6

Why does this work:

$ cat test.sh
#!/bin/bash
cat <(date|awk '{print $1}')

$ ./test.sh
Thu

but not as a command passed to bash:

$ bash -c "cat <(date|awk '{print $1}')"
Thu  2 Apr 2020 12:52:10 BST

I'm running this on macOS

$ bash --version
GNU bash, version 3.2.57(1)-release (x86_64-apple-darwin19)
Copyright (C) 2007 Free Software Foundation, Inc.
22

In your one-liner, the $1 is within a double quoted string:

"cat <(date|awk '{print $1}')"

This means that it will be expanded by the interactive shell. If the first positional parameter, $1, is empty, it will expand it to an empty string, resulting in an awk program that just prints its input:

awk '{print }'

Escape the $ in $1 as \$1 to stop the shell from trying to expand it. Your complete command would then look like

bash -c "cat <(date|awk '{print \$1}')"

In your script variant, the $1 is in a single quoted string. The shell will therefore not try to expand it.

In this specific case though, it would be better to use just

date +%a

to print the abbreviated name of the day. See also the strftime(3) manual (man 3 strftime).

In newer bash versions (release 4.2+, i.e. not with the default bash on macOS, but with the one installed through e.g. Homebrew),

printf '%(%a)T\n' -1

would do the same thing without invoking any external commands.

| improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.