Continue while loop in Bash Script even if a condition fails

Question

#!/bin/bash
var="true"
i=1
while $var
do
  read -p "Enter value (true/false): " var
  if [[ $var == "true" ]]
  then
    echo "Iteration : $i"
    ((i++))
  elif [[ $var == "false" ]]
  then
    echo "Exiting the process"
  elif [[ $? -eq 1 ]]
  then
    echo "Invalid Choice."
    echo "Avaialable Choices are true or false"
    exit
  fi
done

Script is Working Fine. I Enter true the loop will iterate for false the script stops. I want the script will continue asking "Enter Value" if any other value instead of true or false will be entered.

The elif [[ $? -eq 1 ]] isn't doing anything useful -- the $? it's checking is the exit status of the last command, which is [[ $var == "false" ]], which clearly did fail or it wouldn't have gotten to this point. Just use a plain else instead. Also, your quoting reflexes are a bit backward; generally, static single-word strings without any special characters (like true and false) don't need to be quoted, but variable references like $var should be double-quoted to avoid weird parsing if they do contain weird characters. — Gordon Davisson
– Gordon Davisson, Commented Oct 19, 2021 at 19:14
@GordonDavisson : $var nees to be quoted only in the while line. It does not need to be quoted later, because we are inside [[ ... ]]. — user1934428
– user1934428, Commented Oct 20, 2021 at 7:25
@user1934428 True. I was going to point that out, but ran into the comment size limit. The reason I commented was that quoting simple fixed strings but not variable references suggests a basic misunderstanding of what quotes do in shell syntax. BTW, not quoting isn't always safe even in [[ ]] -- in [[ $var1 = $var2 ]], $var2 will be treated as a glob pattern unless it's double-quoted. The rules for when it's safe to leave off double-quotes are complex enough that I generally recommend double-quoting all var references unless there's a specific reason not to. — Gordon Davisson
– Gordon Davisson, Commented Oct 20, 2021 at 7:58

Léa Gris · Accepted Answer · 2021-10-19 19:01:48Z

2

This would do the same with a more academic syntax:

i=0
while :; do
  printf 'Enter value (true/false): '
  read -r var
  case $var in
    true)
      i=$((i + 1))
      printf 'Iteration : %d\n' $i
      ;;
    false)
      printf 'Exiting the process\n'
      break
      ;;
    *)
      printf 'Invalid Choice.\nAvaialable Choices are true or false\n'
      ;;
  esac
done

edited Oct 19, 2021 at 19:01

answered Oct 19, 2021 at 18:56

Léa Gris

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Comments

squidwardsface · Accepted Answer · 2021-10-19 19:42:24Z

2

You might find this to be a cleaner solution:

i=0
while true; do
  read -p "enter value: " myinput
  if [[ $myinput = true ]]; then
    echo "iteration $i"
    i=$((i+1))
  elif [[ $myinput = false ]]; then
    echo "exiting"
    exit
  else
    echo "invalid input"
  fi;
done;

The issue I see with your current code is that it is unclear which command's exit status $? refers to. Does it refer to the echo in the previous elif block? Or the last condition check? Or something else entirely?

edited Oct 19, 2021 at 19:42

answered Oct 19, 2021 at 19:17

squidwardsface

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Comments

kurth4cker · Accepted Answer · 2021-10-19 18:40:17Z

1

I'm new in bash. I tried that:

#!/bin/bash
i=1
while [[ $var != "false" ]]
do
    read -p "Enter value (true/false): " var
    if [[ $var == "true" ]]
    then
    echo "Iteration : $i"
    ((i++))
    elif [[ $var == "false" ]]
    then
    echo "Exiting the process" 
    elif [[ $? -eq 1 ]]
    then
    echo "Invalid Choice."
    echo "Avaialable Choices are true or false"
    fi
done

I changed while $var with while [[ $var ]] because while works like if. It runs the given command. In there it is $var's value.

And I moved exit to first elif expression's end. So if user type false program will exit.

answered Oct 19, 2021 at 18:40

kurth4cker

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