Good! Richard
On Thu, Sep 5, 2013 at 5:55 AM, António Ramos <ramstei...@gmail.com> wrote: > Bingo! > having =(count1 % 2* *==* *1) > > > 2013/9/5 António Ramos <ramstei...@gmail.com> > >> Going with SQLTABLE i have my rows queried like >> >> count1=db.card_logs.id.count() >> >> (db.........)select(db.trabalhador.nome,db.trabalhador.area,count1,groupby=db.trabalhador.nome, >> *having =(count1 % 2 = 1)*) >> >> >> the *having* clause is not accepted , how to do it to query only odd >> counts? >> >> >> I use sqlite! >> >> >> >> >> 2013/9/4 Richard Vézina <ml.richard.vez...@gmail.com> >> >>> Ok, so you need to check against actual time which person are still in... >>> >>> If you really just want the persons/users that are still in a give time, >>> I think you need a group by over user_id... So you could set a limit of 2 >>> records per user and if you have only one and it is a 'check in' stat value >>> you know this user is still in... >>> >>> Though, SQLFORM.grid() doesn't support aggregation ( >>> https://groups.google.com/forum/#!topic/web2py-developers/0pEmptLdND8) since >>> you pass it a query... >>> >>> So, maybe you should think of using the old SQLTABLE or build a custom >>> table yourself with TABLE() and other helpers provide by web2py... >>> >>> First you should try to wrote the SQL query to make sure you can solve >>> this with a single query, if you can, you may use SQLTABLE if you can't and >>> have to iter over record your only remaning option will be a custom table I >>> guess... >>> >>> I can help with the SQL if you need help, but I would need a dumy table >>> with a data sample... >>> >>> :) >>> >>> Richard >>> >>> >>> >>> >>> >>> >>> On Wed, Sep 4, 2013 at 1:17 PM, António Ramos <ramstei...@gmail.com>wrote: >>> >>>> If the user checks his rfid tag within 5 minutes of the last check the >>>> log does not record "check in" or "check out" but "error" >>>> >>>> >>>> 2013/9/4 António Ramos <ramstei...@gmail.com> >>>> >>>>> yes , stat is a string with "chech in" or "check out" i also have a >>>>> timestamp for the check in or check out. >>>>> >>>>> I dont want odd /even hours >>>>> >>>>> I want to know if the user is checked in >>>>> >>>>> I have an RFID app to check in /out outside workers via an rfid tag. >>>>> >>>>> when the user checks firstime, the app records "check in" , after that >>>>> , checking again his rfid tag the app logs "check out" . >>>>> >>>>> During the day the worker can go out to lunch and checks out, after >>>>> lunck checks in again. >>>>> >>>>> >>>>> so for a user i can have >>>>> >>>>> user a check in (time...) >>>>> user a check out (time...) >>>>> user a check in (time...) >>>>> >>>>> >>>>> I want to create a grid to show who is inside, so i need a query of >>>>> odd rfid checks for each user >>>>> >>>>> >>>>> >>>>> 2013/9/4 Richard Vézina <ml.richard.vez...@gmail.com> >>>>> >>>>>> Don't understand what you need exactly... Is stat a string type >>>>>> containing 'check in' or 'check out' and you want just odd number of >>>>>> record >>>>>> or you have an other field with timestamp or something and you want only >>>>>> the odd hours to appear in the grid?? >>>>>> >>>>>> Richard >>>>>> >>>>>> >>>>>> On Wed, Sep 4, 2013 at 12:54 PM, António Ramos >>>>>> <ramstei...@gmail.com>wrote: >>>>>> >>>>>>> >>>>>>> hello i need to use the sqlform.grid(query,etc... to show >>>>>>> records >>>>>>> >>>>>>> *my problem* >>>>>>> the query is not just like >>>>>>> ((db.tab1.stat='check in')|(db.tab1.stat='check out')) >>>>>>> >>>>>>> >>>>>>> >>>>>>> i need to query only records that appear with stat='check in' or >>>>>>> 'check out' odd times in tab1 and not even times >>>>>>> >>>>>>> >>>>>>> For example , this should not be in the grid >>>>>>> >>>>>>> user a check in >>>>>>> user a check out >>>>>>> user a check in >>>>>>> user a check out >>>>>>> >>>>>>> >>>>>>> but this should >>>>>>> >>>>>>> user a check in >>>>>>> user a check out >>>>>>> user a check in >>>>>>> >>>>>>> >>>>>>> how do i create a query for this? to use in sqlform.grid(query, .... >>>>>>> >>>>>>> -- >>>>>>> >>>>>>> --- >>>>>>> You received this message because you are subscribed to the Google >>>>>>> Groups "web2py-users" group. >>>>>>> To unsubscribe from this group and stop receiving emails from it, >>>>>>> send an email to web2py+unsubscr...@googlegroups.com. >>>>>>> For more options, visit https://groups.google.com/groups/opt_out. >>>>>>> >>>>>> >>>>>> -- >>>>>> >>>>>> --- >>>>>> You received this message because you are subscribed to the Google >>>>>> Groups "web2py-users" group. >>>>>> To unsubscribe from this group and stop receiving emails from it, >>>>>> send an email to web2py+unsubscr...@googlegroups.com. >>>>>> For more options, visit https://groups.google.com/groups/opt_out. >>>>>> >>>>> >>>>> >>>> -- >>>> >>>> --- >>>> You received this message because you are subscribed to the Google >>>> Groups "web2py-users" group. >>>> To unsubscribe from this group and stop receiving emails from it, send >>>> an email to web2py+unsubscr...@googlegroups.com. >>>> For more options, visit https://groups.google.com/groups/opt_out. >>>> >>> >>> -- >>> >>> --- >>> You received this message because you are subscribed to the Google >>> Groups "web2py-users" group. >>> To unsubscribe from this group and stop receiving emails from it, send >>> an email to web2py+unsubscr...@googlegroups.com. >>> For more options, visit https://groups.google.com/groups/opt_out. >>> >> >> > -- > > --- > You received this message because you are subscribed to the Google Groups > "web2py-users" group. > To unsubscribe from this group and stop receiving emails from it, send an > email to web2py+unsubscr...@googlegroups.com. > For more options, visit https://groups.google.com/groups/opt_out. > -- --- You received this message because you are subscribed to the Google Groups "web2py-users" group. 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