Python how to get a list of all linux signals but only string parts

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I'm trying to get a list of all signals in linux that are string so i can compare the user input with existing signals if true go on with the program if false end

What I have tried:

i tried to do this:

Python

lista=[]
for i in signal.valid_signals()
{
lista.append(i)
str(i).strip(<Signals.:>)
}
print(lista)

that didn't work
Is there a way to create a fast list of all linux signals without writing them myself into a list?

Posted 25-Jun-21 5:42am

Member 15262726

Updated 26-Jun-21 9:14am

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2 solutions

Solution 1

Your python3 version must be 3.8 or above to support signal.valid_signals()
Then this will give you an array:

import signal
 
valid_signals = signal.valid_signals()
print(valid_signals)
print(len(valid_signals))
arr=str(valid_signals).split(':')
print(arr)

The array will look something like these first few lines:

['{<Signals.SIGHUP', ' 1>, <Signals.SIGINT', ' 2>, <Signals.SIGQUIT', ' 3>, <Signals.SIGILL', ' 4>, <Signals.SIGTRAP',

You should be able to parse as needed.

Posted 25-Jun-21 10:46am

theoldfool

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Member 15262726 26-Jun-21 14:19pm

So yeah this does create a list but i'm trying to get rid of everything but SIGQUIT,SIGILL ect. and strip isn't helping here any solution for that?

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theoldfool · Accepted Answer · 2021-06-26T09:14:00

Well, maybe homework but here is quick and dirty, you can improve:

import signal
valid_signals = signal.valid_signals()
arr=str(valid_signals).split(',')
for i in range(0, len(arr)):
        start=arr[i].find("Signals")+8
        end=arr[i].find(":")
        if(end>0):
                print(arr[i][start:end])