I'm trying to write a RegExp to match only 8 digits, with one optional comma maybe hidden in-between the digits.
All of these should match:
12345678
12,45678
123456,8
Right now I have:
^[0-9,]{8}
but of course that erroneously matches 012,,,67
Example:
https://regex101.com/r/dX9aS9/1
I know optionals exist but don't understand how to keep the 8 digit length applying to the comma while also keeping the comma limited to 1.
Any tips would be appreciated, thanks!
To match 8 char string that can only contain digits and an optional comma in-between, you may use
^(?=.{8}$)\d+,?\d+$
See the regex demo
The lookahead will require the string to contain 8 chars. ,? will make matching a comma optional, and the + after \d will require at least 1 digit before and after an optional comma.
If you need to match a string that has 8 digits and an optional comma, you can use
^(?:(?=.{9}$)\d+,\d+|\d{8})$
See the regex demo
Actually, the string will have 9 characters in the string (if it has a comma), or just 8 - if there are only digits.
Explanation:
^ - start of string
(?:(?=.{9}$)\d+,\d+|\d{8}) - 2 alternatives:
(?=.{9}$)\d+,\d+ - 1+ digits followed with 1 comma followed with 1+ digits, and the whole string matched should be 9 char long (8 digits and 1 comma)
| - or
\d{8} - 8 digits
$ - end of string
See the Java code demo (note that with String#matches(), the ^ and $ anchors at the start and end of the pattern are redundant and can be omitted since the pattern is anchored by default when used with this method):
List<String> strs = Arrays.asList("0123,,678", "0123456", // bad
"01234,567", "01234567" // good
);
for (String str : strs)
System.out.println(str.matches("(?:(?=.{9}$)\\d+,\\d+|\\d{8})"));
NOTE FOR LEADING/TRAILING COMMAS:
You just need to replace + (match 1 or more occurrences) quantifiers to * (match 0 or more occurrences) in the first alternative branch to allow leading/trailing commas:
^(?:(?=.{9}$)\d*,\d*|\d{8})$
See this regex demo
You can use following regex if you want to let trailing comma:
^((\d,?){8})$
Demo
Otherwise use following one:
^((\d,?){8})(?<!,)$
Demo
(?<!,) is a negative-lookbehind.
/^(?!\d{0,6},\d{0,6},\d{0,6})(?=\d[\d,]{6}\d).{8}$/
I guess this cooperation of positive and negative look-ahead does just what's asked. If you remove the start and end delimiters and set the g flag then it will try to match the pattern along decimal strings longer than 8 characters as well.
Please try http://regexr.com/3d63m
Explanation: The negative look ahead (?!\d{0,6},\d{0,6},\d{0,6}) tries not to find any commas side by side if they have 6 or less decimal characters in between while the positive look ahead (?=\d[\d,]{6}\d) tries to find 6 decimal or comma characters in between two decimal characters. And the last .{8} selects 8 characters.
I am trying to get a regex that can get from 1-10 and i have been having trouble
I have tried :
/^[1-9]|10$/ //this will matches out 1-9 but not 10
/^10|[1-9]$/ //this will matches 10 but no 1-9 digits
I feel weird because i have seen this question before and people said either of these expression should work. Any ideas of another way to get any # from 1-10?
Assuming you want to match a string containing a number 1–10 and nothing else, you were close with /^[1-9]|10$/. The problem here is that the alternation | includes the ^ and $ characters, i.e. this expression will match either ^[1-9] (any string beginning with 1–9) or10$` (any string ending with 10). Parentheses solve this neatly:
/^([1-9]|10)$/
See it in action below:
const regex = /^([1-9]|10)$/gm; // `gm` flags for demo only; read below
const str = `1
2
3
4
foo
5
6
7
8
9
10`;
let m;
while (m = regex.exec(str)) {
console.log('Found match:', m[0]);
}
.as-console-wrapper{min-height:100%}
The snippet uses the g and m flags to find all of the matches across multiple lines. m makes ^ and $ match the beginning and end of each line rather than the beginning and end of the string. For your use case you probably don't want either flag.
Without knowing your specific use case, I don't see a reason to include ^ and & here, which requires that the number is surrounded by a start and end of line. Even if your inner regex was correct, it wouldn't match the '6' in: 'I have a 6 year old son.'
Instead, surround the inner regex with \b (a word boundary). So, to use Jordan's example, \b([1-9]|10)\b would match all numbers from 1-10. To use regex in javascript, you need to utilize one of several functions that accept it as an argument, or methods that can be called on it:
let regex = /\b([1-9]|10)\b/;
console.log('I have a 6 year old son.'.match(regex)[0]);
console.log('I have a six year old son.'.match(regex));
console.log('I have a 6 year old son.'.replace(regex, 'six'));
console.log(regex.test('I have a 6 year old son.'));
console.log(regex.test('I have a six year old son.'));
Hi I don't know much about regular expression. But I need it in form validation using angularJs.
Below is the requirement
The input box should accept only if either
(1) first 2 letters alpha + 6 numeric
or
(2) 8 numeric
Below are some correct Inputs :-
(1)SH123456
(2)12345678
(3)sd456565
I tried data-ng-pattern="/(^([a-zA-Z]){2}([0-9]){6})|([0-9]*)?$/" , Its working fine for both the above condition but still it is accepting strings like S2D3E4F5 and may be many other combination as well.
What I am doing wrong I am not able to find it out.
Any help is appreciable !!!
Thanks
In your regex, the two alternative branches are anchored separately:
(^([a-zA-Z]){2}([0-9]){6}) - 2 letters and 6 digits at the start of the string
| - or
([0-9]*)?$ - optional zero or more digits at the end of the string
You need to adjust the boundaries of the group:
data-ng-pattern="/^([a-zA-Z]{2}[0-9]{6}|[0-9]{8})?$/"
^ ^^^^
See the regex demo.
Now, the pattern will match:
^ - start of string
( - start of the grouping:
[a-zA-Z]{2}[0-9]{6} - 2 letters and 6 digits
| - or
[0-9]{8} - 8 digits
)? - end of the grouping and ? quantifier makes it match 1 or 0 times (optional)
$ - end of string.
You can try this DEMO LINK HERE
^(([a-zA-Z]{2}|[0-9]{2})[0-9]{6})?$
It will accept:
ab123456
12345678
aa441236
aw222222
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
Using regular expressions to validate a numeric range
To be clear: When a simple if statement will suffice
if(num < -2055 || 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:
(?<!-)\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.)
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:
(?<!-)\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
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}|(?<!-)0+))\b
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
(?<!-)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
(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;
/**
<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);
}
}
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}|(?<!-)0+))\b
Debuggex Demo
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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Check out this great tool which generates a regex for numeric ranges:
http://gamon.webfactional.com/regexnumericrangegenerator/
For the range requested by OP it generates: -?([0-9]{1,3}|1[0-9]{3}|20[0-4][0-9]|205[0-5])
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.
So many answers, but noone reads (or cares) about the OPs side question in the comments?
I'm writing an interpreter in OCaml .... how can i validate the input
number within the range without using regex ?? – Trung Nguyen Mar 2
at 17:30
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 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
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.
As of OCaml 4.02 this can be cleaned up a bit since exception handling can be done in a match.
let isIntAndInRange s =
match int_of_string s with
| i when i >= -2055 && i <= 2055 -> true
| _
| exception Failure _ -> false
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:
proceed by matching 4-char numbers, then 3-char numbers, etc.
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}$)(?:(?:(?<=^)|(?<=^-)|(?<=1)|(?<=-1))\d|(?:(?<=2)|(?<=-2))0))?(?:(?=\d{2}$)(?:(?:(?<=^)|(?<=^-)|(?<=^\d)|(?<=^-\d)|(?<=^\d{2})(?<!20)|(?<=^-\d{2})(?<!-20))\d|(?:(?<=20)|(?<=-20))[0-5]))?(?:(?=\d$)(?:(?:(?<=^)|(?<=^-)|(?<=^\d)|(?<=^-\d)|(?<=^\d{2})|(?<=^-\d{2})|(?<=^\d{3})(?<!205)|(?<=^-\d{3})(?<!-205))\d|(?:(?<=205)|(?<=-205))[0-5]))$
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}$) # assert that there are three digits left
(?: # non-capturing group
(?:(?<=^)|(?<=^-)|(?<=1)|(?<=-1))\d # if preceding chars are 1, -1 or head of string: value can be any digit
| # or
(?:(?<=2)|(?<=-2))0 # if preceding chars are 2 or -2: value must be 0
) # close non-capturing group
)? # end optional hundreds digits
(?: # OPTIONAL TENS DIGIT
(?=\d{2}$) # 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
$
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:
<?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 />";
?>
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}$)[12])?(?:(?=\d{3}$)(?:(?:(?<=^)|(?<=^-)|(?<=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: ".$diff." percent longer<br />";
Why use Regex only to check a number?
int n = -2000;
if(n >= -2055 && n <= 2055)
//Do something
else
//Do something else
Try this:
\-?\b0*(205[0-5]|20[0-4]\d|1?\d{3}|\d{1,2})\b
Try with a very simple regex.
^([-0][0-1][0-9][0-9][0-9])$|^([-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])$");
for (int number : numbers) {
String string = String.format("%05d", number);
Matcher m = p.matcher(string);
if (m.find()) {
System.out.println(number + " is in range.");
} else {
System.out.println(number + " is not in range.");
}
}
output:
-10002 is not in range.
-3000 is not in range.
-2056 is not in range.
-2055 is in range.
-2000 is in range.
-1999 is in range.
-20 is in range.
-1 is in range.
0 is in range.
1 is in range.
260 is in range.
1999 is in range.
2000 is in range.
2046 is in range.
2055 is in range.
2056 is not in range.
2955 is not in range.
3000 is not in range.
10002 is not in range.
123456 is not in range.
Try this:
^-?0*(1?[0-9]{1,3}|20[0-4][0-9]|205[0-5])$
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.
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}
)
$
I have a long string in javascript like
var string = 'abc234832748374asdf7943278934haskhjd';
I am trying to match like
abc234832748374 - that is - I have tried like
string.match(\abc[^abc]|\def[^def]|) but that doesnt get me both strings because I need numbers after them ?
Basically I need abc + 8 chars after and def the 8-11 chars after ? How can I do this ?
If you want the literal strings abc or def followed by 8-11 digits, you need something like:
(abc|def)[0-9]{8,11}
You can test it here: http://www.regular-expressions.info/javascriptexample.html
Be aware that, if you don't want to match more than 11 digits, you will require an anchor (or [^0-9]) at the end of the string. If it's just 8 or more, you can replace {8,11} with {8}.
To elaborate on an already posted answer, you need a global match, as follows:
var matches = string.match(/(abc|def)\d{8,11}/g);
This will match all subsets of the string which:
Start with "abc" or "def". This is the "(abc|def)" portion
Are then followed by 8-11 digits. This is the "\d{8,11}" portion. \d matches digits.
The "g" flag (global) gets you a list of all matches, rather than just the first one.
In your question, you asked for 8-11 characters rather than digits. If it doesn't matter whether they are digits or other characters, you can use "." instead of "\d".
I also notice that each of your example matches have more than 11 characters following the "abc" or "def". If any number of digits will do, then the following regex's may be better suited:
Any number of digits - var matches = string.match(/(abc|def)\d*/g);
At least one digit - var matches = string.match(/(abc|def)\d+/g);
At least 8 digits - var matches = string.match(/(abc|def)\d{8,}/g);
You can match abc[0-9]{8} for the string abc followed by 8 digits.
If the first three characters are arbitrary, and 8-11 digits after that, try [a-z]{3}[0-9]{8,11}
Use the below regex to get the exact match,
string.match(/(abc|def)\d{8,11}/g);
Ends with g
"g" for global
"i" for ignoreCase
"m" for multiline