Question

如果用户要输入随机字母而不是数字，＆＃34;这不是数字＆＃34;应该是输出。相反，它推出了'＃34;他们不能进来＆＃34;我需要改变什么？

print "how old are you? "
age = gets.chomp.to_i

if age >= 21
  puts "They can come in"
elsif age < 21
  puts "they can not come in"
else 
  puts "that is not a number"
end

Answer 1

如果您在to_i对象上致电string，您将始终获得0.所以

print "how old are you? "
s_age = gets.chomp
if age.match /\A\d+\Z/
  age = s_age.to_i
  puts "They can come in" if age >= 21
  puts "They can not come in" if age < 21
else
  puts "That is not a number"
end

或

begin
  print "how old are you? "
  age = Integer gets.chomp # Integer(string) raise an exception if the parameter is not an integer
  puts "They can come in" if age >= 21
  puts "They can not come in" if age < 21
rescue
  puts "That is not a number"
end

Answer 2

我实际上会事先检查它是否是有效数字。你的逻辑是有缺陷的，就像他们输入字母一样，然后gets.chomp.to_i只返回0，这在技术上小于21。实际上，根据您想要的限制，立即转换为整数是非常安全的。所以，这样做：

age = gets.chomp.to_i
if age == 0 then puts "that is not a number" end

然后是你的其余代码。并非此解决方案不适用于有效的“0岁”答案，因为0默认为无效。

Answer 3

print "How old are you? "
puts Integer(gets) >= 21 ? 'They can come in' : 'They can not come in' rescue puts 'Not a number'

Answer 4

我能看到的最简单的方法是对elsif条件稍加修改：

print "how old are you? "
age = gets.chomp.to_i

if age >= 21
  puts "They can come in"
elsif age >= 0 && age < 21
  puts "they can not come in"
else 
  puts "that is not a number"
end

制作最后一个else语句需要什么？

4 个答案: