案例陈述不起作用

Question

I'm learning Ruby from "Programming Ruby, The Pragmatic Programmers(2nd, 2005)" and I'm stuck in the Case statement chapter. So i copy-paste some code in my version from book:

def kind
  puts "Type year and I'll tell you genre: "
  ask = gets.chomp

  kind = case ask
  when 1850..1889 then "Blues"
  when 1890..1909 then "Ragtime"
  when 1910..1929 then "New Orleans Jazz"
  when 1930..1939 then "Swing"
  when 1940..1950 then "Bebop"
  else "Jazz"
end


puts "You typed year #{ask}. Genre of music in that period is 
       #{kind}."
end

kind

Hence, whatever year I'm put, output is "Jazz"...What am I working incorrectly?

Answer 1

gets.chomp returns a string, and you are comparing that with integers.

You can inspect ask after you assigned it:

ask = gets.chomp
p ask

When you run the script and enter a number (eg 1940), you should see "1940" printed in the terminal. The quotes around the number show you that the variable holds a string, not a number. (FYI don't use puts here, since it won't show the quotes.)

As mudasobwa wrote in his comment, the way to fix this is to cast the input to a number before you compare it:

ask = gets.chomp.to_i

If you add p ask again, you should now see that only the number is printed to the terminal, without any " surrounding it. This shows you that the variable holds an integer.