Mysql如何查询连续的时间次数

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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，连续日期取最大的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。

--------------------------------------------Oracle可以这样做-------------------------------------------------

create table TSQL_TEST_LIANXU_4
(
  ID   NUMBER(4) not null,
  U_ID NUMBER(4),
  STS  TIMESTAMP(6),
  ETS  TIMESTAMP(6)
);

insert into TSQL_TEST_LIANXU_4 (ID, U_ID, STS, ETS)
values (1, 1, to_timestamp('01-01-2014 21:00:00.000000', 'dd-mm-yyyy hh24:mi:ss.ff'), to_timestamp('02-01-2014 07:00:00.000000', 'dd-mm-yyyy hh24:mi:ss.ff'));
insert into TSQL_TEST_LIANXU_4 (ID, U_ID, STS, ETS)
values (2, 1, to_timestamp('02-01-2014 15:37:57.000000', 'dd-mm-yyyy hh24:mi:ss.ff'), to_timestamp('02-01-2014 21:00:00.000000', 'dd-mm-yyyy hh24:mi:ss.ff'));
insert into TSQL_TEST_LIANXU_4 (ID, U_ID, STS, ETS)
values (3, 2, to_timestamp('01-01-2014 09:00:00.000000', 'dd-mm-yyyy hh24:mi:ss.ff'), to_timestamp('01-01-2014 15:00:00.000000', 'dd-mm-yyyy hh24:mi:ss.ff'));
insert into TSQL_TEST_LIANXU_4 (ID, U_ID, STS, ETS)
values (4, 2, to_timestamp('02-01-2014 09:00:00.000000', 'dd-mm-yyyy hh24:mi:ss.ff'), to_timestamp('01-02-2014 16:00:00.000000', 'dd-mm-yyyy hh24:mi:ss.ff'));
insert into TSQL_TEST_LIANXU_4 (ID, U_ID, STS, ETS)
values (5, 1, to_timestamp('04-01-2014 10:00:00.000000', 'dd-mm-yyyy hh24:mi:ss.ff'), to_timestamp('04-01-2014 18:00:00.000000', 'dd-mm-yyyy hh24:mi:ss.ff'));
insert into TSQL_TEST_LIANXU_4 (ID, U_ID, STS, ETS)
values (6, 1, to_timestamp('05-01-2014 12:00:00.000000', 'dd-mm-yyyy hh24:mi:ss.ff'), to_timestamp('05-01-2014 13:00:00.000000', 'dd-mm-yyyy hh24:mi:ss.ff'));
insert into TSQL_TEST_LIANXU_4 (ID, U_ID, STS, ETS)
values (7, 2, to_timestamp('10-01-2014 00:00:00.000000', 'dd-mm-yyyy hh24:mi:ss.ff'), to_timestamp('10-01-2014 06:00:00.000000', 'dd-mm-yyyy hh24:mi:ss.ff'));
insert into TSQL_TEST_LIANXU_4 (ID, U_ID, STS, ETS)
values (8, 2, to_timestamp('11-01-2014 13:00:00.000000', 'dd-mm-yyyy hh24:mi:ss.ff'), to_timestamp('11-01-2014 18:00:00.000000', 'dd-mm-yyyy hh24:mi:ss.ff'));
insert into TSQL_TEST_LIANXU_4 (ID, U_ID, STS, ETS)
values (10, 2, to_timestamp('12-01-2014 12:00:00.000000', 'dd-mm-yyyy hh24:mi:ss.ff'), to_timestamp('12-01-2014 18:00:00.000000', 'dd-mm-yyyy hh24:mi:ss.ff'));
insert into TSQL_TEST_LIANXU_4 (ID, U_ID, STS, ETS)
values (11, 1, to_timestamp('08-01-2014 06:00:00.000000', 'dd-mm-yyyy hh24:mi:ss.ff'), to_timestamp('08-01-2014 16:00:00.000000', 'dd-mm-yyyy hh24:mi:ss.ff'));

本来想使用row_number的，结果没搞定。

select t.u_id,
       to_char(MIN(t.sts), 'yyyy-mm-dd') start_date,
       to_char(MAX(t.sts), 'yyyy-mm-dd') end_date,
       trunc(MAX(t.sts)) - trunc(MIN(t.sts)) + 1
  from (select z.u_id, z.sts, trunc(z.sts) - trunc(z.min_days) - rownum rn
          from (select (select min(sts) from tsql_test_lianxu_4) min_days,
                       t.*
                  FROM tsql_test_lianxu_4 t
                 order by u_id, sts) z) t
 group by u_id, rn
 order by 1, 2

Oracle中这样查询是有问题的，就是uid相同sts在同一天的记录不能有2条，

Oracle另一种方法:

SELECT u_id, MIN(sts) AS STARTDATE, MAX(sts), COUNT(u_id) AS ENDNUM
  FROM (SELECT A.u_id,
               to_date(to_char(A.sts, 'yyyy-mm-dd'), 'yyyy-mm-dd') sts,
               to_date(to_char(A.sts, 'yyyy-mm-dd'), 'yyyy-mm-dd') - ROWNUM AS GNUM
          FROM (SELECT *
                  FROM tsql_test_lianxu_4
                 ORDER BY u_id,sts) A)
 GROUP BY u_id, GNUM
 ORDER BY u_id, MIN(sts)

缺点：uid相同sts在同一天的记录不能有2条

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。