用于创建与数字范围匹配的正则表达式的函数

Question

我正在使用Amazon Mechanical Turk API，它只允许我使用正则表达式来过滤数据字段。

我想向函数输入一个整数范围，例如256-311或45-1233，然后返回一个只与该范围匹配的正则表达式。

匹配256-321的正则表达式为：

\b((25[6-9])|(2[6-9][0-9])|(3[0-1][0-9])|(32[0-1]))\b

这部分相当简单，但我在创建这个正则表达式的循环中遇到了问题。

我正在尝试构建一个定义如下的函数：

function getRangeRegex( int fromInt, int toInt)
{

      return regexString;
}

我看遍了整个网络，我很惊讶，看起来没有人在过去解决过这个问题。这是一个很难解决的问题。

耽误您时间，实在对不起。

Answer 1

这里有一个简单的技巧：

<?php

function regex_range($from, $to) {

  if($from < 0 || $to < 0) {
    throw new Exception("Negative values not supported"); 
  }

  if($from > $to) {
    throw new Exception("Invalid range $from..$to, from > to"); 
  }

  $ranges = array($from);
  $increment = 1;
  $next = $from;
  $higher = true;

  while(true) {

    $next += $increment;

    if($next + $increment > $to) {
      if($next <= $to) {
        $ranges[] = $next;
      }
      $increment /= 10;
      $higher = false;
    }
    else if($next % ($increment*10) === 0) {
      $ranges[] = $next;
      $increment = $higher ? $increment*10 : $increment/10;
    }

    if(!$higher && $increment < 10) {
      break;
    }
  }

  $ranges[] = $to + 1;

  $regex = '/^(?:';

  for($i = 0; $i < sizeof($ranges) - 1; $i++) {
    $str_from = (string)($ranges[$i]);
    $str_to = (string)($ranges[$i + 1] - 1);

    for($j = 0; $j < strlen($str_from); $j++) {
      if($str_from[$j] == $str_to[$j]) {
        $regex .= $str_from[$j];
      }
      else {
        $regex .= "[" . $str_from[$j] . "-" . $str_to[$j] . "]";
      }
    }
    $regex .= "|";
  }

  return substr($regex, 0, strlen($regex)-1) . ')$/';
}

function test($from, $to) {
  try {
    printf("%-10s %s\n", $from . '-' . $to, regex_range($from, $to));
  } catch (Exception $e) {
    echo $e->getMessage() . "\n";
  }
}

test(2, 8);
test(5, 35);
test(5, 100);
test(12, 1234);
test(123, 123);
test(256, 321);
test(256, 257);
test(180, 195);
test(2,1);
test(-2,4);

?>

这会产生：

2-8        /^(?:[2-7]|8)$/
5-35       /^(?:[5-9]|[1-2][0-9]|3[0-5])$/
5-100      /^(?:[5-9]|[1-9][0-9]|100)$/
12-1234    /^(?:1[2-9]|[2-9][0-9]|[1-9][0-9][0-9]|1[0-2][0-3][0-4])$/
123-123    /^(?:123)$/
256-321    /^(?:25[6-9]|2[6-9][0-9]|3[0-2][0-1])$/
256-257    /^(?:256|257)$/
180-195    /^(?:18[0-9]|19[0-5])$/
Invalid range 2..1, from > to
Negative values not supported

未经过适当测试，使用风险自负！

是的，在许多情况下，生成的正则表达式可以编写得更紧凑，但我将其留给读者作为练习:)

Answer 2

对于像我一样正在寻找伟大的@Bart Kier作品的javascript版本的其他人，请参阅上面的

//Credit: Bart Kiers 2011
function regex_range(from, to){
        if(from < 0 || to < 0) {
            //throw new Exception("Negative values not supported"); 
            return null;
        }
        if(from > to) {
            //throw new Exception("Invalid range from..to, from > to"); 
            return null;
        }

        var ranges = [];
        ranges.push(from);
        var increment = 1;
        var next = from;
        var higher = true;

        while(true){
            next += increment;
            if(next + increment > to) {
                if(next <= to) {
                    ranges.push(next);
                }
                increment /= 10;
                higher = false;
            }else{ 
                if(next % (increment*10) == 0) {
                    ranges.push(next);
                    increment = higher ? increment*10 : increment/10;
                }
            }

            if(!higher && increment < 10) {
                break;
            }
        }

        ranges.push(to + 1);
        var regex = '/^(?:';

        for(var i = 0; i < ranges.length - 1; i++) {
            var str_from = ranges[i];
            str_from = str_from.toString();
            var str_to = ranges[i + 1] - 1;
            str_to = str_to.toString();
            for(var j = 0; j < str_from.length; j++) {
                if(str_from[j] == str_to[j]) {
                    regex += str_from[j];
                }
                else {
                    regex += "[" + str_from[j] + "-" + str_to[j] + "]";
                }
            }
            regex += "|";
        }

        return regex.substr(0, regex.length - 1 ) + ')$/';
    }

Answer 3

RegexNumericRangeGenerator的PHP端口

class RegexRangeNumberGenerator {

    static function parse($min, $max, $MatchWholeWord = FALSE, $MatchWholeLine = FALSE, $MatchLeadingZero = FALSE) {
        if (!is_int($min) || !is_int($max) || $min > $max || $min < 0 || $max < 0) {
            return FALSE;
        }
        if ($min == $max) {
            return self::parseIntoPattern($min, $MatchWholeWord, $MatchWholeLine, $MatchLeadingZero);
        }
        $s = [];
        $x = self::parseStartRange($min, $max);
        foreach ($x as $o) {
            $s[] = self::parseEndRange($o[0], $o[1]);
        }
        $n = self::reformatArray($s);
        $h = self::parseIntoRegex($n);
        return self::parseIntoPattern($h, $MatchWholeWord, $MatchWholeLine, $MatchLeadingZero);
    }

    static private function parseIntoPattern($t, $MatchWholeWord = FALSE, $MatchWholeLine = FALSE, $MatchLeadingZero = FALSE) {
        $r = ((is_array($t)) ? implode("|", $t) : $t);
        return (($MatchWholeLine && $MatchLeadingZero) ? "^0*(" . $r . ")$" : (($MatchLeadingZero) ? "0*(" . $r . ")" : (($MatchWholeLine) ? "^(" . $r . ")$" : (($MatchWholeWord) ? "\\b(" . $r . ")\\b" : "(" . $r . ")"))));
    }

    static private function parseIntoRegex($t) {
        if (!is_array($t)) {
            throw new Exception("Argument needs to be an array!");
        }
        $r = [];
        for ($i = 0; $i < count($t); $i++) {
            $e = str_split($t[$i][0]);
            $n = str_split($t[$i][1]);
            $s = "";
            $o = 0;
            $h = "";
            for ($a = 0; $a < count($e); $a++) {
                if ($e[$a] === $n[$a]) {
                    $h .= $e[$a];
                } else {
                    if ((intval($e[$a]) + 1) === intval($n[$a])) {
                        $h .= "[" . $e[$a] . $n[$a] . "]";
                    } else {
                        if ($s === ($e[$a] . $n[$a])) {
                            $o++;
                        }
                        $s = $e[$a] . $n[$a];
                        if ($a == (count($e) - 1)) {
                            $h .= (($o > 0) ? "{" . ($o + 1) . "}" : "[" . $e[$a] . "-" . $n[$a] . "]");
                        } else {
                            if ($o === 0) {
                                $h .= "[" . $e[$a] . "-" . $n[$a] . "]";
                            }
                        }
                    }
                }
            }
            $r[] = $h;
        }
        return $r;
    }

    static private function reformatArray($t) {
        $arrReturn = [];
        for ($i = 0; $i < count($t); $i++) {
            $page = count($t[$i]) / 2;
            for ($a = 0; $a < $page; $a++) {
                $arrReturn[] = array_slice($t[$i], (2 * $a), 2);
            }
        }
        return $arrReturn;
    }

    static private function parseStartRange($t, $r) {
        if (strlen($t) === strlen($r)) {
            return [[$t, $r]];
        }
        $break = pow(10, strlen($t)) - 1;
        return array_merge([[$t, $break]], self::parseStartRange($break + 1, $r));
    }

    static private function parseEndRange($t, $r) {
        if (strlen($t) == 1) {
            return [$t, $r];
        }
        if (str_repeat("0", strlen($t)) === "0" . substr($t, 1)) {
            if (str_repeat("0", strlen($r)) == "9" . substr($r, 1)) {
                return [$t, $r];
            }
            if ((int) substr($t, 0, 1) < (int) substr($r, 0, 1)) {
                $e = intval(substr($r, 0, 1) . str_repeat("0", strlen($r) - 1)) - 1;
                return array_merge([$t, self::strBreakPoint($e)], self::parseEndRange(self::strBreakPoint($e + 1), $r));
            }
        }
        if (str_repeat("9", strlen($r)) === "9" . substr($r, 1) && (int) substr($t, 0, 1) < (int) substr($r, 0, 1)) {
            $e = intval(intval((int) substr($t, 0, 1) + 1) . "" . str_repeat("0", strlen($r) - 1)) - 1;
            return array_merge(self::parseEndRange($t, self::strBreakPoint($e)), [self::strBreakPoint($e + 1), $r]);
        }
        if ((int) substr($t, 0, 1) < (int) substr($r, 0, 1)) {
            $e = intval(intval((int) substr($t, 0, 1) + 1) . "" . str_repeat("0", strlen($r) - 1)) - 1;
            return array_merge(self::parseEndRange($t, self::strBreakPoint($e)), self::parseEndRange(self::strBreakPoint($e + 1), $r));
        }
        $a = (int) substr($t, 0, 1);
        $o = self::parseEndRange(substr($t, 1), substr($r, 1));
        $h = [];
        for ($u = 0; $u < count($o); $u++) {
            $h[] = ($a . $o[$u]);
        }
        return $h;
    }

    static private function strBreakPoint($t) {
        return str_pad($t, strlen(($t + 1)), "0", STR_PAD_LEFT);
    }
}

测试结果

2-8         ^([2-8])$
5-35        ^([5-9]|[12][0-9]|3[0-5])$
5-100       ^([5-9]|[1-8][0-9]|9[0-9]|100)$
12-1234     ^(1[2-9]|[2-9][0-9]|[1-8][0-9]{2}|9[0-8][0-9]|99[0-9]|1[01][0-9]{2}|12[0-2][0-9]|123[0-4])$
123-123     ^(123)$
256-321     ^(25[6-9]|2[6-9][0-9]|3[01][0-9]|32[01])$
256-257     ^(25[67])$
180-195     ^(18[0-9]|19[0-5])$