:00:00', '2014-1-11 18:00:00');INSERT INTO `tmysql_test_lianxu_3` VALUES (10, 2, '2014-1-12 12:00:00', '2014-1-12 18:00:00');INSERT INTO `tmysql_test_lianxu_3` VALUES (11, 1, '2014-1-8 06:00:00', '2014-1-8 16:00:00');INSERT INTO `tmysql_test_lianxu_3` VALUES (12, 2, '2014-1-11 21:00:00', '2014-1-12 06:00:00');

在Oracle中可以使用row_number搞定，mysql中怎么做呢？

可以参考链接：

http://www.explodybits.com/2011/11/mysql-row-number/

首先看原文中给出的答案：

SELECT uid, days, COUNT(*) AS num  FROM (SELECT uid,               @cont_day :=               (CASE                 WHEN (@last_uid = uid AND DATEDIFF(login_dt, @last_dt) = 1) THEN                  (@cont_day + 1)                 ELSE                  1               END) AS days,               (@cont_ix := (@cont_ix + IF(@cont_day = 1, 1, 0))) AS cont_ix,               @last_uid := uid,               @last_dt := login_dt          FROM (SELECT uid, DATE(sts) AS login_dt                  FROM tmysql_test_lianxu_3                 ORDER BY uid, sts) AS t,               (SELECT @last_uid := '',                       @last_dt  := '',                       @cont_ix  := 0,                       @cont_day := 0) AS t1) AS t2 GROUP BY uid, days;

也是使用了mysql模拟oracle的row_number函数。

运行结果是:

我看了半天发现结果好像不是我想要的，我想要的是要有开始时间，结束时间之类的。

看下中间表再说：

SELECT uid,               @cont_day :=               (CASE                 WHEN (@last_uid = uid AND DATEDIFF(login_dt, @last_dt)=1) THEN                  (@cont_day + 1)                 ELSE                  1               END) AS days,               (@cont_ix := (@cont_ix + IF(@cont_day = 1, 1, 0))) AS cont_ix,               @last_uid := uid,               @last_dt := login_dt login_day          FROM (SELECT uid, DATE(sts) AS login_dt                  FROM tmysql_test_lianxu_3                 ORDER BY uid, sts) AS t,               (SELECT @last_uid := '',                       @last_dt  := '',                       @cont_ix  := 0,                       @cont_day := 0) AS t1

结果为：

看了下可以这么做，连续日期去最大的days,开始时间，结束时间去login_day，而是这样写了：

SELECT uid, max(days) lianxu_days, min(login_day) start_date,max(login_day) end_date   FROM (SELECT uid,               @cont_day :=               (CASE                 WHEN (@last_uid = uid AND DATEDIFF(login_dt, @last_dt)=1) THEN                  (@cont_day + 1)                                 ELSE                  1               END) AS days,               (@cont_ix := (@cont_ix + IF(@cont_day = 1, 1, 0))) AS cont_ix,               @last_uid := uid,               @last_dt := login_dt login_day          FROM (SELECT uid, DATE(sts) AS login_dt                  FROM tmysql_test_lianxu_3                 ORDER BY uid, sts) AS t,               (SELECT @last_uid := '',                       @last_dt  := '',                       @cont_ix  := 0,                       @cont_day := 0) AS t1) AS t2 GROUP BY uid, cont_ix;

结果是：

这里存在的问题是：表里面的的sts登录时间不能有2条uid相同时间在同一天内。

解决方法是:在case中添加一个<1 的判断条件

SELECT uid, max(days) lianxu_days, min(login_day) start_date,max(login_day) end_date   FROM (SELECT uid,               @cont_day :=               (CASE                 WHEN (@last_uid = uid AND DATEDIFF(login_dt, @last_dt)=1) THEN                  (@cont_day + 1)                 WHEN (@last_uid = uid AND DATEDIFF(login_dt, @last_dt)<1) THEN                  (@cont_day + 0)                 ELSE                  1               END) AS days,               (@cont_ix := (@cont_ix + IF(@cont_day = 1, 1, 0))) AS cont_ix,               @last_uid := uid,               @last_dt := login_dt login_day          FROM (SELECT uid, DATE(sts) AS login_dt                  FROM tmysql_test_lianxu_3                 ORDER BY uid, sts) AS t,               (SELECT @last_uid := '',                       @last_dt  := '',                       @cont_ix  := 0,                       @cont_day := 0) AS t1) AS t2 GROUP BY uid, cont_ix;

存在的问题：

时间sts的时分秒不见了。

欢迎各位留下更好的查询SQL,如本文中的SQL有问题也请指出，谢谢。

全文完。

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