JavaScript如何编写正则校验含非嵌套括号的大写字母逻辑公式

明确校验规则 #

• 合法字符范围：仅允许A-Z大写英文字母、逻辑连接词and/or、英文半角括号( )、单词间半角空格
• 括号规则：禁止嵌套，左右括号数量必须完全匹配
• 邻接规则：
  • 左括号(后仅允许接A-Z大写字母
  • 右括号)后仅允许接逻辑连接词and/or或直接到字符串末尾
  • 禁止连续出现逻辑连接词、禁止无连接词直接相邻的运算单元、禁止首尾悬空连接词

测试用例参考 #

合法输入 #

A and B and C and D
(A or B) and C
(A or B or C) and D
(A or B or C or D) and E
A and (B or C) and D
A and (B or C) or (C and D)
A or (B and C)
(A and B) or (C and D)

非法输入 #

A and B and C and 
(A or B and C
(A or B or C) and D or
(A or B or C or D and E
A and or (B or C) and D
A and (B or (C and D)))
A (B and C)
(A and B) or C and D)
(A and B or C and D)

现有正则问题 #

你当前使用的正则/^[A-Z(]?[A-Z]| |and|or|[(]|[A-Z]|[)]/gm存在核心逻辑缺陷：

• 未做整串匹配锚定，只会匹配字符串中的合法片段，无法识别片段外的非法内容
• 未实现括号匹配、邻接字符校验、连接词连续性校验的逻辑
• 全局匹配模式会把包含非法片段的字符串误判为合法

最终可用正则 #

以下正则可完全覆盖列明的所有核心校验规则：

/^(?:[A-Z]|\([A-Z](?: (?:and|or) [A-Z])*\))(?: (?:and|or) (?:[A-Z]|\([A-Z](?: (?:and|or) [A-Z])*\)))*$/

逻辑拆解 #

• 首尾用^和$锚定，强制整串完全匹配，避免漏判非法片段
• 先定义两类合法运算单元：
  1. 单个大写字母[A-Z]
  2. 括号包裹的非嵌套表达式：\([A-Z](?: (?:and|or) [A-Z])*\)，从结构上直接禁止括号嵌套，同时保证左括号后必须是大写字母
• 整体表达式结构固定为「单个运算单元 + N组(连接词+运算单元)」，从结构上禁止连续连接词、无连接词相邻单元、首尾悬空连接词的问题，同时保证右括号后只能接连接词或字符串末尾

使用示例 #

const logicReg = /^(?:[A-Z]|\([A-Z](?: (?:and|or) [A-Z])*\))(?: (?:and|or) (?:[A-Z]|\([A-Z](?: (?:and|or) [A-Z])*\)))*$/;

// 合法用例校验
const validList = [
  "A and B and C and D",
  "(A or B) and C",
  "(A or B or C) and D",
  "(A or B or C or D) and E",
  "A and (B or C) and D",
  "A and (B or C) or (C and D)",
  "A or (B and C)",
  "(A and B) or (C and D)"
];
console.log(validList.every(item => logicReg.test(item))); // 输出true

// 非法用例校验
const invalidList = [
  "A and B and C and ",
  "(A or B and C",
  "(A or B or C) and D or",
  "(A or B or C or D and E",
  "A and or (B or C) and D",
  "A and (B or (C and D)))",
  "A (B and C)",
  "(A and B) or C and D)",
  "(A and B or C and D)"
];
console.log(invalidList.every(item => !logicReg.test(item))); // 输出true

如果需要兼容多空格场景，可以将正则中的单个空格替换为\s+，同时校验前先对输入做trim()处理去除首尾空格。

内容的提问来源于stack exchange，提问作者psychozub