在Ruby中，布尔运算符和和条件符之间没有区别

I can't see a practical difference between the boolean operators AND and OR in a Ruby case conditional.

我看不出布尔运算符与AND或Ruby条件句之间的实际区别。

For example, I want to get the user to input the sentence:

例如，我想让用户输入句子:

sudo make me a sandwich

And the case conditional starts as follows:

有条件的情况如下:

case user_selection
when /sudo/ && /sandwich/

However, if the user enters:

但是，如果用户输入:

make me a sandwich

The condition will be a satisfied.

这种情况将会令人满意。

My way around it in this instance is to re-order the conditions:

在这个例子中，我的方法是重新排列条件:

case user_selection
when /sandwich/ && /sudo/

But that pre-supposes that every time a user thinks to use "sudo" they will include the string "sandwich" in their response. However, this is functionally no different from this:

但这一假设前提是，每次用户想要使用“sudo”时，都会在响应中包含字符串“sandwich”。然而，这在功能上与此没有什么不同:

case user_selection
when /sudo/

I looked up boolean operators for Ruby conditionals, but have not found a satisfactory answer.

我查找了Ruby条件语句的布尔运算符，但没有找到满意的答案。

2 个解决方案

#1

You cannot use && in a case block like that, because when /sudo/ && /sandwich/ evaluates to just when /sandwich/.

你不能在这样的情况块中使用&&，因为when /sudo/ & /sandwich/计算为when /sandwich/。

Instead you will need to use one regexp that looks for both words. A simple example might be:

相反，您需要使用一个regexp来查找这两个词。一个简单的例子可能是:

case user_selection
when /sudo.*sandwich/

#2

Just for reference.

仅供参考。

If you absolutely need a logical AND in your case statement, you could use a lambda :

如果你绝对需要一个符合逻辑的案例陈述，你可以使用lambda:

def answer(order)
  case order
  when ->(x) { x =~ /sudo/ && x =~ /sandwich/ }
    puts 'Okay'
  else
    puts 'Do it yourself!'
  end
end

answer 'make me a sandwich'
#=> Do it yourself!
answer 'sudo sing a song'
#=> Do it yourself!
answer 'sudo make me a sandwich'
#=> Okay

If you can compact the boolean logic to a single check (e.g. with a Regex), you probably should, as in @spickermann's answer.

如果您可以将布尔逻辑压缩到单个检查(例如使用Regex)，那么您可能应该这样做，就像@spickermann的答案一样。

#1