bash - Passing a command as a variable to a function in Unix -

can pass command variable function? if yes, syntax?

my code looks like.

function1(){  in $1     echo $a done }  function2(){  function1 "ls -lrt folder/name | grep 'foo' | grep 'bar'"  } 

but doesn't work. tried passing as:

function1 `ls -lrt folder/name | grep 'foo' | grep 'bar'`  

but passes first value of command (and understand why happens).

does know syntax pass command function variable?

as indicated in comments, using $1 contain first parameter. instead, need use $@.

from bash manual → 3.4.2 special parameters:

$@

($@) expands positional parameters, starting one. when expansion occurs within double quotes, each parameter expands separate word. is, "$@" equivalent "$1" "$2" …. if double-quoted expansion occurs within word, expansion of first parameter joined beginning part of original word, , expansion of last parameter joined last part of original word. when there no positional parameters, "$@" , $@ expand nothing (i.e., removed).

then ask

what if function1 takes multiple parameters? , command third parameter being passed, using $@ takes in unnecessary values. how take values $3 onwards?

for this, want use shift:

$ cat myscript.sh #!/bin/bash shift 2  echo "$@" $ ./myscript.sh b c d c d 

shift

 shift [n] 

shift positional parameters left n. positional parameters n+1 … $# renamed $1 … $#-n. parameters represented numbers $# $#-n+1 unset. n must non-negative number less or equal $#. if n 0 or greater $#, positional parameters not changed. if n not supplied, assumed 1. return status 0 unless n greater $# or less zero, non-zero otherwise.