Passing argument to alias in bash [duplicate]

Question

This question already has an answer here:

• Make a Bash alias that takes a parameter? 13 answers

Is it possible to do the following:

I want to run the following:

mongodb bin/mongod

In my bash_profile I have

alias = "./path/to/mongodb/$1"

Answer 1

An alias will expand to the string it represents. Anything after the alias will appear after its expansion without needing to be or able to be passed as explicit arguments (e.g. $1).

$ alias foo='/path/to/bar'
$ foo some args

will get expanded to

$ /path/to/bar some args

If you want to use explicit arguments, you'll need to use a function

$ foo () { /path/to/bar "$@" fixed args; }
$ foo abc 123

will be executed as if you had done

$ /path/to/bar abc 123 fixed args

To undefine an alias:

unalias foo

To undefine a function:

unset -f foo

To see the type and definition (for each defined alias, keyword, function, builtin or executable file):

type -a foo

Or type only (for the highest precedence occurrence):

type -t foo

Answer 2

Usually when I want to pass arguments to an alias in Bash, I use a combination of an alias and a function like this, for instance:

function __t2d {                                                                
         if [ "$1x" != 'x' ]; then                                              
            date -d "@$1"                                                       
         fi                                                                     
} 

alias t2d='__t2d'                                                               

Answer 3

to use parameters in aliases, i use this method:

alias myalias='function __myalias() { echo "Hello $*"; unset -f __myalias; }; __myalias'

its a self-destructive function wrapped in an alias, so it pretty much is the best of both worlds, and doesnt take up an extra line(s) in your definitions... which i hate, oh yeah and if you need that return value, you'll have to store it before calling unset, and then return the value using the "return" keyword in that self destructive function there:

alias myalias='function __myalias() { echo "Hello $*"; myresult=$?; unset -f __myalias; return $myresult; }; __myalias'

so..

you could, if you need to have that variable in there

alias mongodb='function __mongodb() { ./path/to/mongodb/$1; unset -f __mongodb; }; __mongodb'

of course...

alias mongodb='./path/to/mongodb/'

would actually do the same thing without the need for parameters, but like i said, if you wanted or needed them for some reason (for example, you needed $2 instead of $1), you would need to use a wrapper like that. If it is bigger than one line you might consider just writing a function outright since it would become more of an eyesore as it grew larger. Functions are great since you get all the perks that functions give (see completion, traps, bind, etc for the goodies that functions can provide, in the bash manpage).

I hope that helps you out :)

Answer 4

This is the solution which can avoid using function:

alias addone='{ num=$(cat -); echo "input: $num"; echo "result:$(($num+1))"; }<<<'

test result

addone 200
input: 200
result:201

Answer 5

In csh (as opposed to bash) you can do exactly what you want.

alias print 'lpr \!^ -Pps5'
print memo.txt

The notation \!^ causes the argument to be inserted in the command at this point.

The ! character is preceeded by a \ to prevent it being interpreted as a history command.

You can also pass multiple arguments:

alias print 'lpr \!* -Pps5'
print part1.ps glossary.ps figure.ps

(Examples taken from http://unixhelp.ed.ac.uk/shell/alias_csh2.1.html .)

Answer 6

To simplify leed25d's answer, use a combination of an alias and a function. For example:

function __GetIt {
    cp ./path/to/stuff/$* .
}

alias GetIt='__GetIt'