Shell scripting : to print selected text in the string

Question

Log file name: "/home/msubra/WORK/tmo/LOG/BCH1043.9987.log"

From the above string i need to extract the content BCH1043.

The directory structure may differ so the solution should check for the string with BCH until the dot

Erm. Ok. I'll ask first. What _exactly_ have you tried so far? What tools did you use? What exactly are you trying to do? Will the files always start with BCH and end with log? Remember, SO is there for getting help, not to have others do your work for you. — Markus W Mahlberg, Nov 06 '14 at 16:24

Gary_W · Answer 1 · 2014-11-06T17:19:13.977

1

No need to call basename, you can use parameter substitution that is built-in to the shell for the whole thing:

$ cat x.sh
filepath="/home/msubra/WORK/tmo/LOG/BCH1043.9987.log"

# Strip off the path.  Everything between and including the slashes.
filename=${filepath##/*/}

# Then strip off everything after and including the first dot.
part1=${filename%%.*}

echo $part1
$ ./x.sh
BCH1043
$

A dot in the filepath will not cause trouble either.

See section 4.5.4 here for more info: http://docstore.mik.ua/orelly/unix3/korn/ch04_05.htm

Oh and resist the temptation to get tricky and do it all in one line. Breaking into separate components is much easier to debug and maintain down the road, and who knows you may need to use those components too (the path and the rest of the file name).

edited Nov 06 '14 at 17:19

answered Nov 06 '14 at 17:11

Gary_W

9,933
1
22
40

This fails for non-absolute paths and still has some of the caveats of the answer I linked to in my answer. Other than that though it works. – Etan Reisner Nov 06 '14 at 17:25
Absolute paths were not part of the requirements but point taken. – Gary_W Nov 06 '14 at 17:33

Hackaholic · Answer 2 · 2014-11-06T17:45:31.197

0

try like this:

a="/home/msubra/WORK/tmo/LOG/BCH1043.9987.log"
echo ${a##*/} | cut -d "." -f 1

or
 basename $a | cut -d "." -f 1

or
 var=${a##*/}; echo ${var%%.*}

output:

BCH1043

It dosent include dot. Your question is not clear, but you can extract like that

${a##*/} will extract after last / like same as basename

edited Nov 06 '14 at 17:45

answered Nov 06 '14 at 16:34

Hackaholic

19,069
5
54
72

Your order of operations here is backwards. If there is a `.` anywhere in the path this fails. – Etan Reisner Nov 06 '14 at 16:58
@EtanReisner now i think it will not fail – Hackaholic Nov 06 '14 at 17:21
Yes, now it should work but using `cut` isn't necessary the previous parameter expansion was fine. – Etan Reisner Nov 06 '14 at 17:26
@EtanReisner i have given most of the possibilities – Hackaholic Nov 06 '14 at 17:46

score 0 · Answer 3 · answered Nov 06 '14 at 16:34

basename will reduce /home/msubra/WORK/tmo/LOG/BCH1043.9987.log to BCH1043.9987.log

echo /home/msubra/WORK/tmo/LOG/BCH1043.9987.log | basename

You can use regular expressions, awk, perl, sed etc to extract "BCH1043" from "BCH1043.9987.log". First I need to know what the range of possible filenames is before I can suggest a regular expression for you.

score 0 · Answer 4 · edited May 23 '17 at 11:49

Use basename to extract only the filename and then use parameter expansion to strip off the data you don't want.

log=/home/msubra/WORK/tmo/LOG/BCH1043.9987.log
log=$(basename "$log")
echo "${log%%.*}"

The following is almost equivalent but doesn't use the external basename process. However there are cases where it will give different results (though whether those cases are relevant here is up to you and your usage/input). See this answer for examples/details.

log=/home/msubra/WORK/tmo/LOG/BCH1043.9987.log
log=${log#*/}
echo "${log%%.*}"

Shell scripting : to print selected text in the string

4 Answers4