How to read multiple words as one argument

Question

I am trying to write a shell script that takes the first word given to it as one argument and anything that comes after it as a second argument.

The user must call my shell script and provide their arguments in one line from the terminal: shellscript.sh word1 word2 word3 ...... wordn How can I write my script such that

arg1 = word1

arg2 = word2 - wordn?

I would recommend making it the caller's responsibility to construct the 2nd argument: `shellscript.sh word1 "word2 word3 ..."`. — chepner, Feb 19 '19 at 21:49
See [Bash, argument list segment](https://stackoverflow.com/q/2390738/4154375). — pjh, Feb 20 '19 at 19:08

choroba · Answer 1 · 2019-02-19T22:03:59.463

1

Use shift to remove the first argument, and use "$*" to concatenate the remaining ones.

#!/bin/bash
first=$1
shift
rest="$*"  # Assuming the first character of $IFS is a space.

printf '<%s>\n' "$first" "$rest"

edited Feb 19 '19 at 22:03

answered Feb 19 '19 at 21:30

choroba

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Note that you are changing `IFS` globally; assignments don't themselves take pre-command modifiers. – chepner Feb 19 '19 at 21:54
@chepner: Thanks, fixed. – choroba Feb 19 '19 at 22:04

How to read multiple words as one argument

1 Answers1