ruby - "string literal in condition"是什么意思？

Question

每当我尝试运行该程序时，都会弹出一条错误消息“条件字符串文字(第 10 行)”。我做错了什么？

puts "Welcome to the best calculator there is. Would you like to (a) calculate the area of a geometric shape or (b) calculate the equation of a parabola? Please enter an 'a' or a 'b' to get started."
response = gets.chomp

if response == "a" or "A"

       puts "ok."

elsif response == "b" or "B"

       puts "awesome."

else

       puts "I'm sorry. I did not get that. Please try again."

end

Answer 1

您必须在 或 的两边指定完整的条件。

if response == "a" or response == "A"

或的两端不相连； Ruby 不会根据左边的内容假设右边的内容。如果右侧是裸字符串 "A"，那么，除了 false 或 nil 之外的任何内容都被视为“true”，因此整个expression 始终评估为“true”。但是 Ruby 注意到它是一个字符串而不是一个 bool 值，怀疑您可能没有指定您的意思，因此在问题中发出警告。

您还可以使用 case 表达式来简化针对单个值的多个测试；如果您在单个 when 中提供多种可能性的列表，它们将有效地 or 组合在一起:

case response
  when "a","A"
    puts "ok"
  when "b","B"
    puts "awesome."
  else
    puts "I'm sorry. I did not get that.  Please try again."
end

对于忽略字母大小写的具体情况，也可以在测试前转换为大写或小写:

case response.upcase 
  when "A"
    puts "ok"
  when "B"
    puts "awesome."
  else
    puts "I'm sorry, I did not get that.  Please try again."
 end