I've read through a couple of similar questions, notably this one about imp.load_module which seems to be close to what I want, but I can't understand why I'm still getting ImportErrors. Here is my folder hierarchy:
program\
__init__.py
main.py
thirdparty\
__init__.py
css\
__init__.py
css.py
utils\
__init__.py
http.py
In main.py I have the following code. This is intended to search the thirdparty\ directory and load each module it finds. Each module is in its own separate directory.
import os
import imp
for root, dirs, files in os.walk("thirdparty"):
for source in (s for s in files if s.endswith(".py")):
name = os.path.splitext(os.path.basename(source))[0]
m = imp.load_module(name, *imp.find_module(name, [root]))
The problem is that css.py happens to use its own subfolder that it loads stuff off of, utils. It has a line in it that says:
from utils import http
And that is where it fails. I get this error when I run main.py.
Traceback (most recent call last):
File "main.py", line 7, in <module>
m = imp.load_module(name, *imp.find_module(name, [root]))
File "thirdparty/css/css.py", line 1, in <module>
from utils import http
ImportError: No module named utils
I'm stumped. css.py is self contained in its own folder, and when I run css.py separately it imports utils just fine. What is causing this?
Maybe you can solve this by changing the import to:
from .utils import http
Or by adding the folder you import to the Python Path:
sys.path.append(os.path.join(root, source))
When you import modules in thirdparty, the place Python looks for modules is still the main directory. The initial import works, because you give the right path to imp.find_module, but after that Python has no idea where to look for the modules.
Related
I am currently having an issue with importing files from other directories in my python project.
My current file structure is
Project
- Backend
- Config
+ __init__.py
+ databaseConfig.py
- DataAccess
+ __init__.py
+ sqlConns.py
- __init__.py
- api.py
- main.py
- setup.py
What I am trying to do is import /Config/databaseConfig.py file into /DataAccess/sqlConns.py file. I get the following error when trying to run the sqlConns.py file
PS C:\source\repos\aaStats\aaStats> py .\Backend\DataAccess\sqlConns.py
Traceback (most recent call last):
File "C:\source\repos\aaStats\aaStats\Backend\DataAccess\sqlConns.py", line 2, in <module>
import Config.databaseConfig
ModuleNotFoundError: No module named 'Config'
I have also tried using relative imports, but I am met with another error.
PS C:\source\repos\aaStats\aaStats> py .\Backend\DataAccess\sqlConns.py
Traceback (most recent call last):
File "C:\source\repos\aaStats\aaStats\Backend\DataAccess\sqlConns.py", line 2, in <module>
from ..Config import databaseConfig as dbcfg
ImportError: attempted relative import with no known parent package
Config/databaseConfig.py contains database configuration parameters that I want to reference is various places in my project. It isn't a huge deal if I had to move this single file in order to get it to be referenced properly, but I will want to use structures like this for files later on in my project.
Here are some details about my files:
/Config/__init__.py
from . import databaseConfig
/DataAccess/__init__.py
from . import sqlConns
Backend/__init__.py
from . import DataAccess
from . import Config
Backend/setup.py
from setuptools import setup, find_packages
setup(
name='aaStatsApi',
version='0.1.0',
packages= ['DataAccess','Config'],
install_requires=[
'fastapi==0.63.0',
'uvicorn==0.13.4',
'requests==2.25.1',
'pyodbc==4.0.30',
]
)
Check out this post.
The fact that you can't perform relative imports so easily is by design, for better or for worse. The ideal way is have your main script in the root (Backend) directory and do all your calls from there. The function that has __name__ == __main__ is your calling function. If you do not directly call Calls.py or Configs.py from a console, but are calling them from another main function within your root directory, you should be able to place the following into Conns.py:
# FILE: DataAcess\sqlConns.py
from Config.dataBaseConfig import * # or whatever you need to import
Again, the key is to ensure that your starting point in from your root project directory.
NOT RECOMMENDED:
For risk of getting downvoted, and I do not recommend this, but you could append the relative path to your calling class:
import sys, os
sys.path.append(os.path.abspath("../Config"))
from sqlConns import * # or whatever
sys.path.pop() # clear sys.path
I have no idea why my import doesnt work.
My folder structure
Garage_Parking
__init__.py
database.py
rpi_components
__init__.py
NFC_Security.py
I want to import database in NFC_Security.py.
I typed
from Garage_Parking import database
It just keep giving me this error
Traceback (most recent call last):
File "NFC_Security.py", line 6, in <module>
from Garage_Parking import database
ImportError: No module named Garage_Parking
I appreciate any help.
The idea is to add the path of the parent folder to the python path so that the interpreter knows that it should look for files and modules even in the parent directory.
import os,sys
current_directory = os.getcwd()
parent_directory = os.path.dirname(current_directory)
sys.path.insert(0, parent_directory)
The code above adds the parent directory to the python path. Now you can freely import all modules from the parent folder.
Complete code for your specific case:
import os,sys
current_directory = os.getcwd()
parent_directory = os.path.dirname(current_directory)
sys.path.insert(0, parent_directory)
import database
For more information visit
Importing modules from parent folder
You could try putting the main file with the rest of the files and then doing import database. It is the easy way. If you would like to keep your folder organization, I can’t help you.
I have multiple modules in my project and specify some execution point. But when I try to import files from submodules, it doesn't work.
So, how to specify submodules to execute from selected execution file?
project
--bin
---- executeFile
--modules
---- __init__.py
----fileA.py
in executeFile, I try:
from ..modules.fileA import *
but get the error:
Traceback (most recent call last):
File "./bin/muexecute", line 10, in <module>
from ..modules.os import *
SystemError: Parent module '' not loaded, cannot perform relative import
I found solution.
The problem was in my opinion about using init.py.
I placed in executable scripts path to including and it works fine
PACKAGE_PARENT = '..'
SCRIPT_DIR = os.path.dirname(os.path.realpath(os.path.join(os.getcwd(), os.path.expanduser(__file__))))
sys.path.append(os.path.normpath(os.path.join(SCRIPT_DIR, PACKAGE_PARENT)))
So you are having trouble defining your relative path, correct? Try the following:
from sys import path
path.append('C:\\realative_path')
from function_file import required_function
Hope that helps.
All modules you want to import should in in your PYTHONPATH. Therefore there is no hierarchy.
In your case It seems to me that an __init__.py is missing from your project's main folder (with all the models included), so executefile doesn't know about your modules.
I have:
project
|--__init__.py
|--...
+--package1
| |--__init__.py
| |--...
+--dbs
|--...
Where project is a collection of packages to be imported and used by the user. The names are just an example.
Now, some files under project/package1/ need to access some files in project/dbs/, but depending on where the user placed the project folder and from where he or she imported it, path/to/project/dbs/file is not the same. Because of this issue and for other purposes I thought I should have a variable projectroot defined as the root of the project, relative to where it was imported from. For instance:
If the user has ~/Desktop/project and imported it from ~/ then projectroot would be Desktop/project, and path/to/project/dbs/file would be "%s/dbs/file" % (projectroot). Is this really needed? If so, what is the best way of doing it?
I tried using this in project/pathtfinder.py and used project/__init__.py as
import pathfinder
projectroot=pathfinder.module_path()
import package1
But when I run import project I get
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "project/__init__.py", line 3, in <module>
import package1
File "project/package1/__init__.py", line 5, in <module>
db = sqlite3.connect("%s/dbs/main.db" % projectroot)
NameError: name 'projectroot' is not defined
And if use print projectroot just after defining it I get Desktop/project, as expected.
You need to add the path of project to the PYTHONPATH
This is a pattern I use for my Flask apps, but the same should be ideal for you:
Assume the following is in a file under helloworld/app.py
import sys
import os
# Add this app to the path so imports work as expected
APP_PATH = os.path.join(os.path.dirname(__file__), '../')
sys.path.append(APP_PATH)
from helloworld.utils import a_utility
The convention explains itself, but it allows for the imports to occur from the "root" of the project, regardless of where the file is placed. You'll want to place that APP_PATH code perhaps in your init.py, or similar.
This is my folder structure:
src/
__init__py
Lowlevel/
__init__.py
ModuleToCheck.Py
Test/
__init__.py
ModuleToCheck_test.py
(__init__.py are empty files)
Now I want to import ModuleToCheck.py in ModuleToCheck_test.py
How am I able to do this without appending anything to sys.path?
Update:
from ..Lowlevel import ModuleToCheck leads to:
src$ python Test/ModuleToCheck_test.py
Traceback (most recent call last):
File "Test/ModuleToCheck_test.py", line 6, in <module>
from ..Lowlevel import ModuleToCheck
ValueError: Attempted relative import in non-package
The following is from http://docs.python.org/tutorial/modules.html#intra-package-references
Note that both explicit and implicit
relative imports are based on the name
of the current module. Since the name
of the main module is always
"__main__", modules intended for use
as the main module of a Python
application should always use absolute
imports.
You're running your module ModuleToCheck_test.py as the main module, hence the exception.
One solution is to create a test.py module in your src directory containing the following:
import Test.ModuleToCheck_test
You can then run that module using python test.py