我正在使用亚马逊众包API，并且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，并返回仅匹配该范围的正则表达式。
匹配256-321的正则表达式是：
\b((25[6-9])|(2[6-9][0-9])|(3[0-1][0-9])|(32[0-1]))\b
那一部分相当容易，但我在循环创建正则表达式时遇到了麻烦。
我正在尝试构建一个定义如下的函数:
function getRangeRegex( int fromInt, int toInt)
{
return regexString;
}
我在网上搜索了很久，但惊讶地发现似乎没有人以前解决过这个问题。这是一个棘手的问题...
感谢您的时间。
这是一个快速的技巧：
<?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 $e->getMessage() . "\n";
}
}
test(2, 8);
test(5, 35);
test(5, 100);
test(12, 1234);
test(123, 123);
test(256, 321);
test(256, 257);
test(180, 195);
test(2,1);
test(-2,4);
?>
生成如下结果：
2-8 /^(?:[2-7]|8)$/
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])$/
Invalid range 2..1, from > to
Negative values not supported
测试不充分，使用需谨慎！
是的，在许多情况下，生成的正则表达式可以更紧凑地编写，但是我将其留作读者的练习 :)
256-321中也可以看到这一点。正则表达式的最后一部分3 [0-2] [0-1]不匹配例如309，但它应该匹配，因此正则表达式应为3 [0-1] [0-9] | 32 [0-1]。如果结束数字是318，则最终范围应为30 [0-9] | 31 [0-1]。我会尝试修复它，但可能有人会提出比我的更好的解决方案。 :) - tukusejssirs^(?:[2-7]|8)$ 是有些别扭的。 - mickmackusa如果有其他人和我一样在寻找与@Bart Kiers大师所创作的内容相似的javascript版本，那么以下内容可能会对您有所帮助。
//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 ) + ')$/';
}
PHP版本的RegexNumericRangeGenerator
class RegexRangeNumberGenerator {
static function parse($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])$
有必要使用正则表达式吗？不能像这样做：
if ($number >= 256 && $number <= 321){
// do something
}
更新：
有一种简单但不太美观的方法，可以使用range函数来实现：
function getRangeRegex($from, $to)
{
$range = implode('|', range($from, $to));
// returns: 256|257|...|321
return $range;
}
请注意，@Bart Kiers的优秀代码（以及Travis J的JS版本）在某些情况下会失败。例如：
最初的回答:
12-1234 /^(?:1[2-9]|[2-9][0-9]|[1-9][0-9][0-9]|1[0-2][0-3][0-4])$/
does not match "1229", "1115", "1[0-2][0-2][5-9]"
这个其实已经做过了。
看一下this网站。它包含一个链接到一个自动生成这些正则表达式的Python脚本的链接。
#include <stdio.h>
#include <iostream>
#include <vector>
#include <string>
std::string regex_range(int from, int to);
int main(int argc, char **argv)
{
std::string regex = regex_range(1,100);
std::cout << regex << std::endl;
return 0;
}
std::string regex_range(int from, int to) //Credit: Bart Kiers 2011
{
if(from < 0 || to < 0)
{
std::cout << "Negative values not supported. Exiting." << std::endl;
return 0;
}
if(from > to)
{
std::cout << "Invalid range, from > to. Exiting." << std::endl;
return 0;
}
std::vector<int> ranges;
ranges.push_back(from);
int increment = 1;
int next = from;
bool higher = true;
while(true)
{
next += increment;
if(next + increment > to)
{
if(next <= to)
{
ranges.push_back(next);
}
increment /= 10;
higher = false;
}
else if(next % (increment*10) == 0)
{
ranges.push_back(next);
increment = higher ? increment*10 : increment/10;
}
if(!higher && (increment < 10))
{
break;
}
}
ranges.push_back(to + 1);
std::string regex("^(?:");
for(int i = 0; i < ranges.size() - 1; i++)
{
int current_from = ranges.at(i);
std::string str_from = std::to_string(current_from);
int current_to = ranges.at(i + 1) - 1;
std::string str_to = std::to_string(current_to);
for(int j = 0; j < str_from.length(); j++)
{
if(str_from.at(j) == str_to.at(j))
{
std::string str_from_at_j(&str_from.at(j));
regex.append(str_from_at_j);
}
else
{
std::string str_from_at_j(&str_from.at(j));
std::string str_to_at_j(&str_to.at(j));
regex.append("[");
regex.append(str_from_at_j);
regex.append("-");
regex.append(str_to_at_j);
regex.append("]");
}
}
regex.append("|");
}
regex = regex.substr(0, regex.length() - 1);
regex.append(")$");
return regex;
}
明确一点：当简单的if语句足以满足要求时
if(num < -2055 || num > 2055) {
throw new IllegalArgumentException("num (" + num + ") must be between -2055 and 2055");
}
不建议使用正则表达式来验证数字范围。
此外，由于正则表达式分析字符串，因此在测试之前必须先将数字转换为字符串（一个例外是当数字恰好已经是字符串时，例如从控制台获取用户输入时）。
（为确保字符串本身就是数字，您可以使用org.apache.commons.lang3.math.NumberUtils#isNumber(s)）
尽管如此，了解如何使用正则表达式验证数字范围是有趣且有教育意义的。
规则：一个数字必须恰好为15。
The simplest range there is. A regex to match this is
\b15\b
Word boundaries are necessary to avoid matching the 15 inside of 8215242.
The rule: The number must be between 15 and 16. Three possible regexes:
\b(15|16)\b
\b1(5|6)\b
\b1[5-6]\b
The rule: The number must be between -12 and 12.
Here is a regex for 0 through 12, positive-only:
\b(\d|1[0-2])\b
Free-spaced:
\b( //The beginning of a word (or number), followed by either
\d // Any digit 0 through 9
| //Or
1[0-2] // A 1 followed by any digit between 0 and 2.
)\b //The end of a word
Making this work for both negative and positive is as simple as adding an optional dash at the start:
-?\b(\d|1[0-2])\b
(This assumes no inappropriate characters precede the dash.)
To forbid negative numbers, a negative lookbehind is necessary:
(?<!-)\b(\d|1[0-2])\b
Leaving the lookbehind out would cause the 11 in -11 to match. (The first example in this post should have this added.)
\d versus [0-9]In order to be compatible with all regex flavors, all \d-s should be changed to [0-9]. For example, .NET considers non ASCII numbers, such as those in different languages, as legal values for \d. Except for in the last example, for brevity, it's left as \d.
(With thanks to TimPietzcker at stackoverflow)
Rule: Must be between 0 and 400.
A possible regex:
(?<!-)\b([1-3]?\d{1,2}|400)\b
Free spaced:
(?<!-) //Something not preceded by a dash
\b( //Word-start, followed by either
[1-3]? // No digit, or the digit 1, 2, or 3
\d{1,2} // Followed by one or two digits (between 0 and 9)
| //Or
400 // The number 400
)\b //Word-end
Another possibility that should never be used:
\b(0|1|2|3|4|5|6|7|8|9|10|11|12|13|14|15|16|17|18|19|20|21|22|23|24|25|26|27|28|29|30|31|32|33|34|35|36|37|38|39|40|41|42|43|44|45|46|47|48|49|50|51|52|53|54|55|56|57|58|59|60|61|62|63|64|65|66|67|68|69|70|71|72|73|74|75|76|77|78|79|80|81|82|83|84|85|86|87|88|89|90|91|92|93|94|95|96|97|98|99|100|101|102|103|104|105|106|107|108|109|110|111|112|113|114|115|116|117|118|119|120|121|122|123|124|125|126|127|128|129|130|131|132|133|134|135|136|137|138|139|140|141|142|143|144|145|146|147|148|149|150|151|152|153|154|155|156|157|158|159|160|161|162|163|164|165|166|167|168|169|170|171|172|173|174|175|176|177|178|179|180|181|182|183|184|185|186|187|188|189|190|191|192|193|194|195|196|197|198|199|200|201|202|203|204|205|206|207|208|209|210|211|212|213|214|215|216|217|218|219|220|221|222|223|224|225|226|227|228|229|230|231|232|233|234|235|236|237|238|239|240|241|242|243|244|245|246|247|248|249|250|251|252|253|254|255|256|257|258|259|260|261|262|263|264|265|266|267|268|269|270|271|272|273|274|275|276|277|278|279|280|281|282|283|284|285|286|287|288|289|290|291|292|293|294|295|296|297|298|299|300|301|302|303|304|305|306|307|308|309|310|311|312|313|314|315|316|317|318|319|320|321|322|323|324|325|326|327|328|329|330|331|332|333|334|335|336|337|338|339|340|341|342|343|344|345|346|347|348|349|350|351|352|353|354|355|356|357|358|359|360|361|362|363|364|365|366|367|368|369|370|371|372|373|374|375|376|377|378|379|380|381|382|383|384|385|386|387|388|389|390|391|392|393|394|395|396|397|398|399|400)\b
Rule: Must be between -2055 and 2055
This is from a question on stackoverflow.
Regex:
-?\b(20(5[0-5]|[0-4][0-9])|1?[0-9]{1,3})\b
Free-spaced:
-? //Optional dash
\b( //Followed by word boundary, followed by either of the following
20( // "20", followed by either
5[0-5] // A "5" followed by a digit 0-5
| // or
[0-4][0-9] // A digit 0-4, followed by any digit
)
| //OR
1?[0-9]{1,3} // An optional "1", followed by one through three digits (0-9)
)\b //Followed by a word boundary.
Here is a visual representation of this regex:
And here you can try it out yourself: Debuggex demonstration
(With thanks to PlasmaPower on stackoverflow for the debugging assistance.)
Depending on what you are capturing, it is likely that all sub-groups should be made into non-capture groups. For example, this:
(-?\b(?:20(?:5[0-5]|[0-4][0-9])|1?[0-9]{1,3})\b)
Instead of this:
-?\b(20(5[0-5]|[0-4][0-9])|1?[0-9]{1,3})\b
import java.util.Scanner;
import java.util.regex.Matcher;
import java.util.regex.Pattern;
import org.apache.commons.lang.math.NumberUtils;
/**
<P>Confirm a user-input number is a valid number by reading a string an testing it is numeric before converting it to an it--this loops until a valid number is provided.</P>
<P>{@code java UserInputNumInRangeWRegex}</P>
**/
public class UserInputNumInRangeWRegex {
public static final void main(String[] ignored) {
int num = -1;
boolean isNum = false;
int iRangeMax = 2055;
//"": Dummy string, to reuse matcher
Matcher mtchrNumNegThrPos = Pattern.compile("-?\\b(20(5[0-5]|[0-4][0-9])|1?[0-9]{1,3})\\b").matcher("");
do {
System.out.print("Enter a number between -" + iRangeMax + " and " + iRangeMax + ": ");
String strInput = (new Scanner(System.in)).next();
if(!NumberUtils.isNumber(strInput)) {
System.out.println("Not a number. Try again.");
} else if(!mtchrNumNegThrPos.reset(strInput).matches()) {
System.out.println("Not in range. Try again.");
} else {
//Safe to convert
num = Integer.parseInt(strInput);
isNum = true;
}
} while(!isNum);
System.out.println("Number: " + num);
}
}
Output
[C:\java_code\]java UserInputNumInRangeWRegex
Enter a number between -2055 and 2055: tuhet
Not a number. Try again.
Enter a number between -2055 and 2055: 283837483
Not in range. Try again.
Enter a number between -2055 and 2055: -200000
Not in range. Try again.
Enter a number between -2055 and 2055: -300
Number: -300
由于我遇到了与@EmilianoT已经报告的相同问题encountered，我尝试修复它，但最终我选择移植PHP port RegexNumericRangeGenerator（由@EmilianoT移植），虽然不是在类中。我对这个JS端口不太满意，因为所有toString()和parseInt()方法仍然可以优化（它们可能在某些地方是不必要的），但它适用于所有情况。
parse(min, max, width = 0, prefix = '', suffix = '')替换了parse($min, $max, $MatchWholeWord = FALSE, $MatchWholeLine = FALSE, $MatchLeadingZero = FALSE)，这样它就具有了更多选项（有些人可能想把正则表达式放入斜杠中，其他人想匹配行[prefix='^'，suffix='$']等）。此外，我希望能够配置数字的宽度（width=3 → 000、001、052、800、1000，...）。function parse(min, max, width = 0, prefix = '', suffix = '') {
if (! Number.isInteger(min) || ! Number.isInteger(max) || min > max || min < 0 || max < 0) {
return false
}
if (min == max) {
return parseIntoPattern(min, prefix, suffix)
}
let x = parseStartRange(min, max)
let s = []
x.forEach(o => {
s.push(parseEndRange(o[0], o[1]))
})
let n = reformatArray(s)
let h = parseIntoRegex(n, width)
return parseIntoPattern(h, prefix, suffix)
}
function parseIntoPattern(t, prefix = '', suffix = '') {
let r = Array.isArray(t) ? t.join('|') : t
return prefix + '(' + r + ')' + suffix
}
function parseIntoRegex(t, width = 0) {
if (! Array.isArray(t)) {
throw new Error('Argument needs to be an array!')
}
let r = []
for (let i = 0; i < t.length; i++) {
let e = t[i][0].split('')
let n = t[i][1].split('')
let s = ''
let o = 0
let h = ''
for (let a = 0; a < e.length; a++) {
if (e[a] === n[a]) {
h += e[a]
} else if (parseInt(e[a]) + 1 === parseInt(n[a])) {
h += '[' + e[a] + n[a] + ']'
} else {
if (s === e[a] + n[a]) {
o++
}
s = e[a] + n[a]
if (a == e.length - 1) {
h += o > 0 ? '{' + (o + 1) + '}' : '[' + e[a] + '-' + n[a] + ']'
} else if (o === 0) {
h += '[' + e[a] + '-' + n[a] + ']'
}
}
}
if (e.length < width) {
h = '0'.repeat(width - e.length, '0') + h
}
r.push(h)
}
return r
}
function reformatArray(t) {
let arrReturn = []
for (let i = 0; i < t.length; i++) {
let page = t[i].length / 2
for (let a = 0; a < page; a++) {
arrReturn.push(t[i].slice(2 * a))
}
}
return arrReturn
}
function parseStartRange(t, r) {
t = t.toString()
r = r.toString()
if (t.length === r.length) {
return [[t, r]]
}
let breakOut = 10 ** t.length - 1
return [[t, breakOut.toString()]].concat(parseStartRange(breakOut + 1, r))
}
function parseEndRange(t, r) {
if (t.length == 1) {
return [t, r]
}
if ('0'.repeat(t.length) === '0' + t.substr(1)) {
if ('0'.repeat(r.length) == '9' + r.substr(1)) {
return [t, r]
}
if (parseInt(t.toString().substr(0, 1)) < parseInt(r.toString().substr(0, 1))) {
let e = parseInt(r.toString().substr(0, 1) + '0'.repeat(r.length - 1)) - 1
return [t, strBreakPoint(e)].concat(parseEndRange(strBreakPoint(e + 1), r))
}
}
if ('9'.repeat(r.length) === '9' + r.toString().substr(1) && parseInt(t.toString().substr(0, 1)) < parseInt(r.toString().substr(0, 1))) {
let e = parseInt(parseInt(parseInt(t.toString().substr(0, 1)) + 1) + '0'.repeat(r.length - 1)) - 1
return parseEndRange(t, strBreakPoint(e)).concat(strBreakPoint(e + 1), r)
}
if (parseInt(t.toString().substr(0, 1)) < parseInt(r.toString().substr(0, 1))) {
let e = parseInt(parseInt(parseInt(t.toString().substr(0, 1)) + 1) + '0'.repeat(r.length - 1)) - 1
return parseEndRange(t, strBreakPoint(e)).concat(parseEndRange(strBreakPoint(e + 1), r))
}
let a = parseInt(t.toString().substr(0, 1))
let o = parseEndRange(t.toString().substr(1), r.toString().substr(1))
let h = []
for (let u = 0; u < o.length; u++) {
h.push(a + o[u])
}
return h
}
function strBreakPoint(t) {
return t.toString().padStart((parseInt(t) + 1).toString().length, '0')
}
180-195返回了一个不正确的正则表达式）。我已经修复了它（同样的答案）。 - Bart Kiers