Hi
I need to delete first two words from the arguments my script receives .. no matter how many arguments are passed .

Thanks
Praveen

should give you your third argument

Hyy , I need all the remaining argument , not only the third , for eg:- if the total arguments are 7 the output shuld be the last 5 arguments .. and if the total argumetns are 10 thn the o/p shuld be 8 .. hope you got my requirements

Hope this should help you or else would give u an idea on how to proceed...

echo "one two three four" | cut -d " " -f3-
three four

Please correct me if i understood ur problem in a wrong way.... Also give some sample i/p and o/p so that it will be easy to work on it....

shift does exactly what you require. "shift 2" deletes the first two arguments from the argument list, eg. $3 becomes $1, $4 becomes $2, etc. If you want to output all your arguments sans the first two, do this

Hey

You can get the count of total arguments being passed. and then you can preserve all of'em except first 2 arguments using for loop.

echo 'num of args: '$#

following for loop will display all the arguments being passed:
for arg in "$@";
do
echo 'arg: '$arg;
done

You can run the while loop 2 less then the count of $#
var=`$# - 2`

you can then run the loop in reverse order, till value of var is achieved.

Hope you got what I am trying to convey.

Thanks
Varun Gupta