[SOLVED] Method to check if symlink from the current linux header directory to /usr/src/linux exists?

vysero · 08-21-2018, 10:42 AM

So I was given a script as text. Apparently the script will generate a symlink from the current linux header directory to /usr/src/linux. I took the text and created a file copied it into the file and saved it as script.py. I then tried running the script with a sudo ./script command but that doesn't work. So I tried saving the file as script without the .py extension and tried running ./script but that did not work either.

I am not sure why those attempts did not work but I tried just running it with an interpreter via a sudo python script and it might have worked? I have no way of knowing because I don't even know what a symlink is.

rtmistler · 08-21-2018, 11:08 AM

Number one is that you should consider learning what symbolic links are.

Freely available on the web and my personal explanation is that they are just a pointer to a file that is somewhere else in your system. Therefore you see the file in whichever directory you are visiting, and can view it, run it, etc, just the physical file is somewhere else.

You should post this script and place it within [code] tags to retain the formatting of it and then people can help you better. I'm not sure if the whole purpose of that script is just to create a symbolic link, it might be. Meanwhile there is a command line command which makes symbolic links in a single call, ln(1)

Finally. Since you were given this script. Ask the person who gave it to you, and make sure you trust the origin of anything before you try to run it.

vysero · 08-21-2018, 11:24 AM

The script:

Code:

#!/usr/bin/env python 
import os 
import sys 
import subprocess
def _runningKernel():
    result = subprocess.check_output(['uname', '-r']).strip()
    return result
def _kernelHeaderPath(kernelName):
    return os.path.join('/', 'usr', 'src', 'linux-headers-%s' % kernelName)
_linkName = os.path.join('/', 'usr', 'src', 'linux')
_headersPath = _kernelHeaderPath(_runningKernel())
if not os.path.isdir(os.path.join(_headersPath, '.')):
    sys.stderr("Error: Did not the header files directory at %s\n" % _headersPath)
    sys.exit(1)
if (not os.path.islink(_linkName)) and os.path.exists(_linkName):
    sys.stderr("Error: %s is not a symlink.  (Cowardly refusing to overwrite it.)\n" % _headersPath)
    sys.exit(1)
if os.path.islink(_linkName):
    os.remove(_linkName);
os.symlink(_headersPath, _linkName)

Is there a way to check if the link was created?

rtmistler · 08-21-2018, 02:23 PM

Yes, you perform an "ls -l" in the directory where the link is supposed to be and look at the result.

Keith Hedger · 08-21-2018, 02:34 PM

Use readlink

Code:

keithhedger@LFSBomb20:/media/BigBackup/TestRestore-> readlink "/usr/src/linux-4.15.3" 
/LFSSourceArchives/8.2/KERNEL/kernel-4.15.3_1/linux-4.15.3
keithhedger@LFSBomb20:/media/BigBackup/TestRestore-> readlink "/usr/src/linux-4.15.4" 
keithhedger@LFSBomb20:/media/BigBackup/TestRestore->

The 1st link above exists the second don't.

vysero · 08-21-2018, 02:54 PM

Quote:

Originally Posted by rtmistler

Yes, you perform an "ls -l" in the directory where the link is supposed to be and look at the result.

rob@linux038:/usr/src$ ls -l
total 24
lrwxrwxrwx 1 root root 40 Aug 21 09:41 linux -> /usr/src/linux-headers-4.15.0-32-generic
drwxr-xr-x 27 root root 4096 Jul 20 15:13 linux-headers-4.15.0-29
drwxr-xr-x 8 root root 4096 Jul 20 15:13 linux-headers-4.15.0-29-generic
drwxr-xr-x 27 root root 4096 Aug 7 06:52 linux-headers-4.15.0-30
drwxr-xr-x 8 root root 4096 Aug 7 06:52 linux-headers-4.15.0-30-generic
drwxr-xr-x 27 root root 4096 Aug 15 06:27 linux-headers-4.15.0-32
drwxr-xr-x 8 root root 4096 Aug 15 06:27 linux-headers-4.15.0-32-generic

thanks