[SOLVED] Alias with Argument in Bash - Mac

Issue

Hi there trying to add an argument to an alias in bash. I'm using a Mac. I've searched a number of related questions: Alias in Bash, Alias in Bash with autocomplete, and a few others yet still can't sort this out. I know I need to use a function to create an alias that takes an input but it's unclear if its meant to look like any of the below options. All inputs in .bash_profile.

function mins_ago () { `expr $(date +%s) - 60 \* "$1"`; }

alias mins_ago = function mins_ago () { `expr $(date +%s) - 60 \* "$1"`; }

alias mins_ago = "function mins_ago () { `expr $(date +%s) - 60 \* "$1"`; }"

alias mins_ago = function mins_ago () { `expr $(date +%s) - 60 \* $1`; }

None of these seem to work. I either get a syntax error or it doesn't recognize the argument.

What is the actual line you'd put in .bash_profile to sort this properly? Thank you in advance. Is the alias bit included or do I just proceed to the function definition?

Solution

Bash aliases don't support arguments, so you need to use a bash function, and use the bash arithmetic operator $(())

function mins_ago() {
    printf "%s" "$(( $(date +%s) - (60 * $1) ))"
}

Add the above function in .bash_profile, and now testing it in the command-line,

date +%s
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value="$(mins_ago 3)"
printf "%s\n" "$value"
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(or) without a temporary variable to convert to readable format in GNU date, do

printf "%s\n" "$(date -d@$(mins_ago 3))"

Answered By - Inian