用贪婪的量词

Question

正则表达式先行问题需要支持下列格式

3位数字可选空间后跟以下字符集ACERV内指定了三个非重复的字符（空格只在两个字符之间是有效的）

有效格式：

123 
123 A 
123 A v 
123 CER 

无效格式：

123A 
123 AA 
123 A - when followed by a space 

我到目前为止 - 我可能是在与不一定需要向前看符号复杂的：

^([0-9]{3})           # - first 3 digits 
(\s(?=[ACERV]))([ACERV])       # - allow space only when followed by ACERV 
(?!\3)(?=[ACERV ]{0,1})([ACERV ]{0,1})    # - do not allow 1st char to repeat 
(?!\3)            # - do not allow 1st char to repeat 
(?!\4)            # - do not allow 2nd to repeat 
(?!\s)            # - do not allow trailing space 
(?=[ACERV]{0,1})([ACERV]{0,1})|[0-9]{3}$ 

当先行加入它不匹配的有效格式123（\ 4？！） A - 将（？！\ 4）上的量词修改为（？！\ 4）*或（？！\ 4）？允许123 A匹配，但允许重复第一个或第二个字符。

Answer 1

不完全确定的要求，这适用于您的示例。

# ^(?i)\d{3}(?:[ ](?:([ACERV])[ ]?(?![ACERV ]*\1)){1,3}(?<![ ]))?$ 

^      # BOL 
(?i)     # Case insensitive modifier 
\d{3}     # 3 digits 
(?:     # Cluster grp, character block (optional) 
     [ ]     # Space, required 
     (?:     # Cluster grp 
      ([ACERV])   # (1), Capture single character [ACERV] 
      [ ]?     # [ ], optional 
      (?!     # Negative lookahead 
       [ACERV ]*    # As many [ACERV] or [ ] needed 
       \1      # to find what is captured in group 1 
             # Found it, the assertion fails 
      )      # End Negative lookahead 
    ){1,3}     # End Cluster grp, gets 1-3 [ACERV] characters 
     (?<! [ ])    # No dangling [ ] at end 
)?      # End Cluster grp, character block (optional) 
$      # EOL 

更新 -调整更换回顾后。

# ^(?i)\d{3}(?!.*[ ]$)(?:[ ](?:([ACERV])[ ]?(?![ACERV ]*\1)){1,3})?$ 

^      # BOL 
(?i)     # Case insensitive modifier 
\d{3}     # 3 digits 
(?! .* [ ] $)   # No dangling [ ] at end 
(?:     # Cluster grp, character block (optional) 
     [ ]     # Space, required 
     (?:     # Cluster grp 
      ([ACERV])   # (1), Capture single character [ACERV] 
      [ ]?     # [ ], optional 
      (?!     # Negative lookahead 
       [ACERV ]*    # As many [ACERV] or [ ] needed 
       \1      # to find what is captured in group 1 
             # Found it, the assertion fails 
      )      # End Negative lookahead 
    ){1,3}     # End Cluster grp, gets 1-3 [ACERV] characters 
)?      # End Cluster grp, character block (optional) 
$      # EOL 

Answer 2

如何对正则表达式

^\d{3}(?:$|\s)(?:([ACERV])(?!\1)|\s(?!$|\1))*$ 

将匹配字符串

123 
123 A 
123 A V 
123 CER 

见正则表达式在http://regex101.com/r/mW5qZ9/9

• ^锚正则表达式如何符合项目在字符串的开头

场

• \d{3}比赛3 occurence任何数字的

• (?:$|\s)匹配串$或空间的端部，\s

• (?:\s?([ACERV])(?!\1)){0,3}匹配非重复字符从

  • (?:)非捕获组

  • \s?可选空间

  • ([ACERV])中的字符匹配的类

  • (?:([ACERV])(?!\1)|\s(?!$|\1))断言，正则表达式后面没有\1，最近被捕获的。确保字符不重复。

    • (?!\1)断言，字符类不能被接着重复字符

    • \s(?!$|\1))断言，如果它是一个空间，那么它不能被随后柱的端部或重复字符从\1

  • {0,3}量词指定最小出现次数为零和最大发生次数为3

• $将正则表达式锚定在字符串末尾。

Answer 3

一个计划是一个简单的正则表达式来拉开字符串，那么第二步骤，以验证字符不重复。

// check all characters in a string are unique, 
// by ensuring that each character is its own first appearance 
function unique_characters(str) { 
    return str.split('').every(function(chr, i, chrs) { 
     return chrs.indexOf(chr) === i; 
    }); 
} 

// check that the code is valid 
function valid_code(str) { 
    var spacepos = str.indexOf(' '); 
    return unique_characters(str) && 
     (spacepos === -1 || (spacepos === 1 && str.length === 3)); 
} 

// check basic format and pull out code portion 
function check_string(str) { 
    var matches = str.match(/^\d{3} ?([ACERV ]{0,3})$/i); 
    valid = matches && valid_code(matches[1]); 
    return valid; 
} 

>> inputs = ['123', '123 A', '123 A v', '123 CER', '123A', '123 AA', '123 A '] 
[true, true, true, true, true, false, false] 

第四个测试案例表明是有效的，因为如果空间确实是可选的，那么如果123 A是有效的，那么它似乎123A将是有效的为好。

这种方法的一个可能的好处是，如果引入额外的验证规则，他们可以更容易地比一个巨大的正则表达式内碴左右实现。