对于逗号分隔

Question

我是 RegEx 和 JavaScript 的新手，我想知道是否有人知道 RegEx 用于检测输入字段是否包含以下类型的格式：

• 至少一个可以包含空格的字母数字标签（例如“Test Tag”，但不能包含“Test@Tag”）

• 每个标签用一个逗号分隔（例如“汽车、车辆、大型狗、床”，但不是“汽车、车辆、老虎）

我的意思是这样的一个例子，这些是 valid 标签：

boy, man,girl, woman,tyrannosaurus rex, lion

这些将是无效标签：

hat, cat, rat, c3po, @gmail

因为“@gmail”中有无效字符。

它也应该能够只接受一个标签，只要字符是字母数字。

Answer 1

假设您要允许 _ 并且不允许开头或结尾的空格，这将是最短的解决方案：

/^\w(\s*,?\s*\w)*$/

在末尾引入空格：

/^\s*\w(\s*,?\s*\w)*\s*$/

从允许的字符中删除_：

/^\s*[a-z0-9](\s*,?\s*[a-z0-9])*\s*$/

---

这是我最初发布的蛮力正则表达式。它将您的要求转换为正则表达式语法。我想把它留在这里以供参考。

/^\s*([a-z0-9]+(\s[a-z0-9]+)*)(\s*,\s*([a-z0-9]+(\s[a-z0-9]+)*))*\s*$/

Answer 2

试试这样的：

var re = /^(\w+,? ?)+$/;
var str1 = "boy, man,girl, woman,tyrannosaurus rex, lion";
var str2 = "hat, cat, rat, c3po, @gmail";
alert(str1.test(re));  // true
alert(str2.test(re));  // false

分解它... \w 匹配单词字符，\w+ 匹配 1 个或多个单词字符。 ，？ ?匹配可选的逗号和空格。 （两个逗号将被拒绝。）所有内容周围的 ()+ 表示一次或多次。最后 ^ 和 $ 将其锚定到字符串的开头和结尾，以确保所有内容都匹配。

Answer 3

假设下划线 (_) 不无效：

/^(\w+\s?[\w\s]*)(,\s*\w+\s?[\w\s]*)*$/

Assert position at the beginning of a line (at beginning of the string or after a line break character) «^»
Match the regular expression below and capture its match into backreference number 1 «(\w+\s?[\w\s]*)»
   Match a single character that is a “word character” (letters, digits, and underscores) «\w+»
      Between one and unlimited times, as many times as possible, giving back as needed (greedy) «+»
   Match a single character that is a “whitespace character” (spaces, tabs, and line breaks) «\s?»
      Between zero and one times, as many times as possible, giving back as needed (greedy) «?»
   Match a single character present in the list below «[\w\s]*»
      Between zero and unlimited times, as many times as possible, giving back as needed (greedy) «*»
      A word character (letters, digits, and underscores) «\w»
      A whitespace character (spaces, tabs, and line breaks) «\s»
Match the regular expression below and capture its match into backreference number 2 «(,\s*\w+\s?[\w\s]*)*»
   Between zero and unlimited times, as many times as possible, giving back as needed (greedy) «*»
   Note: You repeated the capturing group itself.
   The group will capture only the last iteration.
   Put a capturing group around the repeated group to capture all iterations. «*»
   Match the character “,” literally «,»
   Match a single character that is a “whitespace character” (spaces, tabs, and line breaks) «\s*»
      Between zero and unlimited times, as many times as possible, giving back as needed (greedy) «*»
   Match a single character that is a “word character” (letters, digits, and underscores) «\w+»
      Between one and unlimited times, as many times as possible, giving back as needed (greedy) «+»
   Match a single character that is a “whitespace character” (spaces, tabs, and line breaks) «\s?»
      Between zero and one times, as many times as possible, giving back as needed (greedy) «?»
   Match a single character present in the list below «[\w\s]*»
      Between zero and unlimited times, as many times as possible, giving back as needed (greedy) «*»
      A word character (letters, digits, and underscores) «\w»
      A whitespace character (spaces, tabs, and line breaks) «\s»
Assert position at the end of a line (at the end of the string or before a line break character) «$»


Created with RegexBuddy

Answer 4

这里正在回答一个单独的问题。 如何做同样的事情但允许标签最多包含两个单词？

/^\s*[a-z0-9]+(\s+[a-z0-9]+)?(\s*,\s*[a-z0-9]+(\s+[a-z0-9]+)?)*\s*$/

经过测试。