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Get all arguments except those matching pattern

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# 1  
Old 04-13-2012
Get all arguments except those matching pattern
I am working in Ubuntu 11.10 writing some bash scripts

I have a tcsh script containing the following code that I am converting to a bash script.

It reads all the arguments passed to the tcsh script (without the script name and fields matching certain patterns):

Code:
strLst=`echo $argv[$firstArg-$lastArg] | tr ' ' '\n'  \
  | grep -ivE --   '(--check)' \
  | grep -ivE --   '(--sort)'  \
  | grep -ivE --   '(--group)' \
  | grep -ivE --   '(--shift)' \
  | grep -ivE --   '(-v|--vrb-level)' \
  | grep -ivE --   '(-q|--quiet)'     \
  | grep -ivE --   '(-u|--usage)'     \
  | grep -ivE --   '(-e|--examples)'  \
  | grep -ivE --   '(-h|--help)'      \
  | grep -ivE --   '(--shift=[0-9]+)' \
  | grep -ivE --   '--((sort|group)=[0-9]+/[0-9]+(/[0-9]+)*)$' | tr '\n' ' '`

My problem is how to get all arguments, an equivalent to
Code:
$argv[$firstArg-$lastArg]

# 2  
Old 04-13-2012
$* or $@ (they have slightly different behaviour when quoted - check the bash man page).
# 3  
Old 04-13-2012
running the below did not return anything

Code:
echo "Args $*"          # returns all arguments
echo "Args $@"

---------- Post updated at 09:52 AM ---------- Previous update was at 08:56 AM ----------

Problem fixed
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