Bash: eliminating redundant path components?

Question

How can I eliminate redundant components in a path?

For example, I would like to transform

/foo/../foo/bar

to

/foo/bar

See [How do you normalize a file path in Bash?](http://stackoverflow.com/q/284662/4154375). — pjh, Apr 05 '16 at 19:44
Note that on Unix-like systems that support symbolic links, '/X/../Y' is not always the same as '/Y'. In the example in this question, if you have directories '/bar/foo/bar' and '/bar/qux/bar', and '/foo' is a symlink to '/bar/qux', then '/foo/../foo/bar' is actually '/bar/foo/bar' and '/foo/bar' is actually '/bar/qux/bar'. All attempted solutions that just do textual substitutions on paths are wrong. (But it is ok to do that kind of thing on Windows because '..' means something different there.) — pjh, Apr 05 '16 at 20:12

anubhava · Accepted Answer · 2016-04-05T20:21:55.820

2

p='/foo/../foo/bar'

realpath -m "$p"

Output:

/foo/bar

As per realpath --help:

-m, --canonicalize-missing   no components of the path need exist

You can also use more commonly available readlink (thanks to @pjh):

readlink -m "$p"

edited Apr 05 '16 at 20:21

answered Apr 05 '16 at 16:42

anubhava

2

This answer is good, but note that `readlink` does the same thing and is more commonly available. – pjh Apr 05 '16 at 20:16
1

Thanks @pjh, I have added it in my answer. – anubhava Apr 05 '16 at 20:22
If you use `-m` option as I suggested then it won't check for existence of path/link – anubhava Apr 05 '16 at 20:27

score 0 · Answer 2 · answered Apr 05 '16 at 16:47

0

You might pipe through something like: sed 's-/../foo/-/-g' to replace up/down reference in path names.

answered Apr 05 '16 at 16:47

Muehlkreuz

2 Answers2