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 <nip...@gmail.com 
> <javascript:>>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, theoffi...@gmail.com wrote:
>>>
>>> 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!
>>>
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