Python os.path.islink() Method

1. Syntax
2. Parameters
3. Returns
4. Examples

The islink() method belongs to the os.path module. The os module is an in-built module available with Python that offers various procedures to interact with the operating system.

However, the path module is a part of the os module, which contains utilities to work with file system paths. The islink() method checks if a path is a symbolic link and leads to an existing resource.

Syntax

os.path.islink(path)

Parameters

It also accepts an object (of type string or bytes) implementing the os.PathLike protocol if we are using Python 3.6 or higher.

Returns

The islink() method returns a Boolean value. If the path is a valid symbolic link, it returns True. On the contrary, it returns False for invalid paths.

Additionally, if the Python runtime environment does not support symbolic links, this method always returns False.

Examples

Symbolic Link to a Resource

In Unix systems, we can create symbolic links or soft links using the ln command. This command has the following syntax. To learn more about this command, refer to the official manual pages here.

ln [-sf] [source] [destination]

Create a text file, data.txt, and add random data to it. Next, fire up the command line and run the following command to create a symbolic link for the file in the current directory.

ln -s 'data.txt' 'data-symlink.txt'

Now let’s understand how to use the islink() method. First, refer to the following Python code.

from os.path import islink

path = "/home/user/documents/files/data-symlink.txt"
print(islink(path))

Output:

True

Since the path is a symbolic link or the new file we just generated moments ago, the method returns True.

Original Path to a Resource

from os.path import islink

path = "/home/user/documents/files/data.txt"
print(islink(path))

Output:

False

Since the path stores address to the original resource, the islink() method returns False.