问题
Is there a way to use a case statement with integer comparisons in ruby? I have found lots of examples comparing strings, but my case example below fails with syntax errors.
def get_price_rank(price)
case price
when <= 40
return 'Cheap!'
when 41..50
return 'Sorta cheap'
when 50..60
return 'Reasonable'
when 60..70
return 'Not cheap'
when 70..80
return 'Spendy'
when 80..90
return 'Expensive!'
when >= 90
return 'Rich!'
end
end
回答1:
In case..when block you can't perform any comparisons except ===. So I'd write your code as below :
def get_price_rank(price)
case price
when 41..50
'Sorta cheap'
when 50..60
'Reasonable'
when 60..70
'Not cheap'
when 70..80
'Spendy'
when 80..90
'Expensive!'
else
if price >= 90
'Rich!'
elsif price <= 40
'Cheap!'
end
end
end
return is implicit, thus no need to mention.
回答2:
Rewrite your case like this:
case price
when 0..40 then
return 'Cheap!'
when 41..50 then
return 'Sorta cheap'
when 50..60 then
return 'Reasonable'
when 60..70 then
return 'Not cheap'
when 70..80 then
return 'Spendy'
when 80..90 then
return 'Expensive!'
else
return 'Rich!'
end
来源：https://stackoverflow.com/questions/22700194/ruby-case-statement-with-comparison