你好我有这样的表:
员工
EmployeeID EmployeeName DepartementName
1234 Seulong Accounting
1235 Jokwon Accounting
1236 Jinwoon IT
1237 Changmin IT
1238 Junho IT
1239 Taecyeon IT
Shiftsceduling
(d代表日期,例子d01 =日期1,d02 =日期2等)
employeeID_shift month d01 d02 d03 d04 d05 d06 d07 d08 d09 d10
1234 7 1 1 2 3 0 1 1 2 1 2
1235 7 1 2 1 2 0 1 2 3 0 1
1236 7 1 2 3 0 2 1 1 1 1 1
1237 7 1 3 1 1 1 0 1 1 0 1
1238 7 0 2 1 3 2 1 1 2 1 1
1239 7 1 1 1 1 1 1 0 0 2 1
ShiftCode
idshift start end information
0 00:00:00 00:00:00 OFF
1 08:00:00 16:00:00 8am - 16pm
2 09:00:00 18:00:00 9am - 18pm
3 16:00:00 04:00:00 20pm - 04am
车
PoliceNumber EmployeeID_car
J 0115 JYP 1234
J 0634 JYP 1235
J 1227 JYP 1236
J 0430 JYP 1237
J 0125 JYP 1238
J 0211 JYP 1239
我有一个功能可以爆炸今天的日期:
$date=date("Y-m-d");
$a = $date;
{
$b = explode('-', $a);
$year = $b[0];
$bulan = $b[1];
$date2 = $b[2];
}
echo $date2;
我想搜索7月2日09:00 - 17:00可用或工作的IT部门的所有员工ID,并显示他们使用的汽车。我已经使用了这段代码,并且与继续搜索他们的班次时间表相混淆:
SELECT Employee.* , Shiftsceduling.*, car.*, ShiftCode.*
FROM Employee
INNER JOIN Shiftsceduling ON Employee.EmployeeID = Shiftsceduling.employeeID_shift
INNER JOIN car ON Employee.EmployeeID = car.EmployeeID_car
WHERE Employee.DepartementName = 'IT' AND d'.$date2.' = '2'
AND ShiftCode.start <= '$starttime' AND ShiftCode.end >= '$endtime'
但它不起作用,你知道问题出在哪里吗?谢谢你的帮助
答案 0 :(得分:1)
我认为主要问题是你没有加入Shiftcode表。但是,通过两次检查时间段,你似乎也会让自己感到困惑。
您似乎在WHERE子句中检查时间段实际时间,检查开始和结束。但是你也检查Shiftsceduling的d01 / d02 / etc字段是2(这是09:00到18:00的时间段)。
未经测试,但我认为你想要这样的东西: -
SELECT Employee.* ,
Shiftsceduling.*,
car.*,
ShiftCode.*
FROM Employee
INNER JOIN Shiftsceduling ON Employee.EmployeeID = Shiftsceduling.employeeID_shift
INNER JOIN car ON Employee.EmployeeID = car.EmployeeID_car
INNER JOIN ShiftCode ON ShiftCode.idshift = Shiftsceduling.d'.$date2.'
WHERE Employee.DepartementName = 'IT'
AND ShiftCode.start <= '$starttime'
AND ShiftCode.end >= '$endtime'