how to cut directory from $file in the bash shell

Question

I'm newbie for a shell script. So I need to cut the directory name from my $file. e.g.

fullpath='/var/log/apache/access.log

Now I need to cutting "apache" to output. How can I do.

Thank in advance.

You're probably looking for `basename`. Have you searched StackOverflow for similar questions? You may find your answer by just searching a little more. Here is one, for example: http://stackoverflow.com/questions/965053/extract-filename-and-extension-in-bash/965072#965072 — vastlysuperiorman, Mar 27 '15 at 06:53
If it is the directory you're after, it's probably `dirname` you want. — Matijs, Mar 27 '15 at 06:55
Valid point. Though from the example, he may want just "apache", not /var/log/apache, which would be the output of dirname. In that case `basename $(dirname $FULLPATH)` would give the desired output. — vastlysuperiorman, Mar 27 '15 at 06:58

score 0 · Accepted Answer · answered Mar 27 '15 at 07:01

0

The combination of basename and dirname will give you the parent directory of the file you provided in $fullpath

For example:

FULLPATH=/var/log/apache/access.log; basename $(dirname $FULLPATH)

will print "apache"

answered Mar 27 '15 at 07:01

vastlysuperiorman

@dimsum If this answer works for you, please check the box next to it (to select as the correct answer for your question). – vastlysuperiorman Mar 27 '15 at 17:46

Ravi Bhushan · Answer 2 · 2015-03-27T06:58:56.790

-1

vim Filename

fullpath='/var/log/apache/access.log

then save and exit

cut -d'/' -f4 Filename

edited Mar 27 '15 at 06:58

answered Mar 27 '15 at 06:53

Ravi Bhushan

This solution isn't very helpful as the number of directories in a full path is not constant. Taking the fourth field will not always yield a filename, pathname, or anything at all. – vastlysuperiorman Mar 27 '15 at 06:59

2 Answers2