如何传递多个 arguments 并读入 shell 脚本

Question

I need to read multiple arugments are assign each to different variable in shell script

example:

.run.sh --argument1=value1 --argument2=value2 --argument3=value3

In my code I need to read this and assign as,

variable1=value1

variable2=value2

variable3=value3

I was trying to read in a loop but not sure how to read the value immediately after the "=" sign

for i in "$@"
do
case $i in
--argument1)
    variable1="..."
    shift
    ;;
esac
done 

Answer 1

Set variable1=$2 then shift 2 .

Although I would recommend instead using getopts for parameter parsing. A good reference would be this blog post by Kevin Sookocheff .

Answer 2

Perhaps you need an associative array.

#!/usr/bin/env bash

n=1
declare -A array

for f; do
  array["variable$((n++))"]=${f##*=}
done

echo "${array[variable1]}"
echo "${array[variable2]}"
echo "${array[variable3]}

Running that against your sample arguments.

./run.sh --argument1=value1 --argument2=value2 --argument3=value3

Output

value1
value2
value3

Although an option parser might be better for this kind of situation.