Question

美好的一天。

我有这样的环境：在Mysql数据库上，每次用户登录某个站点时，都会创建一个新行，包括他的名字和登录时间。由于系统是互斥的，因此在给定时间只有用户，如果新用户到达，则记录的用户将被注销。

现在他们已经要求我计算系统上所有用户的总时间，所以基本上，我必须将登录和下一个用户的所有时差加在一起。

user  |       timestamp     |
------------------------------
alpha | 2013-01-19 03:14:07
beta  | 2013-01-20 11:24:04
alpha | 2013-01-21 02:11:37
alpha | 2013-01-21 03:10:31    <---- a user could login twice, it is normal
gamma | 2013-01-21 11:24:04
beta  | 2013-01-21 11:25:00

我想问你的意见，因为有很多登录，这是计算用户总记录时间的最佳方法？在此示例中，“gamma”将具有56秒的登录时间，并且可以忽略最后一次登录beta，因为它将在执行此检查时在线。所以“beta”只有一个条目。

有没有办法通过查询计算它？或者最好添加一个“在线时间”列，并在每次用户注销在线时间时让系统计算？

Answer 1

如果要在MySQL中执行此操作，则需要自联接。因为MySQL没有内置的rownum功能，所以自我加入是一件痛苦的事情。但它仍然可行。

首先，我们需要创建一个子查询来创建一个模拟SELECT rownum, user, timestamp FROM login的虚拟表，我们可以这样做。 http://sqlfiddle.com/#!2/bf6ef/2/0

SELECT @a:=@a+1 AS rownum, user, timestamp
    FROM (
        SELECT user, timestamp
          FROM login
         ORDER BY timestamp
    ) C,
    (SELECT @a:=0) s

接下来，我们需要将此虚拟表自我连接到自身的副本。我们在这个结果集中想要的是表中所有连续行对的列表。该查询是一个毛球 - 它将结构化放在结构化查询语言中。但它的确有效。这是：http://sqlfiddle.com/#!2/bf6ef/4/0

SELECT first.user AS fuser, 
       first.timestamp AS ftimestamp,
       second.user AS suser,
       second.timestamp as stimestamp,
       TIMESTAMPDIFF(SECOND, first.timestamp, second.timestamp) AS timeloggedin

  FROM (
       SELECT @a:=@a+1 AS rownum, user, timestamp
         FROM (
             SELECT user, timestamp
               FROM login
           ORDER BY timestamp
              ) C,
          (SELECT @a:=0) s
        ) AS first
  JOIN (
       SELECT @b:=@b+1 AS rownum, user, timestamp
         FROM (
             SELECT user, timestamp
               FROM login
           ORDER BY timestamp
              ) C,
          (SELECT @b:=0) s
        ) AS second ON first.rownum+1 = second.rownum

比较连续行的整个技巧是

SELECT (virtual_table) AS first
  JOIN (virtual_table) AS second ON first.rownum+1 = second.rownum

查询模式。 rownum + 1 = rownum的东西收集连续行号的行。

接下来，我们需要总结该查询的结果，以获取每个用户登录的总时间。这将是这样的：

  SELECT user, SUM(timeloggedin) AS timeloggedin
    FROM (
          /* the self-joined query */
         ) AS selfjoin
   GROUP BY user
   ORDER BY user

看起来像这样：http://sqlfiddle.com/#!2/bf6ef/5/0

这是整个查询放在一起。

SELECT user, SUM(timeloggedin) AS timeloggedin
  FROM (
      SELECT first.user AS user, 
             TIMESTAMPDIFF(SECOND, first.timestamp, second.timestamp) AS timeloggedin
        FROM (
             SELECT @a:=@a+1 AS rownum, user, timestamp
         FROM (
                   SELECT user, timestamp
                     FROM login
                 ORDER BY timestamp
                    ) C,
                (SELECT @a:=0) s
              ) AS first
        JOIN (
             SELECT @b:=@b+1 AS rownum, user, timestamp
               FROM (
                   SELECT user, timestamp
                     FROM login
                 ORDER BY timestamp
                    ) C,
                (SELECT @b:=0) s
              ) AS second ON first.rownum+1 = second.rownum
         ) AS selfjoin
   GROUP BY user
   ORDER BY user

对于习惯于程序，算法，思考的人来说，这并不是真正直观的。但这是你在SQL中进行这种连续行比较的方式。

Answer 2

试试这个......你的问题的解决方案越来越少......

    CREATE TABLE `matteo` (
      `user` varchar(20) DEFAULT NULL,
      `timestamp` int(11) DEFAULT NULL
    ) ENGINE=InnoDB DEFAULT CHARSET=latin1;

    INSERT INTO matteo(user, `timestamp`) VALUES ('alpha', 7);
    INSERT INTO matteo(user, `timestamp`) VALUES ('beta', 9);
    INSERT INTO matteo(user, `timestamp`) VALUES ('alpha', 17);
    INSERT INTO matteo(user, `timestamp`) VALUES ('alpha', 27);
    INSERT INTO matteo(user, `timestamp`) VALUES ('gamma', 77);
    INSERT INTO matteo(user, `timestamp`) VALUES ('beta', 97);

    select a.*,b.*,b.`timestamp`-a.`timestamp` as delta
    from
    (SELECT @rownum := @rownum + 1 AS id,t.*
          FROM matteo t,(SELECT @rownum := 0) r) a
    join
    (SELECT @rownum2 := @rownum2 + 1 AS id,t.*
          FROM matteo t,(SELECT @rownum2 := 0) r) b 
    where a.id=b.id-1

:-)看你星期一!!!

在mysql表上计算经过的时间

2 个答案: