How do I pass varaibles to another python file-CodePudding

I have two files and needed to move a variable from one to the other. There is also some shell scripting involved. For example, I have script1.py and a variable called var1.

script1.py

var1 = sys.argv[1]

When I try to import this into another Python file. For e.g.

script2.py

from script1.py import var1
print(var1)

It is actually running the sys.argv[1] again in script2.py. This is how my shell script looks.

shellcode.sh 
python3 script1.py "1st parameter"
python3 script2.py "2nd"

How to get only the "1st parameter" not "2nd". I'm a complete beginner at python and shell scripting. Is it related to pass by value? Any help would be highly appreciated.

CodePudding user response：

pass directly arguments from script2.py it should work.

script1.py

import sys
var1 = sys.argv[1]

in script2.py

from script1 import var1
print(var1)

on shell run like this

python3 script2.py --test

Will print #

test

CodePudding user response：

One way around this would be to use the parameter as a variable in your shell script and just pass both to the second python argument.

shellcode.sh:

#!/bin/bash
PAR1="1st paramter"
PAR2="2nd parmaeter"
python3 script1.py $PAR1
python3 script2.py $PAR1 $PAR2

Otherwise you could write and read the argument something like

script1.py:

import sys
var1 = sys.argv[1]
with open('script1_var.txt', 'w') as f:
    f.write(var1)

script2.py:

import sys
with open('script1_var.txt', 'r') as f:
    var1 = f.readlines()[0].strip()
var2 = sys.argv[1]
print('var1', var1)
print('var2', var2)