At this moment I have:
#!/bin/bash
screen -p 'ScreenName' -x eval 'stuff '"'"$@"'"'\015'
echo eval 'stuff '"'"$@"'"'\015'
But when I call my script as:
# script.sh asd "asd" 'asd'
my arguments passed as: asd asd asd
and I get output:
eval stuff 'asd asd asd'\015
I except a: asd "asd" 'asd'
How I can change my script to pass whole arguments line with all quotes?
Your shell is not passing the quotes through to the script. If you want to pass quotes, escape them with a backslash:
# ./script.sh asd \"asd\" \'asd\'
sh -c "screen -x 'ScreenName' -X eval 'stuff \"$@\"\015'"
The only solution is to indicate to your shell that these are not syntax but actual content, like this :
script.sh $'asd "asd" \'asd\''
# ^^ ^ ^ ^
# || | | |
# 12 3 3 2
1: prefixing a string by $ change its meaning, allowing to backslash-escape quotes, cf https://unix.stackexchange.com/a/30904/383566
2: enclosing the value between single quotes indicate to the shell that it must be treated as a single argument
3: backslash-escaping the single quote, because it is used as enclosing syntax (see 2)
Proof :
# defining a function to test
> function count_and_print_args() { # if not using bash, remove the "function"
echo $# # print the number of arguments
for arg in "$@" # for each of all arguments
do
echo "$arg" # print the argument on a newline
done;
}
> count_and_print_args
0
> count_and_print_args a b c
3
a
b
c
> count_and_print_args $'asd "asd" \'asd\''
1
asd "asd" 'asd'
