Alias with Argument in Bash - Mac [duplicate]

问题:

This question already has an answer here:

• How to pass command line arguments to a shell alias? 12 answers

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?

回答1:

Defining alias is not the right approach when you can easily do it with functions, 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))"

回答2:

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>

回答3:

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)".

回答4:

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"` 
}