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。