Question

我想像这样显示每日报告

          Fulltime Contract Casual
2018/06/04  1       0        0
2018/06/05  1       0         0
2018/06/06  0       1         1
2018/06/07  2       1         0
2018/06/08  1       1         1
2018/06/09  0       1         1

但我拥有的是这样的

Date       Jobtype Meal
2018/06/04 Fulltime 1
2018/06/05 Fulltime 1
2018/06/06 Casual   1
2018/06/06 Contract 1
2018/06/07 Casual   1
2018/06/07 Contract 2
2018/06/08 Casual   1
2018/06/08 Contract 1
2018/06/08 Fulltime 1
2018/06/09 Casual   1
2018/06/09 Contract 1

我尝试过：

select Date, Jobtype,'Meal'=(COUNT(Date)) 
from CanLog 
where WW BETWEEN '2018/06/06' and '2018/06/09' 
group by Date, Jobtype 
order by 1

Answer 1

不确定WW列的来源。我以为它是日期栏。如果不是，请调整。

您需要使用UNPIVOT运算符并丢弃等于0的记录：

select u.date, u.jobtype, u.meal
from canlog
unpivot
(
  meal
  for jobtype in ( fulltime, contract, casual )
) u
where 
  [Date] between '2018/06/06' and '2018/06/09'
  and meal <> 0;

请参阅Live DEMO。

结果：

date        jobtype     meal
2018-06-04  Fulltime    1
2018-06-05  Fulltime    1
2018-06-06  Contract    1
2018-06-06  Casual      1
2018-06-07  Fulltime    2
2018-06-07  Contract    1
2018-06-08  Fulltime    1
2018-06-08  Contract    1
2018-06-08  Casual      1
2018-06-09  Contract    1
2018-06-09  Casual      1

Answer 2

我想你可以试试这个：

SELECT Date,
       (SELECT COUNT(*) FROM CanLog as c WHERE c.WW = clog.WW AND jobtype = 'fulltime') AS Fulltime,
       (SELECT COUNT(*) FROM CanLog as c WHERE c.WW = clog.WW AND jobtype = 'contract') AS Contract,
       (SELECT COUNT(*) FROM CanLog as c WHERE c.WW = clog.WW AND jobtype = 'casual') AS Casual
FROM CanLog AS clog
WHERE WW BETWEEN '2018/06/06' AND '2018/06/09' 
GROUP BY Date, Jobtype 
ORDER BY Date

在括号中选择当天给定字符串的数量。如果你从小的所有放大的字母写，这没关系，sql是不区分大小写的

Answer 3

选择不同的日期，（SELECT COUNT（）FROM CanLog as c WHERE c.WW = clog.WW AND jobtype ='fulltime'）AS Fulltime，（SELECT COUNT（）FROM CanLog as c WHERE c.WW = clog.WW AND jobtype ='contract'）AS Contract，（SELECT COUNT（*）FROM CanLog as c WHERE c.WW = clog.WW AND jobtype ='casual'）AS Casual FROM CanLog AS clog WW BETWEEN'2018 / 06/06'和'2018/06/09' GROUP BY日期，作业类型按日期排序

每日报告日期与mssql的多列

3 个答案: