问题:

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Ruby range: operators in case statement 3 answers

Is there a way to use a case statement with integer comparisons in ruby? I have found lots of examples comparing strings, but my case example below fails with syntax errors.

def get_price_rank(price)
    case price
    when <= 40
        return 'Cheap!'
    when 41..50 
        return 'Sorta cheap'
    when 50..60
        return 'Reasonable'
    when 60..70
        return 'Not cheap'
    when 70..80
        return 'Spendy'
    when 80..90
        return 'Expensive!'
    when >= 90
        return 'Rich!'
    end
end

回答1:

In case..when block you can't perform any comparisons except ===. So I'd write your code as below :

def get_price_rank(price)
    case price
    when 41..50 
        'Sorta cheap'
    when 50..60
        'Reasonable'
    when 60..70
        'Not cheap'
    when 70..80
        'Spendy'
    when 80..90
        'Expensive!'
    else
        if price >= 90
         'Rich!'
        elsif price <= 40
         'Cheap!'
        end
    end
end

return is implicit, thus no need to mention.

回答2:

Rewrite your case like this:

case price
when 0..40 then
    return 'Cheap!'
when 41..50 then 
    return 'Sorta cheap'
when 50..60 then
    return 'Reasonable'
when 60..70 then
    return 'Not cheap'
when 70..80 then
    return 'Spendy'
when 80..90 then
    return 'Expensive!'
else 
    return 'Rich!'
end