Displaying portion of file name using command

Question

i have hundreds of file having name in a specific pattern , i want to get portion of filename only for example

like if filenames are REE_45K.txt or REE_46B.txt or REE_21K.txt or REE_229N.txt ot REE9999G.txt i want to display only 45K 46B ,21K,229N,9999G

if possible please give Linux and windows versions commands

findstr _.*.  *.*  i found this command working for windows , can u verify it

this works for windows but not always .. findstr /s _.*. *.* . >a.txt — user143252, Jun 09 '17 at 20:51
Is it always going to be the last three characters of the file name, or is it the bit between the last underscore and the extension? — SomethingDark, Jun 09 '17 at 23:40

score 1 · Answer 1 · answered Jun 09 '17 at 20:11

1

For bash on linux:

#!/bin/bash
filename="REE_45K.txt"
no_ext="${filename%.*}"  # "REE_45K"
echo "${no_ext##*_}"

gives 45K

answered Jun 09 '17 at 20:11

Kind Stranger

thanks , while googling i found this command grep -h '??.*??' PATH , can u expain this one if possible – user143252 Jun 09 '17 at 20:26
@user143252 See this link: https://stackoverflow.com/questions/22937618/reference-what-does-this-regex-mean – Kind Stranger Jun 09 '17 at 23:58

score 0 · Answer 2 · answered Jun 09 '17 at 21:36

0

For linux bash:

for filename in REE_45K.txt  REE_46B.txt  REE_21K.txt
do
  echo $filename |sed  's/REE_\(.*\).txt/\1/g'
done

It gives:

45K
46B
21K

answered Jun 09 '17 at 21:36

xiawi

score 0 · Answer 3 · answered Jun 10 '17 at 06:01

Here is how you string replace in a windows batch file. {var} = %{var}:{find}={relacement}% Here is an example for you:

@echo off
set filename="REE_45K.txt"
echo %filename%
set filename=%filename:REE_=%
set filename=%filename:.txt=%
echo %filename%

Output:

"REE_45K.txt"
"45K"

3 Answers3