How do I compare two string variables in an 'if' statement in Bash? [duplicate]

Question

I'm trying to get an if statement to work in Bash (using Ubuntu):

#!/bin/bash  s1="hi" s2="hi"  if ["$s1" == "$s2"] then   echo match fi 

I've tried various forms of the if statement, using [["$s1" == "$s2"]], with and without quotes, using =, == and -eq, but I still get the following error:

[hi: command not found

I've looked at various sites and tutorials and copied those, but it doesn't work - what am I doing wrong?

Eventually, I want to say if $s1 contains $s2, so how can I do that?

I did just work out the spaces bit... :/ How do I say contains?

I tried

if [[ "$s1" == "*$s2*" ]] 

but it didn't work.

Answer 1

For string equality comparison, use:

if [[ "$s1" == "$s2" ]] 

For string does NOT equal comparison, use:

if [[ "$s1" != "$s2" ]] 

For the a contains b, use:

if [[ $s1 == *"$s2"* ]] 

(and make sure to add spaces between the symbols):

Bad:

if [["$s1" == "$s2"]] 

Good:

if [[ "$s1" == "$s2" ]]