您可以使用Ruby Gem Facets中的String#随机。

它基本上做到了这一点：

class String
  def self.random(len=32, character_set = ["A".."Z", "a".."z", "0".."9"])
    characters = character_set.map { |i| i.to_a }.flatten
    characters_len = characters.length
    (0...len).map{ characters[rand(characters_len)] }.join
  end
end

使用此方法，您可以传入一个很短的长度。默认设置为6。

def generate_random_string(length=6)
  string = ""
  chars = ("A".."Z").to_a
  length.times do
    string << chars[rand(chars.length-1)]
  end
  string
end

这是基于一些其他答案，但它增加了一点复杂性：

def random_password
  specials = ((32..47).to_a + (58..64).to_a + (91..96).to_a + (123..126).to_a).pack('U*').chars.to_a
  numbers  = (0..9).to_a
  alpha    = ('a'..'z').to_a + ('A'..'Z').to_a
  %w{i I l L 1 O o 0}.each{ |ambiguous_character| 
    alpha.delete ambiguous_character 
  }
  characters = (alpha + specials + numbers)
  password = Random.new.rand(8..18).times.map{characters.sample}
  password << specials.sample unless password.join =~ Regexp.new(Regexp.escape(specials.join))
  password << numbers.sample  unless password.join =~ Regexp.new(Regexp.escape(numbers.join))
  password.shuffle.join
end

本质上，它确保密码长度为8-20个字符，并且至少包含一个数字和一个特殊字符。

以下内容对我很有用

def generate_random_password(min_length, max_length)
    length = SecureRandom.random_number(max_length - min_length) + min_length
    character_sets = [ 
      ('a'..'z').to_a,
      ('A'..'Z').to_a,
      ('0'..'9').to_a,
      "~!@^&*()_-+=[]|:;<,>.?".split('')
    ]   
    retval = []
    #   
    # Add one character from each set
    #   
    character_sets.each do |character_set|
      character = character_set[SecureRandom.random_number(character_set.count)]
      retval.push character
    end 
    #   
    # Fill the rest of the password with a random character from a random set
    #   
    i = character_sets.count - 1 
    while i < length
      character_set = character_sets[SecureRandom.random_number(character_sets.count)]
      character = character_set[SecureRandom.random_number(character_set.count)]
      retval.push character
      i += 1
    end
    retval.shuffle.join
  end

我喜欢使用的另一种方法：

 rand(2**256).to_s(36)[0..7]

如果您真的对正确的字符串长度有疑问，请添加ljust：

 rand(2**256).to_s(36).ljust(8,'a')[0..7]

[*('A'..'Z')].sample(8).join

生成随机8个字母的字符串（例如NVAYXHGR）

([*('A'..'Z'),*('0'..'9')]-%w(0 1 I O)).sample(8).join

生成随机的8个字符串（例如3PH4SWF2），不包括0/1/I/O。红宝石1.9