Using regular expressions to validate a numeric range

匿名 (未验证) 提交于 2019-12-03 01:18:02

问题:

My input number is an int. But the input number must be in a range from -2055 to 2055 and I want to check this by using regular expression.

So is there anyway to write a regular expression to check whether a number is in (-2055, 2055) or not ?

It is easier to use if statement to check whether the number is in range or not. But I'm writing an interpreter so I should use regex to check the input number

回答1:

Using regular expressions to validate a numeric range

To be clear: When a simple if statement will suffice

if(num  2055)  {    throw  new IllegalArgumentException("num (" + num + ") must be between -2055 and 2055"); } 

using regular expressions for validating numeric ranges is not recommended.

In addition, since regular expressions analyze strings, numbers must first be translated to a string before they can be tested. An exception is when the number happens to already be a string, such as when getting user input from the console.

(To ensure the string is a number to begin with, you could use org.apache.commons.lang3.math.NumberUtils#isNumber(s))

Despite this, figuring out how to validate number ranges with regular expressions is interesting and instructive.

(The links in this answer come from the Stack Overflow Regular Expressions FAQ.)

A one number range

Rule: A number must be exactly 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.

A two number range

The rule: The number must be between 15 and 16. Here are three possible regexes:

\b(15|16)\b \b1(5|6)\b \b1[5-6]\b 

(The groups are required for the "or"-ing, but they could be non-capturing: \b(?:15|16)\b)

A number range "mirrored" around zero

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:

(?

Leaving the lookbehind out would cause the 11 in -11 to match. (The first example in this post should have this added.)

Note: \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)

Three digits, with all but the first digit equal to zero

Rule: Must be between 0 and 400.

A possible regex:

(?

Free spaced:

   (?

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 

Final example: Four digits, mirrored around zero, that does not end with zeros.

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]{3}|[1-9][0-9]{0,2}|(?

Debuggex Demo

Free-spaced:

(             //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          (?

(With thanks to PlasmaPower and Casimir et Hippolyte for the debugging assistance.)

Final note

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 

Example Java implementation

  import  java.util.Scanner;   import  java.util.regex.Matcher;   import  java.util.regex.Pattern;   import  org.apache.commons.lang.math.NumberUtils; /**   

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.

{@code java UserInputNumInRangeWRegex}

**/ 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}|(?

}

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 

Original answer to this stackoverflow question

This is a serious answer that fits your specifications. It is similar to @PlasmaPower's answer.

(-?\b(?:20(?:5[0-5]|[0-4][0-9])|1[0-9]{3}|[1-9][0-9]{0,2}|(?

Debuggex Demo



回答2:

Don't ever use it, but this works. :)

\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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


回答3:

So many answers, but noone reads (or cares) about the OPs side question in the comments?

As so many answers - correctly - pointed out that using regex is horrible for this scenario, lets think about other ways in OCaml! It is a while since I used OCaml, but with looking up a few constructs I was able to knock this together:

let isInRange i =      not(i  2055);;  let isIntAndInRange s =      try     let i = int_of_string s in     not(i  2055)     with Failure "int_of_string" -> false;;  let () = print_string "type a number: " in let s = read_line () in isIntAndInRange s 

If anything about is unclear, please read up on its syntax, type conversion functions and exception handling and input-output functions.

The user input part is only used to demonstrate. It might be more convenient to use the read_int function there. But the basic concept of handling the exception stays the same.



回答4:

Analyze the problem

If you "must" use a regex, break the problem down by analysing the accepted permutations.

"a range from -2055 to 2055" can be expressed as:

  • an optional -
  • optional leading zeros
  • followed by a number from 0 to 2055

"A number from 0 to 2055" can be one of a finite number of specific permutations:

  • one digit (0-9)
  • two digits (10-99)
  • three digits (100-999)
  • four digits starting with a 1 (1000-1999)
  • four digits starting with 20 (2000-204*9)
  • four digits starting with 205 (2050-2055*)

Note that for the purpose of this regex, it's not necessary to distinguish between the range "0-9" and "1-9", and only the last two ranges have any restrictions on the range of accepted digits/characters (indicated with a star).

Write component regex expressions

Each of the above component parts are easy to individually express as a regular 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]

Put the expressions together

The relevant expression for the whole match would be:

-?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]) 

Or slightly more concisely:

-?0*([0-9]{1,3}|1[0-9]{3}|20[0-4][0-9]|205[0-5]) 

Assuming the input contains only "the number" and nothing else, the final regex is therefore:

^-?0*([0-9]{1,3}|1[0-9]{3}|20[0-4][0-9]|205[0-5])$ 

If it's necessary to allow for a leading plus sign, this becomes:

^[-+]?0*([0-9]{1,3}|1[0-9]{3}|20[0-4][0-9]|205[0-5])$ 

Here's a js fiddle demonstrating what passes and what would fails the last regex.



回答5:

As an alternate approach to the great expression offered by aliteralmind, here is one that is much longer but interesting in order to see what another approach might look like (and what not to do).

It's an interesting exercise, because you can think of two distinct method: roughly, you can either:

  1. proceed by matching 4-char numbers, then 3-char numbers, etc.
  2. or proceed by matching the thousands digit, then the hundreds digit, etc.

Without trying, how would you know which is best? It turns out that the first approach (aliteralmind's answer) is far more economical.

Lower, I include a series of tests in the PHP language in case you or someone else would like to check the output.

Below, I will give you the regex in "free-spacing mode", which allows comments inside the regex so you can easily understand what it does. However, not all regex engines support free-spacing mode, so before we start with the interesting part, here is the regex as a one-liner.

Note that your question mentions numbers from -2055 to 2055. I assumed that you wanted to match "normal numbers", without leading zeroes. This means that the regex will match 999 but not 0999. If you would like leading zeroes, let me know, that is a very easy tweak.

Also, if you are matching in utf-8 mode, the \d should be replaced by [0-9]. This is the more common form.

The Regex as a One-Liner

^(?!-0$)-?(?:(?=\d{4}$)[12])?(?:(?=\d{3}$)(?:(?:(?

The Regex in Free-Spacing Mode

(?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   (?:(?

Series of Tests

These tests try to match against numbers from -100000 to 100000. They produce the following output:

Successful test: matches from -2055 to 2055 Successful test: NO matches from -100000 to -2056 Successful test: NO matches from 2056 to 100000 

Here is the code:

";     } } echo $success."
"; // Test 2 $success="Successful test: NO matches from -100000 to -2056"; for($i=-100000;$i"; } } echo $success."
"; // Test 3 $success="Successful test: NO matches from 2056 to 100000"; for($i=2056;$i"; } } echo $success."
"; ?>

Speed Tests

Here is the output of my simple speed test, matching from -1M to +1M. As Casimir pointed out, if aliteralmind's expression were anchored, instead of being slower it would be faster by 25%!

zx81: 3.796217918396 aliteralmind: 3.9922280311584 difference: 5.1632998151294 percent longer 

Here is the test code:

$regex="~(?x)^(?!-0$)-?(?:(?=\d{4}$)[12])?(?:(?=\d{3}$)(?:(?:(?"; echo "aliteralmind: ".$alit."
"; echo "difference: ".$diff." percent longer
";


回答6:

Why use Regex only to check a number?

int n = -2000;  if(n >= -2055 && n 


回答7:

Try this:

\-?\b0*(205[0-5]|20[0-4]\d|1?\d{3}|\d{1,2})\b 


回答8:

Check out this great tool which generates a regex for numeric ranges. You can also download the source code in Python:

http://utilitymill.com/utility/Regex_For_Range http://code.activestate.com/recipes/534137-generate-a-regular-expression-to-match-an-arbitrar/

For the range requested by OP it generates: -?([0-9]{1,3}|1[0-9]{3}|20[0-4][0-9]|205[0-5])



回答9:

Try with a very simple regex.

^([-0][0-1][0-9][0-9][0-9])$|^([-0]20[0-4][0-9])$|^([-0]205[0-5])$ 

Visual representation

It's very simple to understand.

  • group 1 [-0][0-1][0-9][0-9][0-9] will cover [-1999, 1999] values
  • group 2 [-0]20[0-4][0-9] will cover [-2000,-2049] and [2000,2049] values
  • group 3 [-0]205[0-5] will cover [-2050, -2055] and [2050, 2055] values

String.format("%05d", number) is doing very well done job here?

Sample code: (Read inline comments for more clarity.)

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.");     } } 

output:

-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. 


回答10:

Try this:

^-?0*(1?[0-9]{1,3}|20[0-4][0-9]|205[0-5])$ 

The regex before the brackets matches an optional - and any leading 0s.

The first part in the brackets (1?[0-9]{1,3}) matches 0-1999.

The second part in the brackets (20[0-4][0-9]) matches 2000-2049.

The third part in the brackets (205[0-5]) matches 2050-2055.



回答11:

Yet another -

 #  ^-?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}   )  $ 


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