Symbolic-Link always points to its parent directory

Question

Placing Symbolic-Links other than the current directory does not work because they always point to their current directory

I'm on Windows 10, what I'm missing?

fs.symlink('./testFile', './testDir/testSymLink', function(err){       // creates a symbolic- link in the 'testDir' subfolder relative to the current directory 
    if(err) console.log(err);
});

fs.readlink('./testDir/testSymlink',function(err, links){              // reads the created symbolic link 
    if(err) console.log(err);
    console.log(links);                                                // -> '.\testFile'   (points to the current directory not to the parent directory)
});

fs.readFile('./testDir/testSymlink.txt', function(err, data){          // file doesn't exist
    if(err) console.log(err);                                          // -> ENOENT no such file or directory 
    console.log(data);                                                 // -> undefined 
});

The symbolic link is created (we can read it) but points to its current directory .\testFile it should point to its parent directory where the reference file is ..\testFile

Answer 1

the symlink() method's 1st argument is the target location used by the created symbolic-link.

It's is important to know that the symbolic-link uses the 1st argument as it was passed! (does not resolve it as absolute path!)

Here is the tricky part, because the above 1st argument is ./testFile the symbolic-link will use this path to reference the 'testFile' in its own directory (not in the parent directory)

Solutions:

1: the above code could be fixed like this (the testSymLink in the testDir directory references the testFile in the parent directory)

fs.symlink('../testFile', './testDir/testSymLink', function(err){       
    if(err) console.log(err);
});

2: passing absolute path as 1st argument in the symlink() method (this will save you a lot of hassle!)

fs.symlink(/* absolute path */, './testDir/testSymLink', function(err){       
    if(err) console.log(err);
});