Bash command not found comparing string with variable

Question

I need to compare a variable with a string (how hard can it be)

#!/bin/bash

a=22

if ["$a" == "23"]
then
    echo yes yes
fi

I get

./x: line 5: [22: command not found

I've tried to remove quotes, or use [[...]]. But whatever I do, I get the command not found error. Any suggestions what I do wrong?

Well, that changes the error @Fabio. Now I get: `./x: line 3: local: can only be used in a function ./x: line 5: [: missing `]'` — Jeanluca Scaljeri, Jun 22 '19 at 21:11
Spaces are important delimiters in shell syntax, and adding or removing them will often change the meaning of a command. — Gordon Davisson, Jun 22 '19 at 22:08
notice that while the fundamental problem was the spacing, you should use `-eq` to compare integers instead of `==`. Furthermore, `==` is understood by Bash, but isn't POSIX compliant; use `=` instead for improved portability. — Benjamin W., Jun 22 '19 at 22:42

score 1 · Accepted Answer · answered Jun 22 '19 at 21:20

1

[ "$a" == "22" ] <--- you need a space before the closing backet

answered Jun 22 '19 at 21:20

gregory

1 Answers1