How to load resources file without giving entire path in Python

Question

I want to read the properties file in Python without giving the entire path. Because if my code is deployed somewhere else then my package would fail if I pass the hardcore value. So below is my code to read the properties file by giving the entire path:

import configparser
config = configparser.RawConfigParser()
config.read('C:\\Users\\s\\IdeaProjects\\PysparkETL\\src\\main\\resources\\configs.properties')
dbname = config['DB']['dbname']
username=config['DB']['user']
password=config['DB']['password']
table=config['DB']['tablename']
driver=config['DB']['driver']
print(dbname)

and below is my configs.properties file:

[DB]
dbname=ourdb
user=root
password=root
tablename=loadtable
driver=com.mysql.cj.jdbc.Driver

I tried different ways like ConfigParser instead of RawConfigParser but didn't work. Is there any way to load files from the classpath?
Also, I tried different ways from this link but it didn't help. All I need is a path to pass it to config.read method but it should not be hardcoded as I did it in the code.

Below is my project structure:

Also, as suggested I tried below code and passed the URL to the config.read method but it's not working.

props = os.path.join(
    os.path.dirname("resources"),  # folder where the python file is
    'src/main/resources/configs.properties'
)
config.read(props)

I get below error:

   raise KeyError(key)
KeyError: 'DB'

Does this answer your question? [How to read a (static) file from inside a Python package?](https://stackoverflow.com/questions/6028000/how-to-read-a-static-file-from-inside-a-python-package) — Tomerikoo, Aug 30 '20 at 08:49
I tried `pkgutil` and `importlib_resources` packages but it didn't work for me..What I need is to get the path of the file to be read so that I can pass it to the `config.read` method — whatsinthename, Aug 30 '20 at 12:47

score 0 · Answer 1 · answered Aug 30 '20 at 13:15

0

if the file will always be in the same place relative to your python file, you can do the following:

props = os.path.join(
    os.path.dirname(__name__),  # folder where the python file is
    'relative/path/to/configs.properties'
)

answered Aug 30 '20 at 13:15

Cyril Jouve

990
5
15

what is the relative path over here? Is it ok if i just pass the file name and it can get the file path for me? – whatsinthename Aug 30 '20 at 14:15
I have edited my question as per your suggestion. Could you please have a glance once? – whatsinthename Aug 30 '20 at 14:21

How to load resources file without giving entire path in Python

1 Answers1