Regex 是否存在用于检测有效正则表达式的正则表达式?

Regex 是否存在用于检测有效正则表达式的正则表达式?,regex,Regex,是否可以用另一个正则表达式检测有效的正则表达式?如果是,请在下面给出示例代码。不太可能 在try..catch或您的语言提供的任何内容中对其进行评估 / ^ # start of string ( # first group start (?: (?:[^?+*{}()[\]\\|]+

是否可以用另一个正则表达式检测有效的正则表达式?如果是,请在下面给出示例代码。

不太可能

try..catch
或您的语言提供的任何内容中对其进行评估

/
^                                             # start of string
(                                             # first group start
  (?:
    (?:[^?+*{}()[\]\\|]+                      # literals and ^, $
     | \\.                                    # escaped characters
     | \[ (?: \^?\\. | \^[^\\] | [^\\^] )     # character classes
          (?: [^\]\\]+ | \\. )* \]
     | \( (?:\?[:=!]|\?<[=!]|\?>)? (?1)?? \)  # parenthesis, with recursive content
     | \(\? (?:R|[+-]?\d+) \)                 # recursive matching
     )
    (?: (?:[?+*]|\{\d+(?:,\d*)?\}) [?+]? )?   # quantifiers
  | \|                                        # alternative
  )*                                          # repeat content
)                                             # end first group
$                                             # end of string
/
压实:

^(?:(?:[^?+*{}()[\]\\|]+|\\.|\[(?:\^?\\.|\^[^\\]|[^\\^])(?:[^\]\\]+|\\.)*\]|\((?:\?[:=!]|\?<[=!]|\?>|\?<[^\W\d]\w*>|\?'[^\W\d]\w*')?(?<N>)|\)(?<-N>))(?:(?:[?+*]|\{\d+(?:,\d*)?\})[?+]?)?|\|)*$(?(N)(?!))
在以下几段代码中,以下以下几段代码::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::)以下以下以下以下以下以下::::::::::::::::::::::::::::::::除了除了除了除了除了除了除了除了::::::::::::::::::::::::::::::::::::::????????????????????????????????????????????????????????????????????(N)(?!) 从评论中:

/^((?:(?:[^?+*{}()[\]\\|]+|\\.|\[(?:\^?\\.|\^[^\\]|[^\\^])(?:[^\]\\]+|\\.)*\]|\((?:\?[:=!]|\?<[=!]|\?>)?(?1)??\)|\(\?(?:R|[+-]?\d+)\))(?:(?:[?+*]|\{\d+(?:,\d*)?\})[?+]?)?|\|)*)$/
这会验证替换和翻译吗

它将只验证替换和转换的正则表达式部分<代码>s//…/

理论上不可能将所有有效的正则表达式语法与正则表达式匹配

如果正则表达式引擎支持递归(如PCRE),那么这是可能的,但这不能再称为正则表达式了

实际上,“递归正则表达式”不是正则表达式。但这是一个经常被接受的对正则表达式引擎的扩展。。。讽刺的是,这个扩展正则表达式与扩展正则表达式不匹配

“在理论上,理论和实践是一样的。在实践中,它们不是。”几乎所有知道正则表达式的人都知道正则表达式不支持递归。但是PCRE和大多数其他实现所支持的远远不止是基本的正则表达式

在grep命令中与shell脚本一起使用,它显示了一些错误。。grep:{}的内容无效。我正在制作一个脚本,可以grep一个代码库来查找所有包含正则表达式的文件

此模式利用了一个名为递归正则表达式的扩展。这是不受POSIX风格的正则表达式支持的。您可以尝试使用-P开关来启用PCRE正则表达式

正则表达式本身“不是正则语言,因此无法由正则表达式解析…”

经典正则表达式也是如此。一些现代的实现允许递归,这使得它成为一种上下文无关的语言,尽管这项任务有点冗长

我知道您在哪里匹配了
[]()/\
。和其他特殊的正则表达式字符。在哪里允许使用非特殊字符?似乎这将匹配
^(?:[\.]+)$
,但不匹配
^abcdefg$
。这是一个有效的正则表达式

[^?+*{}()[\]\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\]
将匹配任何单个字符,而不是任何其他构造的一部分。这包括文字(
a
-
z
)和某些特殊字符(
^
$
)。

问得好

真正的正则语言不能决定任意深度嵌套的格式良好的括号。如果您的字母表包含
”(“
”)”
则目标是确定其中的字符串是否具有格式良好的匹配括号。因为这是正则表达式的必要条件,所以答案是否定的

然而,如果您放松了需求并添加了递归,那么您很可能可以做到这一点。原因是递归可以充当堆栈,让您通过推到该堆栈上“计算”当前嵌套深度


Russ Cox写了一篇关于正则表达式引擎实现的精彩论文

不,如果您严格地说是正则表达式,而不包括一些实际上是上下文无关语法的正则表达式实现

正则表达式有一个局限性,这使得不可能编写一个只匹配所有正则表达式的正则表达式。不能匹配大括号等成对的实现。正则表达式使用许多这样的构造,让我们以
[]
为例。每当有一个
[
时,就必须有一个匹配的
]
,这对于正则表达式
“\[.\]”
来说足够简单

使正则表达式不可能的是它们可以嵌套。如何编写匹配嵌套括号的正则表达式?答案是你不能没有一个无限长的正则表达式。您可以通过蛮力匹配任意数量的嵌套括号,但无法匹配任意长的嵌套括号集

此功能通常称为计数,因为您正在计算嵌套的深度。根据定义,正则表达式不具有计数功能



关于这一点,我最后写了“”。

尽管使用MizardX所发布的递归正则表达式是完全可能的,但对于这类事情,解析器要有用得多。正则表达式最初用于正则语言,递归或具有平衡组只是一个补丁

定义有效正则表达式的语言实际上是一种上下文无关语法,您应该使用适当的解析器来处理它。下面是一个解析简单正则表达式(没有大多数构造)的大学项目的示例。它使用JavaCC。是的,注释是西班牙文的,尽管方法名称是不言自明的

SKIP :
{
    " "
|   "\r"
|   "\t"
|   "\n"
}
TOKEN : 
{
    < DIGITO: ["0" - "9"] >
|   < MAYUSCULA: ["A" - "Z"] >
|   < MINUSCULA: ["a" - "z"] >
|   < LAMBDA: "LAMBDA" >
|   < VACIO: "VACIO" >
}

IRegularExpression Expression() :
{
    IRegularExpression r; 
}
{
    r=Alternation() { return r; }
}

// Matchea disyunciones: ER | ER
IRegularExpression Alternation() :
{
    IRegularExpression r1 = null, r2 = null; 
}
{
    r1=Concatenation() ( "|" r2=Alternation() )?
    { 
        if (r2 == null) {
            return r1;
        } else {
            return createAlternation(r1,r2);
        } 
    }
}

// Matchea concatenaciones: ER.ER
IRegularExpression Concatenation() :
{
    IRegularExpression r1 = null, r2 = null; 
}
{
    r1=Repetition() ( "." r2=Repetition() { r1 = createConcatenation(r1,r2); } )*
    { return r1; }
}

// Matchea repeticiones: ER*
IRegularExpression Repetition() :
{
    IRegularExpression r; 
}
{
    r=Atom() ( "*" { r = createRepetition(r); } )*
    { return r; }
}

// Matchea regex atomicas: (ER), Terminal, Vacio, Lambda
IRegularExpression Atom() :
{
    String t;
    IRegularExpression r;
}
{
    ( "(" r=Expression() ")" {return r;}) 
    | t=Terminal() { return createTerminal(t); }
    | <LAMBDA> { return createLambda(); }
    | <VACIO> { return createEmpty(); }
}

// Matchea un terminal (digito o minuscula) y devuelve su valor
String Terminal() :
{
    Token t;
}
{
    ( t=<DIGITO> | t=<MINUSCULA> ) { return t.image; }
}
跳过:
{
" "
|“\r”
|“\t”
|“\n”
}
代币:
{

|
|
|
|
}
IRegularExpression表达式():
{
IRegularExpression r;
}
{
r=Alternation(){return r;}
}
//红茶:呃|呃
IRegularExpression Alternation():
{
IRegularExpression r1=null,r2=null;
}
{
r1=连接()(“|”r2=交替())?
{ 
if(r2==null){
返回r1;
}否则{
返回createAlternation(r1,r2);
} 
}
}
//火柴:呃
IRegularExpression连接():
{
IRegularExpression r1=null,r2=null;
}
{
r1=repeation()(““r2=repeation(){r1=createConcatenation(r1,r2);})*
{返回r1;}
}
//重复匹配:呃*
IRegularExpression repeation():
{
IRegularExpression r;
}
{
r=Atom()(“*”{r=createRepeation(r);})*
{返回r;}
}
//匹配
SKIP :
{
    " "
|   "\r"
|   "\t"
|   "\n"
}
TOKEN : 
{
    < DIGITO: ["0" - "9"] >
|   < MAYUSCULA: ["A" - "Z"] >
|   < MINUSCULA: ["a" - "z"] >
|   < LAMBDA: "LAMBDA" >
|   < VACIO: "VACIO" >
}

IRegularExpression Expression() :
{
    IRegularExpression r; 
}
{
    r=Alternation() { return r; }
}

// Matchea disyunciones: ER | ER
IRegularExpression Alternation() :
{
    IRegularExpression r1 = null, r2 = null; 
}
{
    r1=Concatenation() ( "|" r2=Alternation() )?
    { 
        if (r2 == null) {
            return r1;
        } else {
            return createAlternation(r1,r2);
        } 
    }
}

// Matchea concatenaciones: ER.ER
IRegularExpression Concatenation() :
{
    IRegularExpression r1 = null, r2 = null; 
}
{
    r1=Repetition() ( "." r2=Repetition() { r1 = createConcatenation(r1,r2); } )*
    { return r1; }
}

// Matchea repeticiones: ER*
IRegularExpression Repetition() :
{
    IRegularExpression r; 
}
{
    r=Atom() ( "*" { r = createRepetition(r); } )*
    { return r; }
}

// Matchea regex atomicas: (ER), Terminal, Vacio, Lambda
IRegularExpression Atom() :
{
    String t;
    IRegularExpression r;
}
{
    ( "(" r=Expression() ")" {return r;}) 
    | t=Terminal() { return createTerminal(t); }
    | <LAMBDA> { return createLambda(); }
    | <VACIO> { return createEmpty(); }
}

// Matchea un terminal (digito o minuscula) y devuelve su valor
String Terminal() :
{
    Token t;
}
{
    ( t=<DIGITO> | t=<MINUSCULA> ) { return t.image; }
}
# 
# invRegex.py
#
# Copyright 2008, Paul McGuire
#
# pyparsing script to expand a regular expression into all possible matching strings
# Supports:
# - {n} and {m,n} repetition, but not unbounded + or * repetition
# - ? optional elements
# - [] character ranges
# - () grouping
# - | alternation
#
__all__ = ["count","invert"]

from pyparsing import (Literal, oneOf, printables, ParserElement, Combine, 
    SkipTo, operatorPrecedence, ParseFatalException, Word, nums, opAssoc,
    Suppress, ParseResults, srange)

class CharacterRangeEmitter(object):
    def __init__(self,chars):
        # remove duplicate chars in character range, but preserve original order
        seen = set()
        self.charset = "".join( seen.add(c) or c for c in chars if c not in seen )
    def __str__(self):
        return '['+self.charset+']'
    def __repr__(self):
        return '['+self.charset+']'
    def makeGenerator(self):
        def genChars():
            for s in self.charset:
                yield s
        return genChars

class OptionalEmitter(object):
    def __init__(self,expr):
        self.expr = expr
    def makeGenerator(self):
        def optionalGen():
            yield ""
            for s in self.expr.makeGenerator()():
                yield s
        return optionalGen

class DotEmitter(object):
    def makeGenerator(self):
        def dotGen():
            for c in printables:
                yield c
        return dotGen

class GroupEmitter(object):
    def __init__(self,exprs):
        self.exprs = ParseResults(exprs)
    def makeGenerator(self):
        def groupGen():
            def recurseList(elist):
                if len(elist)==1:
                    for s in elist[0].makeGenerator()():
                        yield s
                else:
                    for s in elist[0].makeGenerator()():
                        for s2 in recurseList(elist[1:]):
                            yield s + s2
            if self.exprs:
                for s in recurseList(self.exprs):
                    yield s
        return groupGen

class AlternativeEmitter(object):
    def __init__(self,exprs):
        self.exprs = exprs
    def makeGenerator(self):
        def altGen():
            for e in self.exprs:
                for s in e.makeGenerator()():
                    yield s
        return altGen

class LiteralEmitter(object):
    def __init__(self,lit):
        self.lit = lit
    def __str__(self):
        return "Lit:"+self.lit
    def __repr__(self):
        return "Lit:"+self.lit
    def makeGenerator(self):
        def litGen():
            yield self.lit
        return litGen

def handleRange(toks):
    return CharacterRangeEmitter(srange(toks[0]))

def handleRepetition(toks):
    toks=toks[0]
    if toks[1] in "*+":
        raise ParseFatalException("",0,"unbounded repetition operators not supported")
    if toks[1] == "?":
        return OptionalEmitter(toks[0])
    if "count" in toks:
        return GroupEmitter([toks[0]] * int(toks.count))
    if "minCount" in toks:
        mincount = int(toks.minCount)
        maxcount = int(toks.maxCount)
        optcount = maxcount - mincount
        if optcount:
            opt = OptionalEmitter(toks[0])
            for i in range(1,optcount):
                opt = OptionalEmitter(GroupEmitter([toks[0],opt]))
            return GroupEmitter([toks[0]] * mincount + [opt])
        else:
            return [toks[0]] * mincount

def handleLiteral(toks):
    lit = ""
    for t in toks:
        if t[0] == "\\":
            if t[1] == "t":
                lit += '\t'
            else:
                lit += t[1]
        else:
            lit += t
    return LiteralEmitter(lit)    

def handleMacro(toks):
    macroChar = toks[0][1]
    if macroChar == "d":
        return CharacterRangeEmitter("0123456789")
    elif macroChar == "w":
        return CharacterRangeEmitter(srange("[A-Za-z0-9_]"))
    elif macroChar == "s":
        return LiteralEmitter(" ")
    else:
        raise ParseFatalException("",0,"unsupported macro character (" + macroChar + ")")

def handleSequence(toks):
    return GroupEmitter(toks[0])

def handleDot():
    return CharacterRangeEmitter(printables)

def handleAlternative(toks):
    return AlternativeEmitter(toks[0])


_parser = None
def parser():
    global _parser
    if _parser is None:
        ParserElement.setDefaultWhitespaceChars("")
        lbrack,rbrack,lbrace,rbrace,lparen,rparen = map(Literal,"[]{}()")

        reMacro = Combine("\\" + oneOf(list("dws")))
        escapedChar = ~reMacro + Combine("\\" + oneOf(list(printables)))
        reLiteralChar = "".join(c for c in printables if c not in r"\[]{}().*?+|") + " \t"

        reRange = Combine(lbrack + SkipTo(rbrack,ignore=escapedChar) + rbrack)
        reLiteral = ( escapedChar | oneOf(list(reLiteralChar)) )
        reDot = Literal(".")
        repetition = (
            ( lbrace + Word(nums).setResultsName("count") + rbrace ) |
            ( lbrace + Word(nums).setResultsName("minCount")+","+ Word(nums).setResultsName("maxCount") + rbrace ) |
            oneOf(list("*+?")) 
            )

        reRange.setParseAction(handleRange)
        reLiteral.setParseAction(handleLiteral)
        reMacro.setParseAction(handleMacro)
        reDot.setParseAction(handleDot)

        reTerm = ( reLiteral | reRange | reMacro | reDot )
        reExpr = operatorPrecedence( reTerm,
            [
            (repetition, 1, opAssoc.LEFT, handleRepetition),
            (None, 2, opAssoc.LEFT, handleSequence),
            (Suppress('|'), 2, opAssoc.LEFT, handleAlternative),
            ]
            )
        _parser = reExpr

    return _parser

def count(gen):
    """Simple function to count the number of elements returned by a generator."""
    i = 0
    for s in gen:
        i += 1
    return i

def invert(regex):
    """Call this routine as a generator to return all the strings that
       match the input regular expression.
           for s in invert("[A-Z]{3}\d{3}"):
               print s
    """
    invReGenerator = GroupEmitter(parser().parseString(regex)).makeGenerator()
    return invReGenerator()

def main():
    tests = r"""
    [A-EA]
    [A-D]*
    [A-D]{3}
    X[A-C]{3}Y
    X[A-C]{3}\(
    X\d
    foobar\d\d
    foobar{2}
    foobar{2,9}
    fooba[rz]{2}
    (foobar){2}
    ([01]\d)|(2[0-5])
    ([01]\d\d)|(2[0-4]\d)|(25[0-5])
    [A-C]{1,2}
    [A-C]{0,3}
    [A-C]\s[A-C]\s[A-C]
    [A-C]\s?[A-C][A-C]
    [A-C]\s([A-C][A-C])
    [A-C]\s([A-C][A-C])?
    [A-C]{2}\d{2}
    @|TH[12]
    @(@|TH[12])?
    @(@|TH[12]|AL[12]|SP[123]|TB(1[0-9]?|20?|[3-9]))?
    @(@|TH[12]|AL[12]|SP[123]|TB(1[0-9]?|20?|[3-9])|OH(1[0-9]?|2[0-9]?|30?|[4-9]))?
    (([ECMP]|HA|AK)[SD]|HS)T
    [A-CV]{2}
    A[cglmrstu]|B[aehikr]?|C[adeflmorsu]?|D[bsy]|E[rsu]|F[emr]?|G[ade]|H[efgos]?|I[nr]?|Kr?|L[airu]|M[dgnot]|N[abdeiop]?|Os?|P[abdmortu]?|R[abefghnu]|S[bcegimnr]?|T[abcehilm]|Uu[bhopqst]|U|V|W|Xe|Yb?|Z[nr]
    (a|b)|(x|y)
    (a|b) (x|y)
    """.split('\n')

    for t in tests:
        t = t.strip()
        if not t: continue
        print '-'*50
        print t
        try:
            print count(invert(t))
            for s in invert(t):
                print s
        except ParseFatalException,pfe:
            print pfe.msg
            print
            continue
        print

if __name__ == "__main__":
    main()
@preg_match($regexToTest, '');
SyntaxError
// VALID ONE
> /yes[^]*day/
Out: /yes[^]*day/

// INVALID ONE
> /yes[^*day/
Out: VM227:1 Uncaught SyntaxError: Invalid regular expression: missing /
var RegexParser = function(input) {

    // Parse input
    var m = input.match(/(\/?)(.+)\1([a-z]*)/i);

    // Invalid flags
    if (m[3] && !/^(?!.*?(.).*?\1)[gmixXsuUAJ]+$/.test(m[3])) {
        return RegExp(input);
    }

    // Create the regular expression
    return new RegExp(m[2], m[3]);
};
var RegexString = "/yes.*day/"

var isRegexValid = input => {
 try {
 const regex = RegexParser(input);
 }
 catch(error) {
   if(error.name === "SyntaxError") 
    {
      return false;
    }
    else 
    {
     throw error;
    }
 }
 return true;
}