如果方法的参数是符号,如何检查Ruby？

Question

def my_method(parameter)
  if <what should be here?>
    puts "parameter is a string"
  elsif <and here?>
    puts "parameter is a symbol"
  end
end

Answer 1

最简单的形式是:

def my_method(parameter)
  puts "parameter is a #{parameter.class}"
end

但是如果你真的想根据类型做一些处理呢:

def my_method(parameter)
  puts "parameter is a #{parameter.class}"
  case parameter
    when Symbol
      # process Symbol logic
    when String
      # process String logic
    else
      # some other class logic
   end
end

Answer 2

def my_method(parameter)
  if parameter.is_a? String
    puts "parameter is a string"
  elsif parameter.is_a? Symbol
    puts "parameter is a symbol"
  end
end

应该解决你的问题

Answer 3

if parameter.is_a? String
  puts "string"
elsif parameter.is_a? Symbol
  puts "symbol"
end

我希望这有帮助.