RegEx for value Range from 1 - 365 - regex

What is the RegEx for value Range from 1- 365

Try this:
^(?:[1-9]\d?|[12]\d{2}|3[0-5]\d|36[0-5])$
The start anchor ^ and end anchor
$ are to match the whole input and
not just part of it.
(? ) is for grouping.
| is for alternation
[1-9]\d? matches 1 to 99
[12]\d{2} matches 100 to 299
3[0-5]\d matches 300 to 359
36[0-5] matches 360 to 365

You would have to list the possible combinations 1-9, 10-99, 100-299, 300-359, 360-365:
^([1-9]\d?|[12]\d\d|3[0-5]\d|36[0-5])$

Not really a good fit for regex, but if you insist:
^(?:36[0-5]|3[0-5][0-9]|[12][0-9][0-9]|[1-9][0-9]|[1-9])$
This is not allowing leading zeroes. If you wish to allow those, let me know.
The expression above can be shortened a little to
^(?:36[0-5]|3[0-5]\d|[12]\d{2}|[1-9]\d?)$
but I find the first solution to be a bit more readable. YMMV.

A general solution for matching the numbers from 1 to XYZ
^(?!0)(?!\d{4}$)(?![X+1-9]\d{2}$)(?!X[Y+1-9]\d$)(?!XY[Z+1-9]$)\d+$
Notes:
If any of X, Y or Z are 9 that will make X+1 etc. be 10. If that happens the regex part that would require using the 10 should be left out.
This can be extended to numbers with more or less digits following the same principles.
It does not allow left-padding 0es.
Applied to your case:
^(?!0)(?!\d{4}$)(?![4-9]\d{2}$)(?!3[7-9]\d$)(?!36[6-9]$)\d+$
Lets explain:
(?!0\d*) - does not start with 0
(?!\d{4}$) - does not have 4 digits, i.e. between 1000 and infinity
(?![4-9]\d{2}$) - it's not between 400 and 999
(?!3[7-9]\d$) - it's not between 370 and 399
(?!36[6-9]$) - it's not between 366 and 369
Test it.

^36[0-5]|(3[0-5]|[12]?[0-9])[0-9]$

^3(6[0-5]|[0-5]\d)|[12]\d\d|[1-9]\d|[1-9]$
Or if numbers like 05 can not be in input:
^3(6[0-5]|[0-5]\d)|[12]?\d?\d$
P.S.: Anyway no need of regex here. Use ToInt(), <=, >=

It really depends on your regex engine since they may not all be PCRE-style. I usually work to the lowest common denominator unless I know it will be targeting a minimum engine.
To that end, I'd just use something like:
^[1-9]|[1-9][0-9]|[1-2][0-9]{2}|3[0-5][0-9]|36[0-5]$
This will take care of (in order):
1-9.
10-99.
100-299.
300-359.
360-365.
However, unless you're absolutely required to use just a regex, I wouldn't. It's like trying to kill a fly with a thermo-nuclear warhead.
Just use the much simpler ^[0-9]{1,3}$ then use whatever language features you have to convert it to an integer and check it's between 1 and 365 inclusive:
def isValidDayOtherThanLeapYear (s):
if not s.matches ("^[0-9]{1,3}$"):
return false
n = s.toInteger()
if n < 1 or n > 365:
return false
return true
Your code will be more readable that way and I tend to rethink the use of regular expressions the second they start looking like they may be hard to read six months down the track.

This worked for me...
^[1-3][0-6]?[0-5]?$

Related

Optimization of Regular Expression to match numbers bigger or equal to 50

I want to check if a number is 50 or more using a regular expression. This in itself is no problem but the number field has another regex checking the format of the entered number.
The number will be in the continental format: 123.456,78 (a dot between groups of three digits and always a comma with 2 digits at the end)
Examples:
100.000,00
50.000,00
50,00
34,34
etc.
I want to capture numbers which are 50 or more. So from the four examples above the first three should be matched.
I've come up with this rather complicated one and am wondering if there is an easier way to do this.
^(\d{1,3}[.]|[5-9][0-9]|\d{3}|[.]\d{1,3})*[,]\d{2}$
EDIT
I want to match continental numbers here. The numbers have this format due to internal regulations and specify a price.
Example: 1000 EUR would be written as 1.000,00 EUR
50000 as 50.000,00 and so on.
It's a matter of taste, obviously, but using a negative lookahead gives a simple solution.
^(?!([1-4]?\d),)[1-9](\d{1,2})?(\.\d{3})*,\d{2}\b
In words: starting from a boundary ignore all numbers that start with 1 digit OR 2 digits (the first being a 1,2,3 or 4), followed by a comma.
Check on regex101.com
Try:
EDIT ^(.{3,}|[5-9]\d),\d{2}$
It checks if:
there 3 chars or more before the ,
there are 2 numbers before the , and the first is between 5 and 9
and then a , and 2 numbers
Donno if it answer your question as it'll return true for:
aa50,00
1sdf,54
But this assumes that your original string is a number in the format you expect (as it was not a requirement in your question).
EDIT 3
The regex below tests if the number is valid referring to the continental format and if it's equal or greater than 50. See tests here.
Regex: ^((([1-9]\d{0,2}\.)(\d{3}\.){0,}\d{3})|([1-9]\d{2})|([5-9]\d)),\d{2}$
Explanation (d is a number):
([1-9]\d{0,2}\.): either d., dd. or ddd. one time with the first d between 1 and 9.
(\d{3}\.){0,}: ddd. zero or x time
\d{3}: ddd 3 digit
These 3 parts combined match any numbers equals or greater than 1000 like: 1.000, 22.002 or 100.000.000.
([1-9]\d{2}): any number between 100 and 999.
([5-9]\d)): a number between 5 and 9 followed by a number. Matches anything between 50 and 99.
So it's either the one of the parts above or this one.
Then ,\d{2}$ matches the comma and the two last digits.
I have named all inner groups, for better understanding what part of number is matched by each group. After you understand how it works, change all ?P<..> to ?:.
This one is for any dec number in the continental format.
^(?P<common_int>(?P<int>(?P<int_start>[1-9]\d{1,2}|[1-9]\d|[1-9])(?P<int_end>\.\d{3})*|0)(?!,)|(?P<dec_int_having_frac>(?P<dec_int>(?P<dec_int_start>[1-9]\d{1,2}|[1-9]\d|[1-9])(?P<dec_int_end>\.\d{3})*,)|0,|,)(?=\d))(?P<frac_from_comma>(?<=,)(?P<frac>(?P<frac_start>\d{3}\.)*(?P<frac_end>\d{1,3})))?$
test
This one is for the same with the limit number>=50
^(?P<common_int>(?P<int>(?P<int_start>[1-9]\d{1,2}|[1-9]\d|[1-9])(?P<int_end>\.\d{3})+|(?P<int_short>[1-9]\d{2}|[5-9]\d))(?!,)|(?P<dec_int_having_frac>(?P<dec_int>(?P<dec_int_start>[1-9]\d{1,2}|[1-9]\d|[1-9])(?P<dec_int_end>\.\d{3})+,)|(?P<dec_short_int>[1-9]\d{2}|[5-9]\d),)(?=\d))(?P<frac_from_comma>(?<=,)(?P<frac>(?P<frac_start>\d{3}\.)*(?P<frac_end>\d{1,3})))?$
tests
If you always have the integer part under 999.999 and fractal part always 2 digits, it will be a bit more simple:
^(?P<dec_int_having_frac>(?P<dec_int>(?P<dec_int_start>[1-9]\d{1,2}|[1-9]\d|[1-9])(?P<dec_int_end>\.\d{3})?,)|(?P<dec_short_int>[1-9]\d{2}|[5-9]\d),)(?=\d)(?P<frac_from_comma>(?<=,)(?P<frac>(?P<frac_end>\d{1,2})))?$
test
If you can guarantee that the number is correctly formed -- that is, that the regex isn't expected to detect that 5,0.1 is invalid, then there are a limited number of passing cases:
ends with \d{3}
ends with [5-9]\d
contains \d{3},
contains [5-9]\d,
It's not actually necessary to do anything with \.
The easiest regex is to code for each of these individually:
(\d{3}$|[5-9]\d$|\d{3},|[5-9]\d)
You could make it more compact and efficient by merging some of the cases:
(\d{3}[$,]|[5-9]\d[$,])
If you need to also validate the format, you will need extra complexity. I would advise against attempting to do both in a single regex.
However unless you have a very good reason for having to do this with a regex, I recommend against it. Parse the string into an integer, and compare it with 50.

regex find match within the first n items

I have a string of 8 separated hexadecimal numbers, such as:
3E%12%3%1F%3E%6%1%19
And I need to check if the number 12 is located within the first 4 set of numbers.
I'm guessing this shouldn't be all that complex, but my searches turned up empty. Regular expressions are always a trouble for me, but I don't have access to anything else in this scenario. Any help would be appreciated.
^([^%]+%){0,3}12%
See it in action
The idea is:
^ - from the start
[^%]+% - match multiple non % characters, followed by a % character
{0,3} - between 0 and 3 of those
12% - 12% after that
Here you go
^([^%]*%){4}(?<=.*12.*)
This will match both the following if that is what is intended
1%312%..
1%123%..
Check the solution if %123% is matched or not
If the number 12 should stand on its own then use
^([^%]*%){4}(?<=.*\b12\b.*)

*NIX REGEXP number series

Am playing around with regexp's but this is my headache. I have a dynamic number which needs a suffix. The suffix is always 0 to 9, 99 or 999.
Example:
I have the number 461200 and now I want to create an regexp that will match 461200 to 461209. What I've learned it should be ^46120[0-9]$? Is this correct or somewhere to the left of hell?
Ok, let us assume it is correct and I now want to match 461200 - 461299? This is where I get lost.
^4612[0-9]{2}?
It cannot be. I am yet to figure this out.
Any help appreciated.
For 1 digit at the end you need:
^4612[0-9]$
2 digits at the end:
^4612[0-9]{2}$
3 digits at the end:
^4612[0-9]{3}$
The number in braces {} means the number of time the preceding character or set has to be repeated.
Ok, let us assume it is correct and I now want to match 461200 -
461299?
You can either repeat the desired character class by saying [0-9][0-9] or use quantifiers [0-9]{2}.
It can be either:
^4612[0-9][0-9]$
or
^4612[0-9]{2}$
Both would work.
maybe try this regex:
^4612\d{2}$

Regular Expression (RegEx) For Hours with Increments

I need to only accept input that meets these rules...
0.25-24
Increments of .25 (.00, .25, .50, .75)
First digit doesn't have to be required.
Would like trailing zeros to be optional.
Examples of some valid entries:
0.25
.50
.5
1
1.0
5.50
23.75
24 (max allowed)
UPDATE: nothing at all, null/blank, should also be accepted as valid
Example of some invalid entries:
0
.0
.00
0.0
0.00
24.25
-1
I understand that RegEx is a pattern matching language therefore it's not great for ranges, less-than, and great-than checking. So to check if it's less than or equal to 24 means I'd have to find a pattern, right? So there are 24 possible patters which would make this a long RegEx, am I understanding this correctly? I could use ColdFusion to do the check to make sure it's in the 0-24 range. It's not the end of the world if I have use ColdFusion for this part, but it'd be nice to get it all into the RegEx if it doesn't cause it to be too long. This is what I have so far:
^\d{0,2}((\.(0|00|25|5|50|75))?)$
http://regex101.com/r/iS7zM3
This handles pretty much all of it except for the 0-24 range check or the check for just a zero. I'll keep plugging away at it but any help would be appreciated. Thanks!
Change \d{0,2} to (?:1[0-9]?|2[0-4]?|[3-9])? and it'll match from 1 to 24 (or nothing).
You can also simplify the second part to (?:\.(?:00?|25|50?|75))? - you could go further to (?:\.(?:[05]0?|[27]5))? but that might obfuscate the intent a bit too far.
To exclude 24.25 you could perhaps use a negative lookahead (?!24\.[^0]) to prevent anything other than 24.0 or 24.00, but it's probably simpler to just exclude 24 from the main pattern and include a specific check for 24/24.0/24.00 at the start:
(?x)
# checks for 24
^24$|^24\.00?$
|
# integer part
^
(?:1[0-9]?|2[0-3]?|[3-9]|0(?=\.[^0])|(?=\.[^0]))
# decimal part
(?:\.(?:00?|25|50?|75))?
$
That also includes a check for 0(?=\.[^0]) which uses a positive lookahead to only allow an initial 0 if the next char is a . followed by a non-zero (so 0.0 and 0.00 isn't allowed).
The (?x) flag allows whitespace to be ignored, allowing readable regex in your code - obviously preferable to squashing it all onto a single line - and also enables the use of # to start line comments to explain parts of a pattern. (Literal whitespaces and hashes can be escaped with backslash, or encoded via e.g. \x23 for hash.)
For comparison, here's a pure-CFML way of doing it:
IsNumeric(Num)
AND Num GT 0
AND Num LTE 24
AND NOT find('.',Num*4)
Now, are you really sure it's better as a regex...
You could try this regex (broken down):
^
(?:
(?:[1-9]|1\d|2[0-3])(?:\.(?:[05]0?|[27]5))? # Non-zeros with optional decimal
|
0?(?:\.(?:50?|[27]5)) # Decimals under 1
|
24(?:\.00?)? # The maximum
)
$
In one line:
^(?:(?:[1-9]|1\d|2[0-3])(?:\.(?:[05]0?|[27]5))?|0?(?:\.(?:50?|[27]5))|24(?:\.00?)?)$
regex101 demo
^([0-1]?[0-9]|2[0-4])((\.(0|00|25|5|50|75))?)$
This means the one's place can be 0-9 if the tens place is missing, a 0, or 1.
If the tens place is a 2, then the ones place can be 0-4.
The second part is great, it's simple and readable too. It has an extra set of parens though that can be removed, reducing it to this:
^([0-1]?[0-9]|2[0-4])(\.(0|00|25|5|50|75))?$

Regex - Find numbers between 2000 and 3000

I have a need to search all numbers with 4 digits between 2000 and 3000.
It can be that letters are before and after.
I thought I can use [2000-3000]{4}, but doesnt work, why?
thank you.
How about
^2\d{3}|3000$
Or as Amarghosh & Bart K. & jleedev pointed out, to match multiple instances
\b(?:2[0-9]{3}|3000)\b
If you need to match a3000 or 3000a but not 13000, you would need lookahead and lookbefore like
(?<![0-9])(?:2[0-9]{3}|3000)(?![0-9])
Regular expressions are rarely suitable for checking ranges since for ranges like 27 through 9076 inclusive, they become incredibly ugly. It can be done but you're really better off just doing a regex to check for numerics, something like:
^[0-9]+$
which should work on just about every regex engine, and then check the range manually.
In toto:
def isBetween2kAnd3k(s):
if not s.match ("^[0-9]+$"):
return false
i = s.toInt()
if i < 2000 or i > 3000:
return false
return true
What your particular regex [2000-3000]{4} is checking for is exactly four occurrences of any of the following character: 2,0,0,0-3,0,0,0 - in other words, exactly four digits drawn from 0-3.
With letters before an after, you will need to modify the regex and check the correct substring, something like:
def isBetween2kAnd3kWithLetters(s):
if not s.match ("^[A-Za-z]*[0-9]{4}[A-Za-z]*$"):
return false
idx = s.locate ("[0-9]")
i = s.substring(idx,4).toInt()
if i < 2000 or i > 3000:
return false
return true
As an aside, a regex for checking the range 27 through 9076 inclusive would be something like this hideous monstrosity:
^2[7-9]|[3-9][9-9]|[1-9][0-9]{2}|[1-8][0-9]{3}|90[0-6][0-9]|907[0-6]$
I think that's substantially less readable than using ^[1-9][0-9]+$ then checking if it's between 27 and 9076 with an if statement?
Hum tricky one. The dash - only applies to the character immediately before and after so what your regex is actually matching is exactly 4 characters between 0 and 3 inclusive (ie, 0, 1, 2 and 3). eg, 3210, 1230, 3333, etc... Try the expression below.
(2[0-9]{3})|(3000)
Here's explanation why and ways to detect ranges: http://www.regular-expressions.info/numericranges.html
Correct regex will be \b(2\d{3}|3000)\b. That means: match character '2' then exactly three digits (this will match any from 2000 to 2999) or just match '3000'. There are some good tutorials on regular expressions:
http://gnosis.cx/publish/programming/regular_expressions.html
http://immike.net/blog/2007/04/06/the-absolute-bare-minimum-every-programmer-should-know-about-regular-expressions/
http://www.regular-expressions.info/
why don't you check for greater or less than? its simpler than a regex
num >= 2000 and num <=3000