```
anothervar = 1
while [$anothervar -lt 1 ] do
    read a
    if [ 42 = $a ]; then
       $anothervar = 2
    else
        echo $a
    fi
done

```

> 
> get line 9: syntax error near unexpected token `done' error.
> 
> 
> 

What did i do wrong ?