Question

我正在尝试使用正则表达式并在ruby中测试我自己的代码。使用下面的例子，我预计最终的回报会返回true，但事实并非如此。但是，check_password方法会返回＆＃34;你的pw不起作用，因为它太短了＃34;为什么真/假检查不返回true？

def err_message(reason)
  puts "Your pw does not work because " + reason
end

def check_password(password)
    if password.length<6
        return err_message("it is too short")
    elsif password.index(/[A-Z]/)==nil
        return err_message("it does not contain a capital letter")
    elsif password.index(/\d|[!,@,#,$,%,&,*,+,:,?]/)==nil
        return err_message("it needs either a digit or special character")
    elsif password.index(/([!,@,#,$,%,&,*,+,:,?\w])/)>0
        return err_message("nope. !,@,#,$,%,&,*,+,:,? are the only special characters available")
    else
        return "Valid Password!"
    end
end

puts check_password("aaaaa")=="Your pw does not work because it is too short"

Answer 1

err_message将评估puts "Your pw does not work because " + reason并返回nil，因为nil是puts的返回值，将方法定义更改为如下所示：

def err_message(reason)
  "Your pw does not work because " + reason
end

然后：

puts check_password("aaaaa")=="Your pw does not work because it is too short"
# => true

Answer 2

以下是一些改进方法的建议。通常我会用评论来表达，但格式限制使得在这种情况下不切实际。考虑这样写：

SPECIAL_CHARS = '!@#$%&*+:?'

def check_password(password)
  "Your pw does not work because " +
  case
  when password.length < 6
    "it is too short"
  when password.match(/[A-Z]/).nil?
    "it does not contain a capital letter"
  when password.match(/\d|[{SPECIAL_CHARS}]/).nil?
    "it needs either a digit or special char"
  else
    return "Valid Password!"
  end
end  

check_password("abcde")
  #=> "Your pw does not work because it is too short" 
check_password("abcdef")
  #=> "Your pw does not work because it does not contain a capital letter"
check_password("abcdeF")
  #=> "Your pw does not work because it needs either a digit or special char"
check_password("abcdeF@")
  #=> "Valid Password!"
check_password("abcdeF1")
  #=> "Valid Password!"

我删除了

  when password.index(/([!,@,#,$,%,&,*,+,:,?\w])/) > 0
    "nope. !,@,#,$,%,&,*,+,:,? are the only special characters available"

因为我不明白它正在检查什么。

请注意，只有一个return，即密码有效时。如果密码无效，则case语句（字符串）的结果将添加到字符串"Your pw does not work because "并返回（因为它将是该方法执行的最后一个语句 - 否{需要{1}}。

我将return替换为index，但您也可以写：

match

或

when !(password =~ /[A-Z]/)

连接字符串不等于文字字符串？

2 个答案: