2

I need to remove the first directory from a path by sed, (the inverse pf what basename does).

For example

echo "/mnt/VPfig/Amer/AR4/Celtel/files/COM.txt" | sed ...

I should get:

/VPfig/Amer/AR4/Celtel/files/COM.txt
5

Try:

echo "/mnt/VPfig/Amer/AR4/Celtel/files/COM.txt" | sed 's|^/[^/]*||'

which gave me:

/VPfig/Amer/AR4/Celtel/files/COM.txt

It looks for the first / followed by as many non-/s as possible, then replaces them with an empty string.

  • Is there need to escape? sed 's,^/[^/]*,,' works fine to me. – fedorqui Jul 17 '14 at 13:11
  • Probably not - I tend to escape and quote everything to be on the safe side :-) I'll edit it. – garethTheRed Jul 17 '14 at 13:19
2

The simplest way would be to remove the shortest string of non-/ characters from the beginning of the string.

In most regular expression languages, [ ] os a character class that matches anything within the brackets. [abc] will match either a or b or c. [^ ] is a negated character class, matching anything except the characters in it. So, [^/]* will match the longest string of non-/ characters. A ^ at the beginning of a regular expression makes it match only at the beginning of the input string. Combining these, you can do:

$ echo "/mnt/VPfig/Amer/AR4/Celtel/files/COM.txt" | sed 's/^\/[^/]*//'
/VPfig/Amer/AR4/Celtel/files/COM.txt

Or, you can set the input field delimiter of awk to / and print everything but the last field:

$ echo "/mnt/VPfig/Amer/AR4/Celtel/files/COM.txt" | 
    awk -F/ '{for(i=3;i<=NF;i++){printf "/%s",$i};print ""}'

Or you could use some Perl trickery:

$ echo "/mnt/VPfig/Amer/AR4/Celtel/files/COM.txt" | 
    perl -F/ -lane 'print "/",join "/",@F[2..$#F]'
/VPfig/Amer/AR4/Celtel/files/COM.txt

or, some better Perl trickery as suggested by @Joseph R.:

$ echo "/mnt/VPfig/Amer/AR4/Celtel/files/COM.txt" | perl -pe 
   perl -pe 's{^/.*?(?=/)}{}'

or even simply:

$ echo "/mnt/VPfig/Amer/AR4/Celtel/files/COM.txt" | perl -pe 's#/[^/]+/#/#'
/VPfig/Amer/AR4/Celtel/files/COM.txt

Finally, you could also use the power of Perl regular expressions with grep:

$ echo "/mnt/VPfig/Amer/AR4/Celtel/files/COM.txt" | grep -oP '/[^/]+?\K/.*'
/VPfig/Amer/AR4/Celtel/files/COM.txt
  • Shouldn't the Perl trickery just make use of the superior Perl regexes? echo ... | perl -pe 's{^/.*?(?=/)}{}' – Joseph R. Jul 17 '14 at 11:07
  • @JosephR. yes it should, thanks. Answer edited. – terdon Jul 17 '14 at 11:19
3

In a POSIX shell:

$ x='/mnt/VPfig/Amer/AR4/Celtel/files/COM.txt'
$ printf "/%s\n" "${x#/*/}"
/VPfig/Amer/AR4/Celtel/files/COM.txt

Explanation

The parameter expansion ${x#/*/} removes the leading component of the path with the two /s enclosing it and the printf tacks a leading / back on.

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