Pass quoted arguments to shell script and maintain quoting [duplicate]

Question

I'm attempting to re-use parameters sent to my script as parameters for a command I execute within my script. See example below where I execute mailx.

bash

$./myscript.sh "My quoted Argument"

myscript.sh

mailx -s $1

This ends up being executed as: mailx -s My Quoted Argument.

---

• I tried "$1", but my quotes are thrown away. (Incorrect statement, read answer below)
• I tried ""$1"" but my quotes are thrown away.
• I tried to do '$1' but that's strong quoting so $1 never gets interpreted.
• I realize I can do $@, but that gives me every param.
• .... you get the picture

Any help would be appreciated!

Answer 1

mailx -s "$1" correctly passes the value of $1 to mailx as-is, embedded spaces and all.

In the case at hand, this means that My Quoted Argument is passed as a single, literal argument to mailx, which is probably your intent.

In a shell command line, quotes around string literals are syntactic elements demarcating argument boundaries that are removed by the shell in the process of parsing the command line (a process called quote removal).

If you really wanted to pass embedded double-quotes (which would be unusual), you have 2 options:

• either: embed the quotes on invocation ./myscript.sh "\"My quoted Argument\""
• or: embed the quotes inside myscript.sh: mailx -s "\"$1\""