How to extract a string at end of line after a specific word

Question

I have different location, but they all have a pattern:

some_text/some_text/some_text/log/some_text.text

All locations don't start with the same thing, and they don't have the same number of subdirectories, but I am interested in what comes after log/ only. I would like to extract the .text

edited question:

I have a lot of location:

/s/h/r/t/log/b.p
/t/j/u/f/e/log/k.h
/f/j/a/w/g/h/log/m.l

Just to show you that I don't know what they are, the user enters these location, so I have no idea what the user enters. The only I know is that it always contains log/ followed by the name of the file.

I would like to extract the type of the file, whatever string comes after the dot

So given a file path `..../log/something.otherthing` you want `otherthing`? — fedorqui, Oct 29 '13 at 16:02
OK then I guess you can find the answer in http://stackoverflow.com/a/965072/1983854 — fedorqui, Oct 29 '13 at 16:07

Kent · Answer 1 · 2013-10-29T16:06:11.617

1

THe only i know is that it always contains log/ followed by the name of the file.

I would like to extract the type of the file, whatever string comes after the dot

based on this requirement, this line works:

grep -o '[^.]*$' file

for your example, it outputs:

text

edited Oct 29 '13 at 16:06

answered Oct 29 '13 at 16:00

Kent

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score 1 · Answer 2 · answered Oct 29 '13 at 16:08

You can use bash built-in string operations. The example below will extract everything after the last dot from the input string.

$ var="some_text/some_text/some_text/log/some_text.text"
$ echo "${var##*.}"
text

Alternatively, use sed:

$ sed 's/.*\.//' <<< "$var"
text

score 0 · Answer 3 · answered Oct 29 '13 at 16:04

0

Not the cleanest way, but this will work

sed -e "s/.*log\///" | sed -e "s/\..*//"

This is the sed patterns for it anyway, not sure if you have that string in a variable, or if you're reading from a file etc.

You could also grab that text and store in a sed register for later substitution etc. All depends on exactly what you are trying to do.

answered Oct 29 '13 at 16:04

James Gawron

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the string is in a variable – user2864207 Oct 29 '13 at 16:05
Then it would be something like new_variable=$(echo $variable | sed -e "s/.*log\///" | sed -e "s/\..*//") ... that's in KShell ... let me try in Bash .. yep, works in both – James Gawron Oct 29 '13 at 16:13

score 0 · Answer 4 · answered Oct 29 '13 at 16:15

0

Using awk

awk -F'.' '{print $NF}' file

Using sed

sed 's/.*\.//' file

answered Oct 29 '13 at 16:15

jkshah

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score 0 · Answer 5 · answered Oct 29 '13 at 16:29

Running from the root of this structure:

    /s/h/r/t/log/b.p
    /t/j/u/f/e/log/k.h
    /f/j/a/w/g/h/log/m.l

This seems to work, you can skip the echo command if you really just want the file types with no record of where they came from.

    $ for DIR in *; do
    >  echo -n "$DIR  "
    >  find $DIR -path "*/log/*" -exec basename {} \; | sed 's/.*\.//'
    > done
    f  l
    s  p
    t  h

How to extract a string at end of line after a specific word

5 Answers5