Passing Flags to a Piped Script

Question

I have written a shell script to configure a development environment and and retrieving using cURL. The script takes up to 3 flags, -d, -f and -s.

How do I pass the flags to the shell script?

Here is the command to run the bash script:

$ curl -sL https://example.com/setup.sh | bash

Here is my first (failed) attempt to pass flags to the script:

$ curl -sL https://example.com/setup.sh | bash -dfs
  bash: -d: invalid option

Can anyone explain how to do this?

try adding `--` to signal the end of bash arguments and the start of command arguments: `curl -sL https://example.com/setup.sh | bash -- -dfs` — joshmeranda, May 31 '22 at 13:20
@joshmeranda, that causes bash to treat `-dfs` as a filename to execute (that being the default way the first non-option argument is parsed). — Charles Duffy, May 31 '22 at 13:21
Possibly duplicate of [Pass args for script when going thru pipe](https://stackoverflow.com/q/14693100/7939871) — Léa Gris, May 31 '22 at 13:56

score 2 · Answer 1 · answered May 31 '22 at 13:20

2

Use the -s argument:

curl -sL https://example.com/setup.sh | bash -s -- -dfs

answered May 31 '22 at 13:20

Charles Duffy

I think this is a dupe of https://stackoverflow.com/questions/14693100/pass-args-for-script-when-going-thru-pipe – Léa Gris May 31 '22 at 13:57
@LéaGris, agreed, thank you. Converted to community wiki, and lent my dupehammer to the close vote. – Charles Duffy May 31 '22 at 14:57

1 Answers1