How to remove some arguments from the arguments list in bash script?

Question

Suppose I need a script, which removes a few specific arguments from the argument list and echo the rest. For instance:

The script is myscript.bash and the arguments to remove are b and e.

> myscript.bash a b c d e f
a c d f
> myscript.bash a b c
a c
> myscript.bash b e

How would you write such a script ?

It sounds nice. Please show what you tried so we can suggest or help wherever you found some problems. — fedorqui, Sep 24 '13 at 09:28

score 2 · Accepted Answer · edited Mar 19 '18 at 13:04

2

I would recommend using case to accomplish this and then echo -n to put the parameters on the same line. You can get all the arguments with $@. Borrowing the example from this post.

for var in "$@"
do
  case $var in
   "a") echo -n $var ;;
   "b") ;;
   [...]
  esac
done

edited Mar 19 '18 at 13:04

nwinkler

52,665
21
154
168

answered Sep 24 '13 at 09:30

cete3

647
6
19

How to remove some arguments from the arguments list in bash script?

1 Answers1