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I'm trying to create a regex pattern that accepts decimal numbers and the maximum length should be 3.
These are the regex I tried but didn't work
new RegExp('d{1-3}')
new RegExp('^[0-9]{3}$')
I want to achive to allow the decimal numbers between 0-999.
For example 185.5
Thanks in advance.
you can try this: ^[1-9]{0,3}$
it only accept numbers with max length of 3
example: 123
and if you want it with complex numbers use this:
^[1-9]{0,3}([,.][0-9]{0,3})?$
example: 154.234
here is one way to do it, which matches any number b/w 0 and 999 including decimal values (unlimited)
^(\d{,3}(\.)?(?(2)\d*|$))$
# should start and ends with a digit
# \d{,3} : match at most 3 digit
# (\.)? : optional period
# ?(?(2) : if period exist match the following pattern
# \d* : match any number of digits after period
# $ : else it should be end of pattern
https://regex101.com/r/B8jRG6/1
regExp ^\d+((\.)|(\.\d{1,3})?)$
I have a requirement where I need a regex which
should not repeat alphabet
should only contain alphabet and comma
should not start or end with comma
can contain more than 2 alphabets
example :-
A,B --- correct
A,B,C,D,E,F --- correct
D,D,A --- wrong
,B,C --- wrong
B,C, --- wrong
A,,B,C --- wrong
Can anyone help ?
Another idea with capturing and checking by use of a lookahead:
^(?:([A-Z])(?!.*?\1),?\b)+$
You can test here at regex101 if it meets your requirements.
If you don't want to match single characters, e.g. A, change the + quantifier to {2,}.
The statement of the question is incomplete in several respects. I have made the following assumptions:
Considering that D,D,A is incorrect I assume that a letter cannot be followed by a comma followed by the same letter.
The string may contain the same letter more than once as long as #1 is satisfied.
Considering that A,,B,C is incorrect I assume a comma cannot follow a comma.
Since the examples contain only capital letters I will assume that lower-case letters are not permitted (though one need only set the case-indifferent flag (i) to permit either case).
We observe that the requirements are satisfied if and only if the string begins with a capital letter and is followed by a sequence of comma-capital letter pairs, provided that no capital letter is followed by a comma followed by the same letter. We therefore can attempt to match the following regular expression.
^(?:([A-Z]),(?!\1))*[A-Z]$
Demo
The elements of the expression are as follows.
^ # match beginning of string
(?: # begin a non-capture group
([A-Z]) # match a capital letter and save to capture group 1
, # match a comma
(?!\1) # use negative lookahead to assert next character is not equal
# to the content of capture group 1
)* # end non-capture group and execute it zero or more times
[A-Z] # match a capital letter
$ # match end of string
Here is a big ugly regex solution:
var inputs = ['A,B', 'D,D,D', ',B,C', 'B,C,', 'A,,B'];
for (var i=0; i < inputs.length; ++i) {
if (/^(?!.*?([^,]+).*,\1(?:,|$))[^,]+(?:,[^,]+)*$/.test(inputs[i])) {
console.log(inputs[i] + " => VALID");
}
else {
console.log(inputs[i] + " => INVALID");
}
}
The regex has two parts to it. It uses a negative lookahead to assert that no two CSV entries ever repeat in the input. Then, it uses a straightforward pattern to match any proper CSV delimited input. Here is an explanation:
^ from the start of the input
(?!.*?([^,]+).*,\1(?:,|$)) assert that no CSV element ever repeats
[^,]+ then match a CSV element
(?:,[^,]+)* followed by comma and another element, 0 or more times
$ end of the input
This one could suit your needs:
^(?!,)(?!.*,,)(?!.*(\b[A-Z]+\b).*\1)[A-Z,]+(?<!,)$
^: the start of the string
(?!,): should not be directly followed by a comma
(?!.*,,): should not be followed by two commas
(?!.*(\b[A-Z]+\b).*\1): should not be followed by a value found twice
[A-Z,]+: should contain letters and commas only
$: the end of the string
(?<!,): should not be directly preceded by a comma
See https://regex101.com/r/1kGVSB/1
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}
)
$
var string = 'Our Prices are $355.00 and $550, down form $999.00';
How can I get those 3 prices into an array?
The RegEx
string.match(/\$((?:\d|\,)*\.?\d+)/g) || []
That || [] is for no matches: it gives an empty array rather than null.
Matches
$99
$.99
$9.99
$9,999
$9,999.99
Explanation
/ # Start RegEx
\$ # $ (dollar sign)
( # Capturing group (this is what you’re looking for)
(?: # Non-capturing group (these numbers or commas aren’t the only thing you’re looking for)
\d # Number
| # OR
\, # , (comma)
)* # Repeat any number of times, as many times as possible
\.? # . (dot), repeated at most once, as many times as possible
\d+ # Number, repeated at least once, as many times as possible
)
/ # End RegEx
g # Match all occurances (global)
To match numbers like .99 more easily I made the second number mandatory (\d+) while making the first number (along with commas) optional (\d*). This means, technically, a string like $999 is matched with the second number (after the optional decimal point) which doesn’t matter for the result — it’s just a technicality.
A non-regex approach: split the string and filter the contents:
var arr = string.split(' ').filter(function(val) {return val.startsWith('$');});
Use match with regex as follow:
string.match(/\$\d+(\.\d+)?/g)
Regex Explanation
/ : Delimiters of regex
\$: Matches $ literal
\d+: Matches one or more digits
()?: Matches zero or more of the preceding elements
\.: Matches .
g : Matches all the possible matching characters
Demo
This will check if there is a possible decimal digits following a '$'
i am using ^(^[a-z]?[a0-z9]+[.|-][a0-z9]+[a-z]+[.|-][a0-z9]+[a-z]?$)+$ , this force user type 2 hyphen(-) in his expression ....
But i need that regular expression can allow maximum 2 hypen and minimum one hyphen in his expression....
This is a simple re implementation.
/^[^-]+(-[^-]+){1,2}$/
And the test result:
"123".match(/^[^-]+(-[^-]+){1,2}$/)
null
"123-123".match(/^[^-]+(-[^-]+){1,2}$/)
["123-123", "-123"]
"123-123-123".match(/^[^-]+(-[^-]+){1,2}$/)
["123-123-123", "-123"]
"123-123-123-123".match(/^[^-]+(-[^-]+){1,2}$/)
null
And if you just copy one from other, then you should try to write one by yourself.
If not, you should delete it and restart.
^[^-]*-+[^-]*-{0,1}[^-]*$
String1: abc-pqr,
String2: aaa-bbb,
String3 aaa-bbb-ccc-ddd,
where String 1 and 2 are valid and 3 is invalid?
This allows String 1 and 2. But not 3.
This regex should work...
/^[^-]*-[^-]*-[^-]*$/
It works by capturing anything except a hyphen (zero or more times), then a hyphen, then anything else again, then another hyphen, then anything else. It has a start and end marker to ensure that the whole string does not contain any more hyphens.
Breaks down like this...
/ # begin regex
^ # start of string
[^-]* # not hyphen ([^-]), zero or more times (*)
- # hyphen
[^-]* # not hyphen ([^-]), zero or more times (*)
- # hyphen
[^-]* # not hyphen ([^-]), zero or more times (*)
$ # end of string
/ # end regex