python import another python script not working

Question

I have two python scripts which are in the same directory(C:\\Users\\user1\\general). I want to execute a function in one script from the 2nd second script and so I am trying to import script1 in script2. Here is my script2:

import sys
sys.path.insert(0, 'C:\\Users\\user1\\general')
import script1
from flask import Flask
from flask_socketio import SocketIO, send
app = Flask(__name__)
app.config['SECRET_KEY'] = 'mysecret'
socketio = SocketIO(app)
@socketio.on('message')
def handleMessage(msg):
 print('Message: ' + msg)
 # script1 has function called api_call(id,message)
 messsage = api_call('user1', msg)
 send(messsage, broadcast=True)
if __name__ == '__main__':
 socketio.run(app)

Here is my script1:

import sys
def api_call(Id,message):
 # Code processing
if __name__ == '__main__':
 uId = sys.argv[0]
 message = sys.argv[1]
 api_call(userId,message)

When I execute above script2 I get NameError: name 'api_call' is not defined. It seems somehow the script1 is not getting imported and so the function is not getting through.

Note: Earlier I had tried without using sys.path.insert() and same outcome then also.

• saniales: did you try from script1 import *?
• FamousJameous: Or changing api_call in script2 to script1.api_call?
• juanpa.arrivillaga: @saniales that's a very bad practice.
• saniales: yes, it was just to understand if it was an import problem, clearly the correct solution would be to import only what you need.

Answer 1

Try from script1 import api_call: this would allow to import from script1 module the api_call function.

Answer 2

Python 2

Create __init__.py in the same folder.

Import using following command

from script1 import api_call

---

Python 3

Import using following command

from .script1 import api_call

• FabienP: This syntax (from .module import ...) refers to relative import from submodules. It should be used within packages, and does not seems an appropriate solution. Same thing for __init__.py, a repertory containing this file will be recognized as a package, which is not necessary, and probably not desired here.