while loop to change variable number and check for value

Question

im new to bash and im trying to figure out the simplest way to create a while loop that changes the variable number, and checks for a value. this way i can simply add to the variables and the if function knows to keep looping to the end

var='start loop'
var1='something'
var2='something else'
var3='maybe nothing but probably something'
function run() {
 while [ ! -z "$var" == true ]
 do
  x=1
  var=var$x
  if ! command -v $var &> /dev/null
  then
   DO SOMETHING
  fi
}
x=x+1
done

Use an array to store multiple things: linuxhandbook.com/bash-arrays — TessellatingHeckler
– TessellatingHeckler, Commented Jul 29, 2022 at 12:39
while [ ! -z "$var" == true ] ?? Surely this gives errors. — William Pursell
– William Pursell, Commented Jul 29, 2022 at 18:47
@WilliamPursell yea probably. I am the king of errors. Thank you though. The help was already received and i was able to get it to work — Bryan
– Bryan, Commented Jul 29, 2022 at 19:39

Bryan · Accepted Answer · 2022-07-29 18:55:52Z

1

Array was the way. Thank you

var=('thing' 'stuff' 'otherStuff')
function run() {
  x=0
  while [ ! -z "${var[x]}" ]
  do
    if ! command -v ${var[x]} &> /dev/null
    then
      apt-get install ${var[x]} -y
    fi
  sleep 2s
  x=$x+1
  done
}

edited Jul 29, 2022 at 18:55

answered Jul 29, 2022 at 18:36

Bryan

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