strftime () function in SQLlite and applying it for specific definition of the day












0














I try to use strftime('%d',time) function to make Group by () by the day. But the problem that I would like to define each day not like the usual day (from 00:00 to 23:59) but from 5:45 to 5:45 next day.



Is it possible to do in SQLlite?










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    0














    I try to use strftime('%d',time) function to make Group by () by the day. But the problem that I would like to define each day not like the usual day (from 00:00 to 23:59) but from 5:45 to 5:45 next day.



    Is it possible to do in SQLlite?










    share|improve this question

























      0












      0








      0







      I try to use strftime('%d',time) function to make Group by () by the day. But the problem that I would like to define each day not like the usual day (from 00:00 to 23:59) but from 5:45 to 5:45 next day.



      Is it possible to do in SQLlite?










      share|improve this question













      I try to use strftime('%d',time) function to make Group by () by the day. But the problem that I would like to define each day not like the usual day (from 00:00 to 23:59) but from 5:45 to 5:45 next day.



      Is it possible to do in SQLlite?







      sql sqlite time group-by strftime






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked Nov 19 '18 at 15:22









      Keithx

      89421834




      89421834
























          1 Answer
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          I believe that the following may do what you wish :-



          strftime('%d',time) - (strftime('%H:%M',time) < '05:45')


          That is if the time is before 05:45 then 1 (true) is subtracted from the day when determining the GROUP BY argument and thus becomes the previous day.






          share|improve this answer





















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            1 Answer
            1






            active

            oldest

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            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1














            I believe that the following may do what you wish :-



            strftime('%d',time) - (strftime('%H:%M',time) < '05:45')


            That is if the time is before 05:45 then 1 (true) is subtracted from the day when determining the GROUP BY argument and thus becomes the previous day.






            share|improve this answer


























              1














              I believe that the following may do what you wish :-



              strftime('%d',time) - (strftime('%H:%M',time) < '05:45')


              That is if the time is before 05:45 then 1 (true) is subtracted from the day when determining the GROUP BY argument and thus becomes the previous day.






              share|improve this answer
























                1












                1








                1






                I believe that the following may do what you wish :-



                strftime('%d',time) - (strftime('%H:%M',time) < '05:45')


                That is if the time is before 05:45 then 1 (true) is subtracted from the day when determining the GROUP BY argument and thus becomes the previous day.






                share|improve this answer












                I believe that the following may do what you wish :-



                strftime('%d',time) - (strftime('%H:%M',time) < '05:45')


                That is if the time is before 05:45 then 1 (true) is subtracted from the day when determining the GROUP BY argument and thus becomes the previous day.







                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered Nov 19 '18 at 19:30









                MikeT

                15k102441




                15k102441






























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