Double quotes in bash

Question

I am trying to create a bash file which calls a program that expects a parameter to submitted between two double quotes.

Take a look at my file content.

#!/bin/bash
echo ./yowsup/yowsup-cli demos --s $1 \"$2\" -c yowsup/config &
./yowsup/yowsup-cli demos --s $1 \"$2\" -c yowsup/config &

The echo part prints exactly the command I need, but when I comes to execute it (next line), I am messing it up.

How can I get this fixed?

Thanks!

This is how I need to call it:

./yowsup/yowsup-cli demos --s 9999999999 "my parameter" -c yowsup/config

Echo result:

./yowsup/yowsup-cli demos --s 9999999999 "my parameter" -c yowsup/config

Call result:

demos: error: unrecognized arguments: parameter

Copying and pasting echo results:

INFO:yowsup.demos.sendclient.layer:Message sent

Answer 1

You need to escape the quotes in the echo command to see them in the output, but you shouldn't escape them in the actual command:

#!/bin/bash
echo ./yowsup/yowsup-cli demos --s $1 \"$2\" -c yowsup/config &
./yowsup/yowsup-cli demos --s $1 "$2" -c yowsup/config &

The quotes are used in the actual command to prevent splitting of $2 - which contains multiple words.

Answer 2

I (nearly) always quote everything to "echo", so I would do the above like:

#!/bin/bash
echo "./yowsup/yowsup-cli demos --s $1 \"$2\" -c yowsup/config &"
./yowsup/yowsup-cli demos --s $1 "$2" -c yowsup/config &

The difference with @muru solution is, that there the original echo command is run "in the background" - here it's echoing to the screen in the foreground, but run as a background command. (@muru's also doesn't show the ampersand at the end of the command.)