Suppose i have a project structure like this
src
└── app
├── main.py
├── db
│ └── database.py
├── models
│ ├── model_a.py
│ └── model_b.py
└── tests
├── test_x.py
└── test_y.py
I want to check which file uses a class or a function from another file. I have a class called Test in main.py
class Test:
pass
I used that class in model_a,
from ..main import Test
But in model_b i used
from ..main import Test
from ..db.database import Data
I want to to check which file uses another file, just like tree command, just a folder name is enough so i tried an old method but it was inefficient ,dirty and that was not something that i expect. The method was i created a file in src named check.py, i imported all packages
from app.db import database
from app.models import model_a, model_b
from app.tests import test_x, test_y
from app import main
print('__file__={0:<35} | __name__={1:<20} | __package__={2:<20}'.format(__file__,__name__,str(__package__)))
And i added this line in the bottom of all files
print('__file__={0:<35} | __name__={1:<20} | __package__={2:<20}'.format(__file__,__name__,str(__package__)))
So when i run check.py i get this result
__file__=/home/yagiz/Desktop/struct/src/app/main.py | __name__=app.main | __package__=app
__file__=/home/yagiz/Desktop/struct/src/app/db/database.py | __name__=app.db.database | __package__=app.db
__file__=/home/yagiz/Desktop/struct/src/app/models/model_a.py | __name__=app.models.model_a | __package__=app.models
__file__=/home/yagiz/Desktop/struct/src/app/models/model_b.py | __name__=app.models.model_b | __package__=app.models
__file__=/home/yagiz/Desktop/struct/src/app/tests/test_x.py | __name__=app.tests.test_x | __package__=app.tests
__file__=/home/yagiz/Desktop/struct/src/app/tests/test_y.py | __name__=app.tests.test_y | __package__=app.tests
__file__=/home/yagiz/Desktop/struct/src/check.py | __name__=__main__ | __package__=None
The result is dirty and doesn't meet my expectations is there a way to get a output like this?
main.py = app/models/model_a, app/models/model_b # These files imports something from main.py
models_b = None # No file imports from models_b
Update, i tried @Hessam Korki's suggestion it doesn't works.
I looked up the source code of modulefinder and i found it adds a badmodule in every import statement which is not useful for me.
Here is how did it go, first i created a function, also i created an another project structure.
src
├── empty.py
├── __init__.py
├── main.py
├── module_finder.py
├── other
│ └── other.py
├── test
│ └── some_app.py
└── this_imports.py
Here is the module_finder.py that contains my function
from modulefinder import ModuleFinder
file_names = ["this_imports.py", "main.py", "test/some_app.py", "other/other.py", "empty.py"]
def check_imports(file_names):
finder = ModuleFinder()
for file in file_names:
finder.run_script(file)
print("\n", file)
for name, mod in finder.modules.items():
print('%s: ' % name, end='')
print(','.join(list(mod.globalnames.keys())[:3]))
print('\n'.join(finder.badmodules.keys()))
Empty file is empty(as expected), in main.py i have
class Test:
pass
In this_imports.py i only have
from src.main import Test
In other/other.py i have
from src.main import Test
from src.test import DifferentTest
And for the last one in test/some_app.py i have
from src.main import Test
class DifferentTest:
pass
So the result should be:
empty.py = None
main.py = None
other/other.py = src.main , src.test
test/some_app.py = src.main
this_imports.py = src.main
But the function gives a wrong result, here is the output:
Filename: this_imports.py
__main__: Test
src.main
Filename: main.py
__main__: Test,__module__,__qualname__
src.main
Filename: test/some_app.py
__main__: Test,__module__,__qualname__
src.main
Filename: other/other.py
__main__: Test,__module__,__qualname__
src.main
src.test
Filename: empty.py
__main__: Test,__module__,__qualname__
src.main
src.test
grep -rnw . -e "from\s\.*main\simport\sTest"– Tying