Ruby 中的全名正则表达式

Question

我知道有很多类似的问题，但我在任何地方都找不到我的案例。

我试图在Ruby on Rails用户模型中写一个全名正则表达式。它应该验证名字和姓氏是否填充了一个空格。这两个名字都应该包含至少两个字符(例如:马丽)。

作为一个奖励，但不是必须的，我想修剪空白到一个字符的情况下，用户将错误地输入一个以上的空白(例如:李

validates :name,
        presence: true,
        length: {
            maximum: 64,
            minimum: 5,
            message: 'must be a minimum: 5 letters and a maximum: 64 letters'},
        format: {
            # Full Name RegEx
            with: /[\w\-\']+([\s]+[\w\-\']){1}/
        }

这对我有效，但是不检查每个名字的最少2个字符(例如:Peter P现在是正确的)。这也接受多个空格，这是不好的(例如:彼得

我知道这个识别名字的问题是非常以文化为中心的，它可能不是验证全名的正确方法（也许有些人只有一个角色名字），但这是目前的要求。

我不想把这个字段分成两个不同的字段，名字和姓氏，因为这将使用户界面变得复杂。

Answer 1

result = subject.gsub(/\A(?=[\w' -]{5,64})([\w'-]{2,})([\s]{1})\s*?([\w'-]{2,})\Z/, '\1\2\3')

http://regex101.com/r/dT1fJ4

Assert position at the beginning of the string «^»
Assert that the regex below can be matched, starting at this position (positive lookahead) «(?=[\w' -]{5,64})»
   Match a single character present in the list below «[\w' -]{5,64}»
      Between 5 and 64 times, as many times as possible, giving back as needed (greedy) «{5,64}»
      A word character (letters, digits, and underscores) «\w»
      The character “'” «'»
      The character “ ” « »
      The character “-” «-»
Match the regular expression below and capture its match into backreference number 1 «([\w'-]{2,})»
   Match a single character present in the list below «[\w'-]{2,}»
      Between 2 and unlimited times, as many times as possible, giving back as needed (greedy) «{2,}»
      A word character (letters, digits, and underscores) «\w»
      The character “'” «'»
      The character “-” «-»
Match the regular expression below and capture its match into backreference number 2 «([\s]{1})»
   Match a single character that is a “whitespace character” (spaces, tabs, and line breaks) «[\s]{1}»
      Exactly 1 times «{1}»
Match a single character that is a “whitespace character” (spaces, tabs, and line breaks) «\s*?»
   Between zero and unlimited times, as few times as possible, expanding as needed (lazy) «*?»
Match the regular expression below and capture its match into backreference number 3 «([\w'-]{2,})»
   Match a single character present in the list below «[\w'-]{2,}»
      Between 2 and unlimited times, as many times as possible, giving back as needed (greedy) «{2,}»
      A word character (letters, digits, and underscores) «\w»
      The character “'” «'»
      The character “-” «-»
Assert position at the end of the string (or before the line break at the end of the string, if any) «$»

Answer 2

您可以匹配以下正则表达式：

/（[\w\-\']{2，}）（[\s] ）（[\w\-\']{2，}）/

并替换为：（假设它支持捕获组）

'\1\3'或$1$3无论语法是什么：

它去掉了多余的空格，只保留一个，正如你所希望的。

演示：http://regex101.com/r/oQ6aO7