ruby案例陈述与比较[重复]

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Ruby range: operators in case statement 3 answers
Ruby范围：运算符在case语句3中的答案

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.

有没有办法在ruby中使用带有整数比较的case语句？我找到了很多比较字符串的例子，但是下面我的案例示例因语法错误而失败。

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

2 个解决方案

#1

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

如果..阻止你不能执行任何比较，除了===。所以我写下你的代码如下：

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.

返回是隐含的，因此无需提及。

#2

Rewrite your case like this:

像这样改写你的情况：

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

#1