Very Computer

Hello,
how can i extract the 3 last characters of a string ?

Thanks

--
Jean-Paul Saliou
Alcatel Telecom - Brest
Tel     : 02.98.14.31.82

Quote:>Hello,
>how can i extract the 3 last characters of a string ?

>Jean-Paul Saliou

  It isn't possible with standard bourne shell.  

  With the posix shell it is possible.  In the posix shell there are
a couple of string addressing mechanisms that deal with arbitrary
chunks of a given string.  Refer to the ${VAR%pattern) part of your
local man page.

so, with a posix shell, you could get the last three letters of a
variable by something like:

---
read var
until case "$var" in ???|??|?|'') ;; *) false ;; esac
do
  var="${var#?}"
done
echo "$var"
---
(get variable. start loop where we check to see if it is 3 or fewer
characters, if not, chop off first character and check again.  Then
print the variable.)

  Unlike the [[ ]] and extended metacharacters, these structures
are available in any shell that pretends to be posix compliant (bash,
ash, pdksh, probably zsh).

  There is probably a shorter way to do this using sed, awk, perl,
C, fortran, ed, emacs, bliss, and perhaps even cut.

chris

: Hello,
: how can i extract the 3 last characters of a string ?

One method is with expr(1)
  last=`expr "$instr" : '.*\(...\)$'`

The value after the colon operator is a regular expression, which can
allow one grouping (parens).  If a grouping is not included, the number
of characters matched is printed to stdout.  If a grouping is included,
the matched substring is printed.  In either case, the return code is
zero (0) if there was a match, one (1) otherwise.

The regular expression is:
  .*    - match any number of characters
  \(    - start a regular expression grouping
  ...   - (three dots) three of any character
  \)    - end the grouping
  $     - match the end-of-string

expr(1) implicitly matches from the beginning of the string, so you must
have the initial ".*".

I hope this helps.
  -Arcege

'sed' will do it. Some systems have an expanded 'expr' which will
do it. I'm sure you could do it with 'awk'. Start reading man pages.

$ echo "ABCDEF12:345deasdsdf" | sed        's/.*\(...\)$/\1/'
sdf

Quote:>how can i extract the 3 last characters of a string ?

If you're using ksh, bash, POSIX shell etc, try using:

typeset -R3 fred=${jim}

Though, be warned; in a function this will make a local variable so it
won't be publically available outside the function, eg:

function foo
{
   typeset fred=jim

Quote:}

foo
echo ${fred}

dave
--

"Dear God above (if you exist), Hope you see the funny side to this,
 Now don't get cross - don't bite your nails,
 Oh, Son of Man your mission's failed."
 - Skyclad "The Sinful Ensemble"

Use typseset -R3 varname=oldvar for last three chars
ans typeset -L3 varname=oldvar forfirst three chars
panneer

Bash doesn't implement the justification options to typeset.  You can
use
        foo=abcdefgh
        echo ${foo: -3}

Beware -- the space after the `:' is required, to avoid confusion with
the ${foo:-bar} parameter expansion.
--
``The lyf so short, the craft so long to lerne.'' - Chaucer
( ``Discere est Dolere'' -- chet)

[ This is a repost of an article that didn't make it beyond the local
  network.  Sorry if the topic is a bit out of date. ]

POSIX shells are not required to support "typeset".  Those that do
are probably just ksh in disguise.

--

How about this?
expr $string : '.*\(...\)'