sql查询，我希望以正确的顺序安排时间表记录

Question

我执行sql查询时得到的输出是

的形式

date      punchin     punchout

05/20/19  12:00 pm    2:00pm    
05/20/19   2:20pm     4:00pm    
05/20/19   4:30pm     5:30pm    
05/20/19   8:30am     9:30am    
05/20/19   9:45am    12:00 am

但所需的结果是

date     punchin    punchout 

05/20/19   8:30am    9:30am   
05/20/19   9:45am   12:00 am    
05/20/19  12:00 pm   2:00pm   
05/20/19   2:20pm    4:00pm   
05/20/19   4:30pm    5:30pm

这是我到目前为止使用的SQL：

SELECT DISTINCT
       P.PERSONID,
       P.PERSONNUM,
       P.PERSONFULLNAME,
       P.HOMELABORLEVELNM5 AS STOREID,
       CAST(T.EVENTDATE AS varchar(11)) AS Day,
       RIGHT(T.STARTDTM, 8) + ' -' + RIGHT(T.ENDDTM, 8) AS SHIFT,
       T.TIMEINSECONDS / 3600.00 AS [Hours Worked]
FROM VP_EMPLOYEEV42 AS P
     INNER JOIN VP_TIMESHEETITMV42 AS T ON P.EMPLOYEEID = T.EMPLOYEEID
WHERE T.EVENTDATE >= '2016-12-01'
  AND P.EMPLOYMENTSTATUS = 'ACTIVE'
  AND P.HOMELABORLEVELNM5 IN ('8166', '8111', '8168', '8171', '8175', '8197', '8233', '8236', '8255', '8267', '8313', '8320', '8325', '8328''8331', '8333', '8334', '8335', '8336', '8337', '8339', '8340', '8341', '8344', '8346', '8348', '8359', '8360', '8361', '8362', '8363', '8364', '8365', '8366', '8367', '8368', '8369', '8370', '8372', '8374', '8375', '8376', '8377', '8379', '8380', '8386', '8391', '8392', '8393', '8394', '8395', '8396', '8413', '8415', '8416', '8427', '8444', '16992', '21036', '21043', '21066', '21067', '69902', '71021', '72046')
GROUP BY P.PERSONID,
         P.PERSONNUM,
         P.PERSONFULLNAME,
         P.HOMELABORLEVELNM5,
         T.EVENTDATE,
         T.TIMEINSECONDS,
         T.STARTDTM,
         T.ENDDTM
ORDER BY P.HOMELABORLEVELNM5;