Alias with Argument in Bash - Mac

Question

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?

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score 2 · Accepted Answer · edited Oct 19 '20 at 00:08

2

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
1485414114

value="$(mins_ago 3)"
printf "%s\n" "$value"
1485413834

(or) without a temporary variable to convert to readable format in GNU date, do

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

edited Oct 19 '20 at 00:08

Rob Bednark

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answered Jan 26 '17 at 06:45

Inian

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I'm running what you posted above. Ah! `source ~/.bash_profile` didn't take for some reason, but exiting and trying again worked. Thank you sir. – HectorOfTroy407 Jan 26 '17 at 07:21
@HectorOfTroy407: Happy to be of help! – Inian Jan 26 '17 at 07:22

score 1 · Answer 2 · answered Jan 26 '17 at 07:09

add this to your .bashrc and source it

mins_ago() {
  if [[ $@ != "" ]]; then
    command expr $(date +%s) - 60 \* "$@"
  else
    command echo "command error: mins_ago <integer>"
  fi
}

output:

$ mins_ago 1
1485414404
$ mins_ago
command error: mins_ago <integer>

Djura · Answer 3 · 2017-01-26T06:41:54.230

0

Just type

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

It works for GNU bash 4.1.2:

$ mins_ago 5
bash: 1485411776: command not found

To avoid the error, use echo:

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

edited Jan 26 '17 at 06:41

answered Jan 26 '17 at 06:30

Djura

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4

Rather than adding `echo`, remove the backquotes. – Gordon Davisson Jan 26 '17 at 06:42
`-bash: /Users/CloudFlare/.bash_profile: line 13: syntax error near unexpected token `{`expr $(date +%s) - 60 \* $1`' -bash: /Users/CloudFlare/.bash_profile: line 13: `function _hours(){`expr $(date +%s) - 60 \* $1`;} _hours '` when i try `mins_ago () { expr $(date +%s - 60 \* "$1"; }` – HectorOfTroy407 Jan 26 '17 at 06:51
@HectorOfTroy407: Can you try my answer above? – Inian Jan 26 '17 at 06:54

score 0 · Answer 4 · answered Jan 26 '17 at 06:33

0

Remove the backticks from your first try, and you should be fine.

The function will then output the timestamp, and you can write echo "$(mins_ago)".

answered Jan 26 '17 at 06:33

Roland Illig

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Alias with Argument in Bash - Mac

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