使用正则expression式来validation数字范围
我的input号码是一个int。 但input的数字必须在-2055到2055的范围内,我想用正则expression式来检查。
那么反正要写一个正则expression式来检查一个数字是否在(-2055,2055)呢?
如果用语句来检查数字是否在范围内,则更容易使用。 但我正在写一个解释器,所以我应该使用正则expression式来检查input的数字
使用正则expression式来validation数字范围
要清楚:当一个简单的if语句就足够了
if(num < -2055 || num > 2055) { throw new IllegalArgumentException("num (" + num + ") must be between -2055 and 2055"); }
不build议使用正则expression式validation数字范围。
另外,由于正则expression式对string进行分析,因此必须先将数字转换为string才能进行testing。 数字碰巧已经是一个string,例如从控制台获取用户input时,就是一个例外。
(为确保string是一个数字,可以使用org.apache.commons.lang3.math. NumberUtils # isNumber(s)
)
尽pipe如此,弄清楚如何用正则expression式来validation数字范围是有趣和有益的。
(这个答案中的链接来自堆栈溢出正则expression式常见问题 。)
一个数字范围
规则:一个数字必须是15
。
有最简单的范围。 一个正则expression式来匹配这个
\b15\b
字边界是必要的,以避免匹配8215242
内部。
两个数字范围
规则:数字必须在15
到16
之间。 这里有三个可能的正则expression式:
\b(15|16)\b \b1(5|6)\b \b1[5-6]\b
(这些组是“或”-ing所必需的,但是它们可能是非捕获的 : \b(?:15|16)\b
)
数字范围“镜像”在零附近
规则:数字必须介于-12
和12
之间。
这是一个0
到12
的正则expression式,
\b(\d|1[0-2])\b
自由间隔:
\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
为负面和正面做这个工作就像在开始时添加一个可选的短划线一样简单:
-?\b(\d|1[0-2])\b
(这个假设在短划线之前没有不恰当的字符。)
为了禁止负数, 负面的逆序是必要的:
(?<!-)\b(\d|1[0-2])\b
离开后视会导致-11
中的-11
匹配。 (这个post中的第一个例子应该有这个补充。)
注意: \d
与[0-9]
为了兼容所有的正则expression式,所有\d
-s应该被改为[0-9]
。 例如,.NET将非ASCII码(如不同语言中的那些)视为\d
合法值。 除了最后一个例子,为了简洁起见,它保留为\d
。
(感谢@TimPietzcker )
三位数字,除了第一位数字都等于零
规则:必须在0
到400
之间。
一个可能的正则expression式:
(?<!-)\b([1-3]?\d{1,2}|400)\b
自由间隔:
(?<!-) //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
另一个不应该被使用的可能性:
\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
最后一个例子:四个数字,镜像在零附近,不以零结尾。
规则:必须在-2055
和2055
之间
这是从一个问题上的计算器。
正则expression式:
(-?\b(?:20(?:5[0-5]|[0-4][0-9])|1[0-9]{3}|[1-9][0-9]{0,2}|(?<!-)0+))\b
Debuggex演示
自由间隔:
( //Capture group for the entire number -?\b //Optional dash, followed by a word (number) boundary (?:20 //Followed by "20", which is followed by one of (?:5[0-5] //50 through 55 | //or [0-4][0-9]) //00 through 49 | //or 1[0-9]{3} //a one followed by any three digits | //or [1-9][0-9]{0,2} //1-9 followed by 0 through 2 of any digit | //or (?<!-)0+ //one-or-more zeros *not* preceded by a dash ) //end "or" non-capture group )\b //End number capture group, followed by a word-bound
(感谢PlasmaPower和Casimir et Hippolyte的debugging帮助。)
最后的笔记
根据你所捕捉到的情况,所有的小组很可能被制作成非捕获组。 例如,这个:
(-?\b(?:20(?:5[0-5]|[0-4][0-9])|1?[0-9]{1,3})\b)
而不是这个:
-?\b(20(5[0-5]|[0-4][0-9])|1?[0-9]{1,3})\b
示例Java实现
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]{3}|[1-9][0-9]{0,2}|(?<!-)0+))\\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); }
}
产量
[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
这个stackoverflow问题的原始答案
这是一个符合您的要求的严肃的答案。 这和@ PlasmaPower的答案类似。
(-?\b(?:20(?:5[0-5]|[0-4][0-9])|1[0-9]{3}|[1-9][0-9]{0,2}|(?<!-)0+))\b
Debuggex演示
永远不要使用它,但这是有效的。 🙂
\b(-2055|-2054|-2053|-2052|-2051|-2050|-2049|-2048|-2047|-2046|-2045|-2044|-2043|-2042|-2041|-2040|-2039|-2038|-2037|-2036|-2035|-2034|-2033|-2032|-2031|-2030|-2029|-2028|-2027|-2026|-2025|-2024|-2023|-2022|-2021|-2020|-2019|-2018|-2017|-2016|-2015|-2014|-2013|-2012|-2011|-2010|-2009|-2008|-2007|-2006|-2005|-2004|-2003|-2002|-2001|-2000|-1999|-1998|-1997|-1996|-1995|-1994|-1993|-1992|-1991|-1990|-1989|-1988|-1987|-1986|-1985|-1984|-1983|-1982|-1981|-1980|-1979|-1978|-1977|-1976|-1975|-1974|-1973|-1972|-1971|-1970|-1969|-1968|-1967|-1966|-1965|-1964|-1963|-1962|-1961|-1960|-1959|-1958|-1957|-1956|-1955|-1954|-1953|-1952|-1951|-1950|-1949|-1948|-1947|-1946|-1945|-1944|-1943|-1942|-1941|-1940|-1939|-1938|-1937|-1936|-1935|-1934|-1933|-1932|-1931|-1930|-1929|-1928|-1927|-1926|-1925|-1924|-1923|-1922|-1921|-1920|-1919|-1918|-1917|-1916|-1915|-1914|-1913|-1912|-1911|-1910|-1909|-1908|-1907|-1906|-1905|-1904|-1903|-1902|-1901|-1900|-1899|-1898|-1897|-1896|-1895|-1894|-1893|-1892|-1891|-1890|-1889|-1888|-1887|-1886|-1885|-1884|-1883|-1882|-1881|-1880|-1879|-1878|-1877|-1876|-1875|-1874|-1873|-1872|-1871|-1870|-1869|-1868|-1867|-1866|-1865|-1864|-1863|-1862|-1861|-1860|-1859|-1858|-1857|-1856|-1855|-1854|-1853|-1852|-1851|-1850|-1849|-1848|-1847|-1846|-1845|-1844|-1843|-1842|-1841|-1840|-1839|-1838|-1837|-1836|-1835|-1834|-1833|-1832|-1831|-1830|-1829|-1828|-1827|-1826|-1825|-1824|-1823|-1822|-1821|-1820|-1819|-1818|-1817|-1816|-1815|-1814|-1813|-1812|-1811|-1810|-1809|-1808|-1807|-1806|-1805|-1804|-1803|-1802|-1801|-1800|-1799|-1798|-1797|-1796|-1795|-1794|-1793|-1792|-1791|-1790|-1789|-1788|-1787|-1786|-1785|-1784|-1783|-1782|-1781|-1780|-1779|-1778|-1777|-1776|-1775|-1774|-1773|-1772|-1771|-1770|-1769|-1768|-1767|-1766|-1765|-1764|-1763|-1762|-1761|-1760|-1759|-1758|-1757|-1756|-1755|-1754|-1753|-1752|-1751|-1750|-1749|-1748|-1747|-1746|-1745|-1744|-1743|-1742|-1741|-1740|-1739|-1738|-1737|-1736|-1735|-1734|-1733|-1732|-1731|-1730|-1729|-1728|-1727|-1726|-1725|-1724|-1723|-1722|-1721|-1720|-1719|-1718|-1717|-1716|-1715|-1714|-1713|-1712|-1711|-1710|-1709|-1708|-1707|-1706|-1705|-1704|-1703|-1702|-1701|-1700|-1699|-1698|-1697|-1696|-1695|-1694|-1693|-1692|-1691|-1690|-1689|-1688|-1687|-1686|-1685|-1684|-1683|-1682|-1681|-1680|-1679|-1678|-1677|-1676|-1675|-1674|-1673|-1672|-1671|-1670|-1669|-1668|-1667|-1666|-1665|-1664|-1663|-1662|-1661|-1660|-1659|-1658|-1657|-1656|-1655|-1654|-1653|-1652|-1651|-1650|-1649|-1648|-1647|-1646|-1645|-1644|-1643|-1642|-1641|-1640|-1639|-1638|-1637|-1636|-1635|-1634|-1633|-1632|-1631|-1630|-1629|-1628|-1627|-1626|-1625|-1624|-1623|-1622|-1621|-1620|-1619|-1618|-1617|-1616|-1615|-1614|-1613|-1612|-1611|-1610|-1609|-1608|-1607|-1606|-1605|-1604|-1603|-1602|-1601|-1600|-1599|-1598|-1597|-1596|-1595|-1594|-1593|-1592|-1591|-1590|-1589|-1588|-1587|-1586|-1585|-1584|-1583|-1582|-1581|-1580|-1579|-1578|-1577|-1576|-1575|-1574|-1573|-1572|-1571|-1570|-1569|-1568|-1567|-1566|-1565|-1564|-1563|-1562|-1561|-1560|-1559|-1558|-1557|-1556|-1555|-1554|-1553|-1552|-1551|-1550|-1549|-1548|-1547|-1546|-1545|-1544|-1543|-1542|-1541|-1540|-1539|-1538|-1537|-1536|-1535|-1534|-1533|-1532|-1531|-1530|-1529|-1528|-1527|-1526|-1525|-1524|-1523|-1522|-1521|-1520|-1519|-1518|-1517|-1516|-1515|-1514|-1513|-1512|-1511|-1510|-1509|-1508|-1507|-1506|-1505|-1504|-1503|-1502|-1501|-1500|-1499|-1498|-1497|-1496|-1495|-1494|-1493|-1492|-1491|-1490|-1489|-1488|-1487|-1486|-1485|-1484|-1483|-1482|-1481|-1480|-1479|-1478|-1477|-1476|-1475|-1474|-1473|-1472|-1471|-1470|-1469|-1468|-1467|-1466|-1465|-1464|-1463|-1462|-1461|-1460|-1459|-1458|-1457|-1456|-1455|-1454|-1453|-1452|-1451|-1450|-1449|-1448|-1447|-1446|-1445|-1444|-1443|-1442|-1441|-1440|-1439|-1438|-1437|-1436|-1435|-1434|-1433|-1432|-1431|-1430|-1429|-1428|-1427|-1426|-1425|-1424|-1423|-1422|-1421|-1420|-1419|-1418|-1417|-1416|-1415|-1414|-1413|-1412|-1411|-1410|-1409|-1408|-1407|-1406|-1405|-1404|-1403|-1402|-1401|-1400|-1399|-1398|-1397|-1396|-1395|-1394|-1393|-1392|-1391|-1390|-1389|-1388|-1387|-1386|-1385|-1384|-1383|-1382|-1381|-1380|-1379|-1378|-1377|-1376|-1375|-1374|-1373|-1372|-1371|-1370|-1369|-1368|-1367|-1366|-1365|-1364|-1363|-1362|-1361|-1360|-1359|-1358|-1357|-1356|-1355|-1354|-1353|-1352|-1351|-1350|-1349|-1348|-1347|-1346|-1345|-1344|-1343|-1342|-1341|-1340|-1339|-1338|-1337|-1336|-1335|-1334|-1333|-1332|-1331|-1330|-1329|-1328|-1327|-1326|-1325|-1324|-1323|-1322|-1321|-1320|-1319|-1318|-1317|-1316|-1315|-1314|-1313|-1312|-1311|-1310|-1309|-1308|-1307|-1306|-1305|-1304|-1303|-1302|-1301|-1300|-1299|-1298|-1297|-1296|-1295|-1294|-1293|-1292|-1291|-1290|-1289|-1288|-1287|-1286|-1285|-1284|-1283|-1282|-1281|-1280|-1279|-1278|-1277|-1276|-1275|-1274|-1273|-1272|-1271|-1270|-1269|-1268|-1267|-1266|-1265|-1264|-1263|-1262|-1261|-1260|-1259|-1258|-1257|-1256|-1255|-1254|-1253|-1252|-1251|-1250|-1249|-1248|-1247|-1246|-1245|-1244|-1243|-1242|-1241|-1240|-1239|-1238|-1237|-1236|-1235|-1234|-1233|-1232|-1231|-1230|-1229|-1228|-1227|-1226|-1225|-1224|-1223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这么多的答案,但没有人阅读(或关心)在评论中的OP方面的问题?
我在OCaml写一个解释器….我怎样才能validationinput数字的范围内,而不使用正则expression式? – Trung Nguyen 3月2日17:30
正如许多答案 – 正确地指出,使用正则expression式在这种情况下是可怕的,让我们考虑OCaml中的其他方法! 从我使用OCaml开始,有一段时间,但是通过查找一些结构,我可以把它们敲在一起:
let isInRange i = not(i < -2055 or i > 2055);; let isIntAndInRange s = try let i = int_of_string s in not(i < -2055 or i > 2055) with Failure "int_of_string" -> false;; let () = print_string "type a number: " in let s = read_line () in isIntAndInRange s
如果有什么不清楚的地方,请阅读它的语法, types转换函数 , exception处理和input输出函数 。
用户input部分仅用于演示。 在那里使用read_int
函数可能会更方便。 但处理exception的基本概念保持不变。
分析问题
如果您“必须”使用正则expression式,请通过分析接受的排列来解决问题。
“从-2055到2055的范围”可以表示为:
- 一个可选 –
- 可选的前导零
- 随后是从0到2055的数字
“从0到2055的数字”可以是有限数量的特定排列之一:
- 一位数字(0-9)
- 两位数(10-99)
- 三位数字(100-999)
- 四位数以1(1000-1999)
- 四位数以20开始(2000-204 * 9)
- 四位数以205开头(2050-2055 *)
请注意,对于这个正则expression式,没有必要区分“0-9”和“1-9”范围,只有最后两个范围对接受的数字/字符的范围有任何限制(用星)。
编写组件正则expression式
以上每个组件都很容易作为正则expression式单独expression:
- – ?
- 0 *
- [0-9]
- [0-9] [0-9]
- [0-9] [0-9] [0-9]
- 1 [0-9] [0-9] [0-9]
- 20 [0-4] [0-9]
- 205 [0-5]
把expression式放在一起
整个比赛的相关表述是:
-?0*([0-9]|[0-9][0-9]|[0-9][0-9][0-9]|1[0-9][0-9][0-9]|20[0-4][0-9]|205[0-5])
或者更简洁些:
-?0*([0-9]{1,3}|1[0-9]{3}|20[0-4][0-9]|205[0-5])
假设input只包含“数字”,没有别的,最后的正则expression式是:
^-?0*([0-9]{1,3}|1[0-9]{3}|20[0-4][0-9]|205[0-5])$
如果有必要考虑领先的加号,这将变成:
^[-+]?0*([0-9]{1,3}|1[0-9]{3}|20[0-4][0-9]|205[0-5])$
这是一个js小提琴展示什么通过,什么会失败的最后一个正则expression式。
作为aliteralmind所提供的伟大expression的替代方法,为了看看另一种方法可能是什么样子(以及不该做什么) ,这里是一个更长更有趣的方法。
这是一个有趣的练习,因为您可以考虑两种截然不同的方法 :粗略地说,您可以:
- 通过匹配4个字符的数字,然后匹配3个字符的数字等。
- 或者通过匹配千位数字,然后百位数字等来进行
没有尝试,你怎么知道哪个是最好的? 事实certificate,第一种方法(aliteralmind的答案)更为经济。
更低的话,我会在PHP语言中包含一系列testing ,以防您或其他人想要检查输出。
下面,我将给你正则expression式的“自由间距模式”,它允许在正则expression式中的注释,所以你可以很容易地理解它是做什么的。 然而,并非所有的正则expression式引擎都支持自由空间模式,所以在我们开始讨论有趣的部分之前,这里是将正则expression式作为一个单线程。
请注意,你的问题提到从-2055到2055年的数字。我假设你想匹配“正常数字”,没有前导零。 这意味着正则expression式将匹配999而不是0999.如果你想要引导零,让我知道,这是一个非常简单的调整。
另外,如果你在UTF-8模式下匹配, \d
应该被replace为[0-9]
。 这是更常见的forms。
正则expression式作为一个单线
^(?!-0$)-?(?:(?=\d{4}$)[12])?(?:(?=\d{3}$)(?:(?:(?<=^)|(?<=^-)|(?<=1)|(?<=-1))\d|(?:(?<=2)|(?<=-2))0))?(?:(?=\d{2}$)(?:(?:(?<=^)|(?<=^-)|(?<=^\d)|(?<=^-\d)|(?<=^\d{2})(?<!20)|(?<=^-\d{2})(?<!-20))\d|(?:(?<=20)|(?<=-20))[0-5]))?(?:(?=\d$)(?:(?:(?<=^)|(?<=^-)|(?<=^\d)|(?<=^-\d)|(?<=^\d{2})|(?<=^-\d{2})|(?<=^\d{3})(?<!205)|(?<=^-\d{3})(?<!-205))\d|(?:(?<=205)|(?<=-205))[0-5]))$
自由间距模式下的正则expression式
(?x) # free-spacing mode ^ # anchor at beginning of string. (?!-0$)-? # optional minus sign, but not for -0 (?: # OPTIONAL THOUSANDS DIGIT (?=\d{4}$)[12] # assert that the number is 4-digit long, match 1 or 2 )? # end optional thousands digit (?: # OPTIONAL HUNDREDS DIGIT (?=\d{3}$) # assert that there are three digits left (?: # non-capturing group (?:(?<=^)|(?<=^-)|(?<=1)|(?<=-1))\d # if preceding chars are 1, -1 or head of string: value can be any digit | # or (?:(?<=2)|(?<=-2))0 # if preceding chars are 2 or -2: value must be 0 ) # close non-capturing group )? # end optional hundreds digits (?: # OPTIONAL TENS DIGIT (?=\d{2}$) # assert that there are two digits left (?: # start non-capturing group # if preceding char is head of string, single digit, # or two digits that are not 20 # (with or without a minus) # value can be any digit (?:(?<=^)|(?<=^-)|(?<=^\d)|(?<=^-\d)| (?<=^\d{2})(?<!20)|(?<=^-\d{2})(?<!-20))\d | # or (?:(?<=20)|(?<=-20))[0-5] # if preceding chars are 20 or -20: value can be from 0 to 5 ) # end non-capturing group )? # close optional tens digits (?: # FINAL DIGIT (non optional) (?=\d$) # assert that there is only one digit left (?: # start non-capturing group # if preceding char is head of string, single digit, # two digits, or three digits that are not 205 # (with or without a minus) # value can be any digit (?:(?<=^)|(?<=^-)|(?<=^\d)|(?<=^-\d)| (?<=^\d{2})|(?<=^-\d{2})| (?<=^\d{3})(?<!205)|(?<=^-\d{3})(?<!-205)) \d | # or (?:(?<=205)|(?<=-205))[0-5] # if preceding chars are 205 or -205: value can be from 0 to 5 ) # end non-capturing group ) # end final digit $
一系列的testing
这些testing尝试匹配从-100000到100000的数字。它们产生以下输出:
Successful test: matches from -2055 to 2055 Successful test: NO matches from -100000 to -2056 Successful test: NO matches from 2056 to 100000
这里是代码:
<?php $regex="~^(?!-0$)-?(?:(?=\d{4}$)[12])?(?:(?=\d{3}$)(?:(?:(?<=^)|(?<=^-)|(?<=1)|(?<=-1))\d|(?:(?<=2)|(?<=-2))0))?(?:(?=\d{2}$)(?:(?:(?<=20)|(?<=-20))[0-5]|(?:(?<=^)|(?<=^-)|(?<=^\d)|(?<=^-\d)|(?<=^\d{2})(?<!20)|(?<=^-\d{2})(?<!-20))\d))?(?:(?=\d$)(?:(?:(?<=205)|(?<=-205))[0-5]|(?:(?<=^)|(?<=^-)|(?<=^\d)|(?<=^-\d)|(?<=^\d{2})|(?<=^-\d{2})|(?<=^\d{3})(?<!205)|(?<=^-\d{3})(?<!-205))\d))$~"; // Test 1 $success="Successful test: matches from -2055 to 2055"; for($i=-2055;$i<=2055;$i++) { $chari = sprintf("%d",$i); if (! preg_match($regex,$chari)) { $success="Failed test: matches from -2055 to 2055"; echo $chari.": No Match!<br />"; } } echo $success."<br />"; // Test 2 $success="Successful test: NO matches from -100000 to -2056"; for($i=-100000;$i<=-2056;$i++) { $chari = sprintf("%d",$i); if (preg_match($regex,$chari)) { $success="Failed test: NO matches from -100000 to -2056"; echo $chari.": Match!<br />"; } } echo $success."<br />"; // Test 3 $success="Successful test: NO matches from 2056 to 100000"; for($i=2056;$i<=100000;$i++) { $chari = sprintf("%d",$i); if (preg_match($regex,$chari)) { $success="Failed test: NO matches from 2056 to 100000"; echo $chari.": Match!<br />"; } } echo $success."<br />"; ?>
速度testing
这里是我简单的速度testing的输出,匹配从-1M到+ 1M。 正如卡西米尔指出的那样,如果说阿尔泰主义的expression是固定的,而不是慢下来,速度会提高25%!
zx81: 3.796217918396 aliteralmind: 3.9922280311584 difference: 5.1632998151294 percent longer
这里是testing代码:
$regex="~(?x)^(?!-0$)-?(?:(?=\d{4}$)[12])?(?:(?=\d{3}$)(?:(?:(?<=^)|(?<=^-)|(?<=1)|(?<=-1))\d|(?:(?<=2)|(?<=-2))0))?(?:(?=\d{2}$)(?:(?:(?<=^)|(?<=^-)|(?<=^\d)|(?<=^-\d)|(?<=^\d{2})(?<!20)|(?<=^-\d{2})(?<!-20))\d|(?:(?<=20)|(?<=-20))[0-5]))?(?:(?=\d$)(?:(?:(?<=^)|(?<=^-)|(?<=^\d)|(?<=^-\d)|(?<=^\d{2})|(?<=^-\d{2})|(?<=^\d{3})(?<!205)|(?<=^-\d{3})(?<!-205))\d|(?:(?<=205)|(?<=-205))[0-5]))$~"; $regex2 = "~(-?\b(?:20(?:5[0-5]|[0-4][0-9])|1[0-9]{3}|[1-9][0-9]{0,2}|(?<!-)0+))\b~"; $start=microtime(TRUE); for ($i=-1000000;$i<1000000;$i++) preg_match($regex,$i); $zxend=microtime(TRUE); for ($i=-1000000;$i<1000000;$i++) preg_match($regex2,$i); $alitend=microtime(TRUE); $zx81 = $zxend-$start; $alit = $alitend-$zxend; $diff = 100*($alit-$zx81)/$zx81; echo "zx81: ".$zx81."<br />"; echo "aliteralmind: ".$alit."<br />"; echo "difference: ".$diff." percent longer<br />";
为什么只使用正则expression式来检查一个数字?
int n = -2000; if(n >= -2055 && n <= 2055) //Do something else //Do something else
尝试这个:
\-?\b0*(205[0-5]|20[0-4]\d|1?\d{3}|\d{1,2})\b
看看这个伟大的工具,为数字范围生成正则expression式。 你也可以用Python下载源代码:
http://utilitymill.com/utility/Regex_For_Range http://code.activestate.com/recipes/534137-generate-a-regular-expression-to-match-an-arbitrar/
对于OP请求的范围,它生成: -?([0-9]{1,3}|1[0-9]{3}|20[0-4][0-9]|205[0-5])
尝试一个非常简单的正则expression式 。
^([-0][0-1][0-9][0-9][0-9])$|^([-0]20[0-4][0-9])$|^([-0]205[0-5])$
视觉表示
这很容易理解 。
- 第1组
[-0][0-1][0-9][0-9][0-9]
将覆盖[-1999,1999]值 - 组2
[-0]20[0-4][0-9]
将覆盖[-2000,-2049]和[2000,2049]值 - 组3
[-0]205[0-5]
将覆盖[-2050,-2055]和[ 2050,2055 ]值
String.format("%05d", number)
在这里做得很好吗?
示例代码:(更多内容,请阅读内嵌评论。)
int[] numbers = new int[] { -10002, -3000, -2056, -2055, -2000, -1999, -20, -1, 0, 1, 260, 1999, 2000, 2046, 2055, 2056, 2955, 3000, 10002, 123456 }; //valid range -2055 to 2055 inclusive Pattern p = Pattern.compile("^([-0][0-1][0-9][0-9][0-9])$|^([-0]20[0-4][0-9])$|^([-0]205[0-5])$"); for (int number : numbers) { String string = String.format("%05d", number); Matcher m = p.matcher(string); if (m.find()) { System.out.println(number + " is in range."); } else { System.out.println(number + " is not in range."); } }
输出:
-10002 is not in range. -3000 is not in range. -2056 is not in range. -2055 is in range. -2000 is in range. -1999 is in range. -20 is in range. -1 is in range. 0 is in range. 1 is in range. 260 is in range. 1999 is in range. 2000 is in range. 2046 is in range. 2055 is in range. 2056 is not in range. 2955 is not in range. 3000 is not in range. 10002 is not in range. 123456 is not in range.
尝试这个:
^-?0*(1?[0-9]{1,3}|20[0-4][0-9]|205[0-5])$
括号前面的正则expression式匹配一个可选的-
和任何前导0。
括号内的第一部分( 1?[0-9]{1,3}
)匹配0-1999。
括号内的第二部分( 20[0-4][0-9]
)匹配2000-2049。
括号内的第三部分( 205[0-5]
)匹配2050-2055。
完后还有 –
# ^-?0*(?:20(?:[0-4][0-9]|5[0-5])|[0-9]{1,3})$ ^ -? 0* (?: 20 (?: [0-4] [0-9] | 5 [0-5] ) | [0-9]{1,3} ) $