Question

我有一个userRegister表，其中包含以下字段：

    **userRegister table:**

    id | name | created            | login               | logout | userId
------------------------------------------------------------------------
1  |test | 2018-02-19 11:34:00  | 2018-02-19 11:34:00 | NULL  |  1
2  |test | 2018-02-21 14:01:42 | 2018-02-21 14:01:42 | NULL   |  1
3  |test | 2018-02-21 15:24:05 | 2018-02-21 15:24:05 | NULL   |  1
4  |test | 2018-02-22 16:46:15 | 2018-02-22 16:46:15 | NULL   |  1
5  |test2 | 2018-02-27 09:51:19 | 2018-02-27 09:51:19 | NULL  |  2
6  |test2 | 2018-02-28 11:59:24 | 2018-02-28 11:59:24 | NULL  |  2
7  |test  |2018-03-01 10:37:59 | 2018-03-01 10:37:59 | NULL   |  1
8  |test  |2018-03-01 10:39:52 | 2018-03-01 10:39:52 | NULL   |  1
9  |test  |2018-03-01 10:41:49 | 2018-03-01 10:41:49 | NULL   |  1
10  |test  |2018-03-01 15:34:52 | 2018-03-01 15:34:52 | NULL  |  1

在这里，我需要检查用户是否在同一天多次登录而不登出我需要使用之前的登录时间更新logout字段

如何编写一个查询，按创建日期对用户进行分组，以便多次登录而无需注销

我试过的代码：

db.query("select * from userRegister where logout is NULL limit 10", function (err, user) {
        if (!_.size(user)){
            console.log([])
        }
        else{
            var result=_.chain(user).groupBy("created").map(function(v, i) {
                return {
                        created: i,
                        login: _.map(v, 'login'),
                        userId: _.map(v, 'userId')
                    }
            }).value()
        }
})

输出：

[ { created: '2018-02-19 11:34:00',
    login: [ '2018-02-19 11:34:00' ],
    userId: [ '1' ] },
  { created: '2018-02-21 14:01:42',
    login: [ '2018-02-21 14:01:42' ],
    userId: [ '1' ] },
  { created: '2018-02-21 15:24:05',
    login: [ '2018-02-21 15:24:05' ],
    userId: [ '1' ] },
  { created: '2018-02-22 16:46:15',
    login: [ '2018-02-22 16:46:15' ],
    userId: [ '1' ] },
  { created: '2018-02-27 09:51:19',
    login: [ '2018-02-27 09:51:19' ],
    userId: [ '2' ] },
  { created: '2018-02-28 11:59:24',
    login: [ '2018-02-28 11:59:24' ],
    userId: [ '2' ] },
  { created: '2018-03-01 10:37:59',
    login: [ '2018-03-01 10:37:59' ],
    userId: [ '1' ] },
  { created: '2018-03-01 10:39:52',
    login: [ '2018-03-01 10:39:52' ],
    userId: [ '1' ] },
  { created: '2018-03-01 10:41:49',
    login: [ '2018-03-01 10:41:49' ],
    userId: [ '1' ] },
  { created: '2018-03-01 15:34:52',
    login: [ '2018-03-01 15:34:52' ],
    userId: [ '1' ] } ]

输出应按创建日期分组（相同日期），还是有其他方法可以做到这一点？请建议

Answer 1

SELECT name, userId, DATE(login) loginDate
FROM userRegister 
WHERE logout IS NULL
GROUP BY name, userId, DATE(login)
HAVING COUNT(*)>1;

查看Demo on SQL Fiddle。

Answer 2

你可以这样做：编辑：我的坏我已经忘记了小组

select userId, name, created, date(login) as date_login, count(*) as nb_login
from YourTable
WHERE logout IS NULL
GROUP BY userId, date_login
HAVING nb_login > 1

使用注销字段更新进行编辑

SET @last_login:=NULL;

UPDATE userRegister r JOIN (
SELECT name, userId, login, if(logout is NULL, @last_login, logout) as logout
, @last_login :=login as dummy
FROM userRegister 
ORDER BY userId, login desc) d USING(userId, login) SET r.logout = d.logout;

在mysql中按相同日期分组

2 个答案: