While loop in BASH script causing syntax error

Question

I recently started writing BASH scripts, and I am currently trying to practice using while loops. However, when I run the following block of code, the command prompt responds with:

run.command: line 12: syntax error near unexpected token `done'
run.command: `done'

Then the program shuts off. This is the code I am running.

#!/bin/bash
echo -e "text"
c=false
while true; do
    printf ">> "
    i=read 
    if [$i = "exit"]; then
        exit
    else if [$i = "no"]; then
        echo "no"
    else
        echo -e "Error: $i is undefined"
    fi
done

I did some research on while loops, however my loop syntax seems correct. When I remove the done at the end, and Unexpected end of file error occurs. Any help would be appreciated!

Please take a look: shellcheck.net

Cyrus
– Cyrus

2016-01-27 17:29:20 +00:00
Commented Jan 27, 2016 at 17:29 — Cyrus
– Cyrus, Commented Jan 27, 2016 at 17:29

Walter A · Accepted Answer · 2016-01-27 21:25:23Z

1

You can use the -p option of read for the prompt and a case ... esac construction:

while true; do
  read -r -p ">> " i
  case "$i" in
     "exit") exit 0  ;;
     "no") echo "no" ;;
     *) echo -e "Error: $i is undefined";;
  esac
done

answered Jan 27, 2016 at 21:25

Walter A

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1 Comment

Jerrybibo Over a year ago

Thanks! I'll do more research on this.

Jerrybibo · Accepted Answer · 2018-06-28 08:44:46Z

1

I fixed it myself!

#!/bin/bash
echo -e "text"
c=false
while true; do
  printf ">> "
  read i
  if [ "$i" = "exit" ]; then
    exit
  elif [ "$i" = "no" ]; then
     echo "no"
  else
     echo -e "Error: $i is undefined"
  fi
done

edited Jun 28, 2018 at 8:44

answered Jan 27, 2016 at 17:35

Jerrybibo

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6 Comments

chepner Over a year ago

You should quote $i, to avoid a "too many arguments" error from [.

Walter A Over a year ago

Explanation: Test it with an input with two or more words. You can use "$i"

Jerrybibo Over a year ago

Thanks for the tip! Is there any more things I can improve upon?

Charles Duffy Over a year ago

@Jerrybibo, using echo -e is actually best avoided: Supporting that option to do anything other than print -e on output is contrary to the POSIX spec for echo. If you want format string support, use printf.

Charles Duffy Over a year ago

@Jerrybibo, ...see pubs.opengroup.org/onlinepubs/009604599/utilities/echo.html for the aforementioned specification (and note all the caveats: Behavior when given a string with any backslash is undefined in POSIX echo; behavior when given -n is undefined; behavior when given -e is defined, but defined in such a way as to require -e to be printed on output; etc).

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