get the last 2 char from a variable

tads98 · #1 (**permalink**) 10-12-2005

i have 2 variables
VAR1=12345_SNT.in1
VAR2=123456_LXP.in2

how can i get the last 2 characters before the .
for VAR1, i only want to get the NT
for VAR2, i only want to get the XP

the length of the variable varies, so does its extension. the only thing that is consistent is the "."

thanks!

tmarikle · #2 (**permalink**) 10-12-2005

Code:

echo $VAR1 | sed 's/.*\(..\)\..*/\1/'

NT

echo $VAR2 | sed 's/.*\(..\)\..*/\1/'

XP

tads98 · #3 (**permalink**) 10-12-2005

Quote:

Originally Posted by tmarikle

Code:

echo $VAR1 | sed 's/.*\(..\)\..*/\1/'

NT

echo $VAR2 | sed 's/.*\(..\)\..*/\1/'

XP

thanks! that's great! i tested it and it worked!

just to understant things... what does this mean? sed 's/.*\(..\)\..*/\1/'

futurelet · #4 (**permalink**) 10-12-2005

Code:

echo $VAR1 | ruby -ne 'puts $_[/(..)\./,1]'

tmarikle · #5 (**permalink**) 10-12-2005

Code:

sed = stream editor

general form = s/search expression/replace expression/

search expression = 
   .*     matches any string of characters
   \(..\) matches exactly two characters and preserves the matching values in a place holder
   \.     matches exactly one period; your constant
   .*     matches the remainder of the string

replace expression =
  \1 only inserts the entire string that was contained in \( and \)

Ygor · #6 (**permalink**) 10-12-2005

Perhaps just use expr instead of echo|sed...

Code:

expr $VAR1 : ".*\(..\)\..*"

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