remove an arg from a command line argument in bash

bash_trim_array.sh

#!/bin/bash
# this is a demo of how to remove an argument given with the [-arg value] notation for a specific
# [arg] (-T in this case, but easy to modify)

echo $@
echo $#

i=0
ORIGINAL_ARGS=("$@")
TRIMMED_ARGS=()
while [ $i -lt ${#ORIGINAL_ARGS[@]} ]; do
  arg=${ORIGINAL_ARGS[$i]}
  echo "i = $i; oa[i] = $arg"
  if [ $arg == "-T" ]; then
    # we want to remove both the "-T" *AND* the following arg.  So we advance i here,
    # and also once more outside of the for loop.
    i=$((i + 1)) # careful!  ((i++)) will kill your script if you use "set -e"
  else
    TRIMMED_ARGS+=($arg)
  fi
  i=$((i + 1)) # careful!  ((i++)) will kill your script if you use "set -e"
done
echo "TRIMMED_ARGS = ${TRIMMED_ARGS[@]}; length = ${#TRIMMED_ARGS[@]}"

I had to use the double square bracket if on line 14 and also had to wrap $arg in double quotes on line 19.

oh thanks for pointing this out @kczx3 . I assume you had an argument with a space in it? I think quotes around $arg in both places would have also worked (but I admit, my bash isn't great). Probably double brackets are fine too.