Concatenate strings in multiple lines using bash

Question

Given a list of arguments, I am trying to produce a string looks like the following using bash script:

 - arg_1
 - arg_2
 - arg_3
 - arg_4

And here is my script trying to do this:

seeds=
for i in $*; do
  seeds=`printf "$seeds    - $i\n"`
done;
echo "$seeds"

However, when I run it with input arg_1 arg_2 arg_3 arg_4, the above script generates the following output, which apparently ignores my spacing and new-line character:

- arg_1 - arg_2 - arg_3 - arg_4

Can I know what did I do wrong in my script? or is there a better way to do this?

Chris Seymour · Accepted Answer · 2013-08-02 20:11:35Z

5

Quoting and whitespace is very important in the shell:

#!/usr/bin/env bash

for arg in "$@"; do
    seed="$seed    - $arg\n"
done
echo -ne "$seed"

Demo:

$ ./script.sh arg_1 arg_2 arg_3 arg_4
    - arg_1
    - arg_2
    - arg_3
    - arg_4

edited Aug 2, 2013 at 20:11

answered Aug 2, 2013 at 20:02

Chris Seymour

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2 Comments

keelar Over a year ago

In addition, I am not trying to print them, but more importantly storing them in a string since I will later invoke sed to use them to modify a file.

Chris Seymour Over a year ago

@keelar stackoverflow.com/questions/12314451/…

that other guy · Accepted Answer · 2013-08-02 20:28:33Z

3

Like sudo_O says (and ShellCheck would have pointed out), you should use "$@" rather than $* for the minimal fix.

You asked if there was a better way of doing it, though, and there definitely is. When given more arguments than necessary to fulfill the format string, printf will repeat the output as necessary. This lets us simply write:

printf "- %s\n" "$@"

answered Aug 2, 2013 at 20:28

that other guy

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2 Comments

Chris Seymour Over a year ago

+1 for SpellCheck however check the OP comments on my answer I am not trying to print them, but more importantly storing them in a string.

Gordon Davisson Over a year ago

@sudo_O: That's easily changed: seeds=$(printf "- %s\n" "$@" ) (although note that the final linefeed will be trimmed before it's stored in the variable -- if this is a problem, there are solutions but a bit messy).

user1146332 · Accepted Answer · 2013-08-02 20:07:23Z

1

I think your script is too complex. I don't think you need command substitution here. May you try this one

seeds=
for i in $*; do
  seeds="$seeds\n- $i"
done;
echo -e ${seeds#\\n}

edited Aug 2, 2013 at 20:07

answered Aug 2, 2013 at 20:00

user1146332

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3 Comments

keelar Over a year ago

Thank you, it seems work except it introduces one additional new line at the beginning (although it's easy to fix that).

user1146332 Over a year ago

I changed it according to your needs

user1146332 Over a year ago

no only the first one. From bash's reference manual:

If the pattern matches the beginning of the expanded value of parameter, then the result of the expansion is the expanded value of parameter with the shortest matching pattern deleted

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