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我有以下用于考勤系统的数据库架构：

数据库模式

我将如何编写 SQL 查询来生成第 X 天的良好条目报告？我需要它来生成一份报告

员工姓名 | 时间 | 超时
鲍勃 | 10:00 | 11:00
山姆 | 10:30 | 18:00
鲍勃 | 11:30 | 15:00

但是定义是进入还是退出的行由 entryType 设置（1 进入，0 退出），所以我将别名 TimeIn 和 TimeOut。

我的尝试是

`SELECT firstName, time from log INNER JOIN users on log.employeeID = users.employeeID WHERE date = GETDATE()`

但这并不能解决有些时候是进入，有些是退出的事实。

请注意，每个日期可以有多个登录。

更新：

另一个尝试，但子查询返回多行

select firstName, (select time as timeIn from log where entryType = 1), (select time as timeOut from log where entryType = 0) inner join users on log.uID = users.uID from log group by uID
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3 回答 3

1

简单地说，这将起作用。试试这个

SELECT FORENAME,SURNAME,LG.IN_TIME,LG.OUT_TIME FROM EMPLOYEES EMP INNER JOIN 
(SELECT MIN(TIME) IN_TIME,MAX(TIME) OUT_TIME,EMPLOYEE_ID FROM LOG
GROUP BY EMPLOYEE_ID) LG ON EMP.EMPLOYEE_ID=LG.EMPLOYEE_ID

注意：我没有包括条目类型，因为在任何时候都会刷入最短时间，刷出最长时间

更新

要显示没有登录和退出尝试这样的事情，

 SELECT FORENAME,SURNAME,LG.IN_TIME,LG.OUT_TIME,LG.no_of_ins,
LG.no_of_outs FROM EMPLOYEES EMP INNER JOIN 
(SELECT MIN(TIME) IN_TIME,MAX(TIME) OUT_TIME,EMPLOYEE_ID,
COUNT( CASE WHEN ENTRY_TYPE='I' THEN 1 ELSE O END noi) no_of_ins,
COUNT( CASE WHEN ENTRY_TYPE='O' THEN 1 ELSE O END nou) no_of_outs,
GROUP BY EMPLOYEE_ID) LG ON EMP.EMPLOYEE_ID=LG.EMPLOYEE_ID
于 2013-07-05T09:24:05.387 回答
1

这在 Oracle 中有效（为非 ANSI 风格道歉，但你应该明白这一点）..

SELECT FORENAME,SURNAME,L1.TIME IN_TIME,L2.TIME OUT_TIME
FROM EMPLOYEES EMP, LOG L1, LOG L2
WHERE EMP.EMPLOYEE_ID = L1.EMPLOYEE_ID
AND EMP.EMPLOYEE_ID = L2.EMPLOYEE_ID
AND L1.ENTRYTYPE = 1
AND L2.ENTRYTYPE = 0
AND L2.TIME = (SELECT MIN(TIME) FROM LOG WHERE EMPLOYEE_ID = L2.EMPLOYEE_ID AND L2.ENTRYTYPE = 0 AND TIME > L1.TIME)

更新：

啊，是的，没有考虑到这一点。在这种情况下，您需要外部连接。像这样的东西（未经测试）：

SELECT FORENAME,SURNAME,L1.TIME IN_TIME,L2.TIME OUT_TIME
FROM EMPLOYEES EMP
INNER JOIN LOG L1 ON EMP.EMPLOYEE_ID = L1.EMPLOYEE_ID AND L1.ENTRYTYPE = 1
LEFT OUTER JOIN LOG L2 ON EMP.EMPLOYEE_ID = L2.EMPLOYEE_ID AND L2.ENTRYTYPE = 0
                       AND L2.TIME = (SELECT MIN(TIME) FROM LOG WHERE EMPLOYEE_ID = L2.EMPLOYEE_ID AND L2.ENTRYTYPE = 0 AND TIME > L1.TIME)
于 2013-07-05T11:56:47.853 回答
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此查询将为您提供员工的最早入职时间和最晚离职时间。

 SELECT E.FORENAME, 
(SELECT MIN(TIME) FROM LOG WHERE EMPLOYEEID = E.EMPLOYEEID AND ENTRYTYPE = 1 AND DATE = <YOUR DAYE>) AS "TIME_IN", 
(SELECT MAX(TIME) FROM LOG WHERE EMPLOYEEID = E.EMPLOYEEID AND ENTRYTYPE = 0 AND DATE = <YOUR DAYE>) AS "TIME_OUT" 
FROM EMPLOYEE E WHERE E.EMPLOYEEID = <EMPLOYEE ID>
于 2013-07-05T08:54:03.410 回答