Combine two command defined as String in variable and run as one [duplicate]

Question

Can I join two commands defined as String and run as one. E.g.

var="ls -alt"
var2="| grep usersfile"
var3="| grep usersfolder"

Following example for joining commands does not work.

a. '{$var & $var2;}'

b. '{$var & $var3;}'

What I what is:

a. run "$var $var2" and get "ls -alt | grep usersfile"

or

b. run "$var $var3" and get "ls -alt | grep usersfolder"

Answer 1

Yes, you can do this by using eval.

$ var="ls -alt"
$ var2="| grep usersfile"
$ var3="| grep usersfolder"
$ eval "$var$var2"
< Output of ls -alt | grep userfile >
$ eval "$var$var3"
< Output of ls -alt | grep usersfolder >

Answer 2

You probably shouldn't do it this way. Depending on what you are actually trying to accomplish there are likely other options:

If you are just trying to have a dynamic option to select your search string you could do something like this instead:

#!/bin/bash
pattern=$1

ls -alt | grep "${pattern:-.}"

This will search for whatever your first positional parameter is, or anything if none is provided.

Or if you're trying to act on some condition:

#!/bin/bash

cmd=(ls -alt)
p1=usersfile
p2=usersfolder

if [[ condition1 ]]; then
    p=$p1
elif [[ condition2 ]]; then
    p=$p2
fi

"${cmd[@]}" | grep "$p"

---

Or in a function:

#!/bin/bash

ls_func () {
    local p=$1
    ls -alt | grep "${p:-.}"
}

while getopts p:c: opt; do
    case $opt in
        c)  command=$OPTARG;;
        p)  pattern=$OPTARG;;
    esac
done

if [[ $command == 'ls' ]]; then
    ls_func "$pattern"
fi