PySiril Message output

Software Siril
1

I have started working with PySiril to automate some processing via Python. I have been able to get the basic implementation running but I would like to be able to control the message output. I looked at the code but I can’t find a way to access the message logger to reduce the level and only see basic progress of the program. I would also like to redirect messages to a log file but not silence the command window. Anyone have an idea how to do this?

2

Hi,

there’s a Configure method for trace class.

So smthg like:

from pysiril.siril import Siril
app=Siril()
app.tr.Configure(0,1,1,1)

Should shut Info messages and keep only normal log, errors and warnings.

Now to redirect to both a log file and stdout, you can use a class like this one.
Then in your code, just redirect stdout to a Logger instance (and redirect back when it ends)

So smthg like:

saveout = sys.stdout
sys.stdout = Logger()

try:
    #do smthg
except:
    #catch exceptions
finally:
    sys.stdout = saveout

should do what you want. There may be more elegant ways, this one has worked for me.

Cheers

C.

3

Thanks for the advice. I’ll give it a try.

GB