How to Refer to Relative Paths of Resources When Working with a Code Repository

How to refer to relative paths of resources when working with a code repository

Try to use a filename relative to the current files path. Example for './my_file':

fn = os.path.join(os.path.dirname(__file__), 'my_file')

In Python 3.4+ you can also use pathlib:

fn = pathlib.Path(__file__).parent / 'my_file'

Reading file using relative path in python project

Relative paths are relative to current working directory.
If you do not your want your path to be, it must be absolute.

But there is an often used trick to build an absolute path from current script: use its __file__ special attribute:

from pathlib import Path

path = Path(__file__).parent / "../data/test.csv"
with path.open() as f:
    test = list(csv.reader(f))

This requires python 3.4+ (for the pathlib module).

If you still need to support older versions, you can get the same result with:

import csv
import os.path

my_path = os.path.abspath(os.path.dirname(__file__))
path = os.path.join(my_path, "../data/test.csv")
with open(path) as f:
    test = list(csv.reader(f))

[2020 edit: python3.4+ should now be the norm, so I moved the pathlib version inspired by jpyams' comment first]

open resource with relative path in Java

Supply the path relative to the classloader, not the class you're getting the loader from. For instance:

resourcesloader.class.getClassLoader().getResource("package1/resources/repository/SSL-Key/cert.jks").toString();

How to read file from relative path in Java project? java.io.File cannot find the path specified

If it's already in the classpath, then just obtain it from the classpath instead of from the disk file system. Don't fiddle with relative paths in java.io.File. They are dependent on the current working directory over which you have totally no control from inside the Java code.

Assuming that ListStopWords.txt is in the same package as your FileLoader class, then do:

URL url = getClass().getResource("ListStopWords.txt");
File file = new File(url.getPath());

Or if all you're ultimately after is actually an InputStream of it:

InputStream input = getClass().getResourceAsStream("ListStopWords.txt");

This is certainly preferred over creating a new File() because the url may not necessarily represent a disk file system path, but it could also represent virtual file system path (which may happen when the JAR is expanded into memory instead of into a temp folder on disk file system) or even a network path which are both not per definition digestable by File constructor.

If the file is -as the package name hints- is actually a fullworthy properties file (containing key=value lines) with just the "wrong" extension, then you could feed the InputStream immediately to the load() method.

Properties properties = new Properties();
properties.load(getClass().getResourceAsStream("ListStopWords.txt"));

Note: when you're trying to access it from inside static context, then use FileLoader.class (or whatever YourClass.class) instead of getClass() in above examples.

How to resolve relative paths in python?

import os
dir = os.path.dirname(__file__)
path = raw_input()
if os.path.isabs(path):
    print "input path is absolute"
else:
    path = os.path.join(dir, path)
    print "absolute path is %s" % path

Use os.path.isabs to judge if input path is absolute or relative, if it is relative, then use os.path.join to convert it to absolute

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