Bash Scripting if statement

Question

I don't know why this isn't working. Please help!

#!/bin/bash
clear
echo "Enter an option"
read $option
if ("$option" == 1) then
  echo "Blah"
fi

I tried like this

if ("$option" -eq 1) then

I can't see why the if statement isn't being run. All I want to do is check what the user entered and do something depending on the value entered.

Another problem with that code is the read command should be `read option`. — PM 2Ring, Nov 19 '14 at 08:40

score 0 · Answer 1 · edited Nov 19 '14 at 00:03

0

Bash if uses [ or [[ as test constructs, instead of parenthesis which are used in other languages.

edited Nov 19 '14 at 00:03

Pokechu22

answered Nov 18 '14 at 23:18

David C. Rankin

score 0 · Answer 2 · answered Nov 18 '14 at 23:19

0

That is not the syntax for an if statement in bash. Try this:

if [ "$option" = "1" ]; then
    echo "Blah"
fi

answered Nov 18 '14 at 23:19

DanielGibbs

score 0 · Accepted Answer · answered Nov 18 '14 at 23:19

0

The syntax for an equality check is:

if [[ $option == 1 ]]; then
  echo "Blah"
fi

Or, for compatibility with older non-bash shells:

if [ "$option" = 1 ]; then
  echo "Blah"
fi

In either one, the whitespace is important. Do not delete the spaces around the square brackets.

answered Nov 18 '14 at 23:19

John Kugelman

1

`[` uses a single `=` to do string comparison. – DanielGibbs Nov 18 '14 at 23:19

score 0 · Answer 4 · answered Nov 18 '14 at 23:21

0

I'm not sure where you got your syntax from...try this:

if [ "$option" -eq 1 ]; then

In the shell, [ is a command, not a syntactic construct.

Alternatively, you can use an arithmetic context in bash:

if (( option == 1 )); then

answered Nov 18 '14 at 23:21

Tom Fenech

4 Answers4