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I think u will need to find the diff of current that with the start and end date of the month u want backup.
and u may use the find . -mtime -60 -mtime +30
-mtime -60,...less than 60 days ago...

well show what u have done..so far.......may be we can help , once u get started with it

You need to understand this isn't a tutor class..there are lots of tutorials available on unix scripting u should try them :-)

Input 1- 100
Input 2 -1000
hope thats what u wanna do....

#! /bin/ksh
typeset -i ranger=$1
typeset -i mark=$2
typeset -i range=`expr $ranger - 1`
while [[ $ranger -lt $mark ]]
do
echo...

try
df -k

This is the prelim and u will have to look for tighter time checking ..but thats the start....
#! /bin/ksh
input=$1
echo $input
if [[ $input = [0-2][0-9]:[0-6][0-9] ]]
then
echo "Correct input"...

I dont have any clue for how to do this either..but may be this is the way forward... may be someone else from the group can help u sorting out this messy script to tell how to do this :-)
#!...

or awk solution

awk ' NR>=3&&NR<=6 {print}' file

I think this will do the trick..but whats the requirements is..still not sure about..
sed -e 's/[^~]*\/\(~[^/]*\).*/\1/' data2

use
sort -t "U" +1nr data1

see if it solves urs purpose.....

have u tried dir command
type:- I cant understand what u looking for??? but i guess thats it...or u want to see files within dir also???


dir

sed -e '1,/pattern/s/pattern/REPLACEMENT/' <INPUTFILE

dont ask me why it works...cause that will sound like homework....just figure it out :-)
regards

can u provide the data..and by example what u really want.....can't get urs requirements

can please someone explain me the diff between
It works perfectly fine to give last word
sed -e 's/.*,\([^,]\)/\1/' dat1
xyz
rty
tsrt

Shouldn't this also give the last word..i accidently...

u can use the cut to extract the field required..this case the fifth field(try to check to make sure that urs system is using the same format for passwd file)and then u can use the grep command to...

I think u want this........
sed -e '/^$/d' -e '/^ \{3,\}/d' Inputfilename
but beware it may not give the same result as the sequential application of the same commands....
Regards.

yah....but as in the file was contaning one space before "...but yah it would be better as sed -e 's/[ ]*"/"/g' inputfile>outputfile

complicating the uncomplicated things.....
sed -e 's/[ ]"/"/g' inputfile>outputfile...
in case u have read the request is to act on file and not on echo statemtn...
as for info ...echo removes the...

sed -e 's/[ ]"/"/g' oldfile>newfile

I have a CC native compiler on solaris10..it works perfectly on it..and i dont think that ..const int max =20;" instead of "#define max 20 is required...C++ ANSI standrads dont prohibits it.

I think u have some confusion regarding variable and file name..hope this will clarify it.

#! /bin/ksh
ls -t tmp/backup/$filename.* | sed 1,2d | while read FILEVAR ; do rm $FILEVAR
done

sed 's/|[ ]*/|/g' data
this will solve urs purpose

Do you want the older file deletion...u can achieve it by ctime of find command..
find /usr/local/apache/htdocs/website/zip_backup/ -name -ctime +2 -exec rm {} \;
This will remove the files which...

Hope this is what u looking for
sed -e "s/\(.*\)/'\1'/" inputfile > outputfile

Thanks man...