it's fine, but what you need to check is for the user who started the web2py process to be able to use your /dev/serialwhatever interface.... How did you install web2py ?
PS: I have a raspberry too :P On Friday, March 15, 2013 9:28:09 AM UTC+1, The Organisation of Secret Shoppers wrote: > > so do you mean i must edit my controller and views from the terminal? > (currently i'm editing them through the web) and the current application > was created using application wizard, which i think it said that it have to > be modified in the wizard itself. > > > On Fri, Mar 15, 2013 at 3:33 PM, Christian Foster Howes > <[email protected]<javascript:> > > wrote: > >> the "user" for web2py depends on how you launch it... >> >> if you call: >> >> python web2py.py >> >> then it runs as you >> >> if you: >> >> sudo su bob >> python web2py.py >> >> it runs a bob. >> >> if you use apache or some other server i believe by default it runs as >> the user that is running apache. >> >> >> On Thursday, March 14, 2013 8:41:25 PM UTC-7, The Organisation of Secret >> Shoppers wrote: >> >>> hmm how do i check? i think running the python script has to be "root" >>> user? but i'm not sure what is the user type for the web2py... >>> >>> >>> On Thu, Mar 14, 2013 at 4:24 PM, Niphlod <[email protected]> wrote: >>> >>>> did you check that the user running your "working" script is the same >>>> as the one running the "I'm going wrong" web2py ? >>>> >>>> >>>> On Thursday, March 14, 2013 8:30:38 AM UTC+1, [email protected]: >>>>> >>>>> Hello guys, >>>>> >>>>> >>>>> i hope someone can help me with this! i have been stuck here for days. >>>>> I just created a web app using web2py and i want to use it to send some >>>>> serial commands to a device through Raspberry Pi. >>>>> >>>>> I have successfully sent the commands from R-Pi to the device by >>>>> running a Python script in the terminal. >>>>> >>>>> But i thought web2py is using python as well so i did something >>>>> similar but i get the error below: >>>>> *<class 'serial.serialutil.SerialException'> could not open port >>>>> /dev/ttyAMA0: [Errno 13] Permission denied: '/dev/ttyAMA0'* >>>>> >>>>> >>>>> my codes are: >>>>> >>>>> in Controller: >>>>> >>>>> >>>>> def test(): >>>>> >>>>> import serial >>>>> >>>>> serialport= serial.Serial("/dev/ttyAMA0", 9600, timeout=0.5) >>>>> >>>>> return dict() <-- not sure what to return >>>>> in view/default/index.html: >>>>> >>>>> <a href = "{{=URL >>>>> <https://10.0.0.132/examples/global/vars/URL>(c='default', >>>>> f='test')}}"><img src ="/Comfort2/static/images/off.****jpg" width ="75" >>>>> height="75"</a> >>>>> >>>>> >>>>> i created a default/test.html as well. >>>>> >>>>> >>>>> i have no idea what's wrong! :( give me some hints please. thank you! >>>>> >>>> -- >>>> >>>> --- >>>> You received this message because you are subscribed to a topic in the >>>> Google Groups "web2py-users" group. >>>> To unsubscribe from this topic, visit https://groups.google.com/d/** >>>> topic/web2py/tiWZkXMoo6E/**unsubscribe?hl=en<https://groups.google.com/d/topic/web2py/tiWZkXMoo6E/unsubscribe?hl=en> >>>> . >>>> To unsubscribe from this group and all its topics, send an email to >>>> web2py+un...@**googlegroups.com. >>>> >>>> For more options, visit >>>> https://groups.google.com/**groups/opt_out<https://groups.google.com/groups/opt_out> >>>> . >>>> >>>> >>>> >>> >>> -- >> >> --- >> You received this message because you are subscribed to a topic in the >> Google Groups "web2py-users" group. >> To unsubscribe from this topic, visit >> https://groups.google.com/d/topic/web2py/tiWZkXMoo6E/unsubscribe?hl=en. >> To unsubscribe from this group and all its topics, send an email to >> [email protected] <javascript:>. >> 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 [email protected]. For more options, visit https://groups.google.com/groups/opt_out.
