I am trying to delete certain text that appears before / on every line.

I have something like:


and I just want to end up with:


Can anyone help me with a regular expression? Everything I have found is deleting after /, not before.

  • 2
    pipe it through something like: sed -e 's|[^/]*/||' Jul 1 '15 at 18:19
  • awk -F/ '{print $2}' file > newfile
    – jasonwryan
    Jul 1 '15 at 18:32
  • @Theophrastus that's the most solid solution proposed so far. Turn it into an answer and I for one will upvote it.
    – roaima
    Jul 1 '15 at 20:45
  • @Theophrastus your regEx looks good but its doing the opposite its stripping everthing and just leaving com/ Jul 1 '15 at 22:53
  • @user3738022 not when i do it: echo "testing.db.com/8a8589bf-49e3-4cd7-af15-6753067355c6" | sed -e 's|[^/]*/||' ends up returning this: 8a8589bf-49e3-4cd7-af15-6753067355c6 Jul 1 '15 at 23:24

Using cut :

$ cut -sd'/' -f2 file.txt   ##This will print only the lines containing /

The following suggestions assumes that / appears only once in a line :

Using grep :

$ grep -o '[^/]*$' file.txt  ##This will print the lines not having / too

If you have / in all of the lines, you can use these too:

Using bash parameter expansion:

$ while read line; do echo "${line#*/}"; done <file.txt 

Or python :

#!/usr/bin/env python2
with open('file.txt') as f:
    for line in f:
        print line.split('/')[1].rstrip()

Note that as far as your example is concerned all of the above suggestions are valid.

  • I'm upvoting this for cut - because good work and that's what I came here to say. But you've got it in mixed company - for this task these commands are nowhere near equivalent. cut should be the only thing here - to make it act like grep -o use cut -sd. Those other two commands are merely a distraction. Run some time tests on large inputs and you'll see what I mean.
    – mikeserv
    Jul 1 '15 at 20:23
  • @mikeserv Thank you very much as always..i was taking the OP's examples literally..anyway i am clarifying it now..
    – heemayl
    Jul 1 '15 at 20:29
  • Let's hope that / doesn't appear more than once per line. Try cut -d/ -f2- and ignore the other suggestions.
    – roaima
    Jul 1 '15 at 20:45

easy enough with sed:

echo "testing.db.com/7fad416d-f2b3-4259-b98d-2449957a3123" | sed -e "s/.*\///"

basic syntax is "s/search/replace/" and here we search for .*\/ which means everything ending with slash (escaped) and we replace it with nothing.

  • 1
    Let's hope that / doesn't appear more than once per line
    – roaima
    Jul 1 '15 at 20:42

Using Perl:

< file.txt perl -pe 's/.*\///'
  • 1
    Let's hope that / doesn't appear more than once per line. You might want to use 's|.*?/||' to bullet-proof the suggestion.
    – roaima
    Jul 1 '15 at 20:47
  • @roaima We don't have enough elements to judge; according to the example either matching greedily or lazily it's the same, because there's always one single /; if there are more than one in the part to exclude than one should match greedily like I did, otherwise one should match lazily like you suggested. Not knowing this, since the string looks pretty regular (only composed by numbers and -), I'd prefer to bullet-proof the solution against possible subdirs in the link (e.g. testing.db.com/subdir)
    – kos
    Jul 1 '15 at 21:04
  • 1
    good argument; I'm convinced!
    – roaima
    Jul 1 '15 at 21:23

Two methods using sed:

sed -i "s|testing.db.com/||g" file_path_here


echo "testing.db.com/7fad416d-f2b3-4259-b98d-2449957a3123" | sed -e "s|testing.db.com/||g"
  • I have tried this too but it is doing the opposite :( It's leaving com/ instead of the other way round Jul 1 '15 at 22:57
  • I don't understand. I tested it on my server and it works fine, it leaves "7fad416d-f2b3-4259-b98d-2449957a3123" only.
    – Artyum
    Jul 2 '15 at 5:18
  • I dont know why it didnt work for me but i ended up using this one [^/]*$ Jul 2 '15 at 16:28

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