Question

MySQL查询有点问题：

表就像这样

id  timedate            user  action
1   2010-01-01 10:01:01 AB991 login
2   2010-01-01 12:01:01 AB991 logout
3   2010-01-01 14:01:01 AB991 login
4   2010-01-01 18:01:01 AB991 logout
5   2010-01-01 10:01:01 ZM991 login
6   2010-01-01 10:01:01 ZM991 logout
7   2010-01-02 10:01:01 AB991 login
8   2010-01-02 18:01:01 AB991 logout
9   2010-01-02 10:01:01 ZM991 login

现在，我需要一个查询或一些PHP代码来逐日计算每个用户登录和注销之间的分钟差异。但是可能存在一些问题，例如：

可能没有注销登录
有一天我可能会有2次登录和2次注销操作。

有智能解决方案（可能只有MySQL）或者我必须使用PHP并制作代码墙吗？

Answer 1

可能的解决方案如下：

select id, timedate, action
from mytable
where user = 'AB991'
order by timedate

您的PHP代码将按日期获得有序列表。该算法应如下所示：

$user = 'AB991';
$stmt = $db->prepare("select id, timedate, action from mytable where user = ? order by timedate");
$stmt->execute(array($user));
$rows = $stmt->fetchAll(PDO::FETCH_ASSOC);
$previousTime = "";
foreach ($rows as $row) {
    if ($row["action"] === "login") {
        $previousTime = strtotime($row["timedate"]);
    } else if ($row["action"] === "logout") {
        if ($previousTime === "") {
            //duplicate logout
        } else {
            //add the remaining time to the day when the user logged in
            //add the whole days passed between log in and log out
            //add the start of the last day when the user logged in
            $previousTime = "";
        }
    }
}

<强>问题：

用户可以单独使用不同的会话登录
用户可能无法注销
用户可能会使用其他会话退出

建议：我会修改表格以包含会话ID和最后一个操作。当用户执行操作时，最后一个操作的时间戳将被更新，因此您将拥有基于会话的数据，其中每一行都会说明一切，显示登录时间和上次操作时间。

Answer 2

我认为这个解决方案有所帮助。作为基础每次登录由logout流动来解释这里的代码我加入所有登录记录（排名）和所有登出记录（排名）因此，如果一次登录将计数1并且如果由登出计数1流动，则将被选中

select
logIn. *, logOut.* ,
TIMESTAMPDIFF(MINUTE, logIn.timedate, logOut.timeDate) as howLongUserStaysInMinutes

from (SELECT @row_counter1 := 0) x,
     (SELECT @row_counter2 := 0) y,
     (select userLogs.* , @row_counter1 := @row_counter1 +1 as row 
      from userLogs 
      where userLogs.action ='login') as logIn

 left join (select * ,   @row_counter2 :=  @row_counter2 + 1 as row 
             from  userLogs 
             where  userLogs.action = 'logout') as logOut 
             on  logOut.user = logIn.user
             and ( logIn.timedate <= logOut.timedate ||   logOut.timedate is null) 
             and logIn.row = logOut.row

我从内到左进行了更改连接，所以我一次登录并且没有注销，并且从行号到按日排名所以每天都会计入登录/注销从1开始，我曾经用Microsoft SQL Server执行此操作。让我们开始解释查询如何工作我需要的第一个数据但首先你需要准备好以下思考的步骤：答：我需要数据（＆＃39;退出记录＆＃39;以及＆＃39;登录记录＆＃39;）＆lt; - 这里您可以应用每日选择或最大值或相同值的逻辑 B.然后我想应用条件来获取这些数据并正确过滤它 - 这里完成加入登录和注销记录，以便你可以正确匹配它们

1.我希望获得每个用户的所有登录操作，并使用变量@row_counter为每一行提供一个数字，并为每一行增加1

  select id, 
         timedate as timedate, # here you can use max() to get max login in day
         user,
         action, # action will be login since it specified at where clause
         @row_counter1 := if(@perv_day1 <> DAY( userLogs.timedate ) , 1 , @row_counter1 +1) rankedByDay
        ,@perv_day1 := if(@perv_day1 <> DAY( userLogs.timedate ) , DAY( userLogs.timedate ) ,  @perv_day1) _day
  from userLogs,
       (select  @row_counter1 :=0) x # here I define the row with 0
       (select @perv_day1 :=0) y # to track rank for each row 
  where userLogs.action ='login'
  group by user # here i group each user
          , DAY( userLogs.timedate ) # also you can group by day

出：

id  timedate            user    action  rowRankedByDay
1   2010-01-01 10:01:01 AB991   login   1
3   2010-01-01 14:01:01 AB991   login   2
5   2010-01-01 10:01:01 ZM991   login   3
7   2010-01-02 10:01:01 AB991   login   1
9   2010-01-02 10:01:01 ZM991   login   2

2.我对注销行动

也这样做

  select id, 
         timedate as timedate, # here you can use max() to get max login in day
         user,
         action, # action will be logout since it specified at where clause
         @row_counter2 := if(@perv_day2 <> DAY( userLogs.timedate ) , 1 , @row_counter1 +1) rankedByDay,
         @perv_day1 := if(@perv_day1 <> DAY( userLogs.timedate ) , DAY( userLogs.timedate ) ,  @perv_day1) _day
  from userLogs , 
       (select  @row_counter2 :=0) y # here I define the row with 0
       (select @perv_day2 :=0) y # to track rank for each row 
  where userLogs.action ='logout'
  group by user # here i group each user

out

id  timedate            user    action  rowRankedByDay
2   2010-01-01 12:01:01 AB991   logout  1
4   2010-01-01 18:01:01 AB991   logout  2
6   2010-01-01 10:01:01 ZM991   logout  3
8   2010-01-02 18:01:01 AB991   logout  1

现在我有了数据，我想加入它们，这样我就可以找到时间上的差异但是需要做些什么条件，

select TIMESTAMPDIFF(MINUTE, logIn.timedate, logOut.timeDate) as howLongUserStaysInMinutes
from logIn
left join logOut on (logOut.user =  logIn.user) # same user
              and DAY( logIn.timedate )  =  DAY( logOut.timedate ) # action at the same day
              and logIn.rowRankedByDay= logOut.rowRankedByDay # login and logout actions rows
              and ( logIn.timedate <= logOut.timedate ||   logOut.timedate is null) # login was before logout or user did not logout

很抱歉，如果这里有更多代码，;）我希望这对你有帮助。

带有子句的TIMESTAMPDIFF

2 个答案: