I'm not a developer or scripter. I can't contribute much to this forum but I do use it to get guidance for my staff of developers. That's my disclaimer because the last time I was on this site, someone reamed me out for asking questions and not contributing. For this, I do apologize.
If anyone is willing to assist, or at least give me a kick-start, how would I find the version of a file if the version has #.#.###; i.e., 6.1.3890?
So, my goal is to find a number that is equal to or greater than 1 and equal to or greater than 389. I am only concerned with the digits after the first l'.' and the second '.'
Thanks to any and all.
A regex for a number greater than 389:
(39[0-9]|[4-9][0-9][0-9]|[1-9][0-9][0-9][0-9]+)
A regex for a number greater than 1:
([2-9]|[1-9][0-9]+)
A combined regex for version above 6.1.389:
(6\.1\.(39[0-9]|[4-9][0-9][0-9]|[1-9][0-9][0-9][0-9]+)|6\.([2-9]|[1-9][0-9]+)\.[0-9]+|([7-9]|[1-9][0-9]+)\.[0-9]+\.[0-9]+)
Non zero numbers should not start with a 0.
If the version number format is limited to #.#.### or possibly fewer digits for the last part, the regex can be simplified to:
(6\.1\.(39[0-9]|[4-9][0-9][0-9])|6\.[2-9]\.[0-9]+|[7-9]\.[0-9]\.[0-9]+)
Related
I'm attempting to create a Regex that finds only 2-digit integers or numbers with a precision of 2 decimal points.
In the example string at the bottom, I want to find only the following:
21 and 10.50
Using this expression, 100% is getting captured, in addition to the strings I desire to capture:
(\d){1,2}(\.?)([0-9]?[0-9]?){1,2}
I know I need to use ^% somewhere, but I can't figure out where it goes. Any suggestions are greatly appreciated.
Here's my sample string:
Earn Up to $21 Per Hour - Deliver Food with !!
Delivery Drivers work when they want and make great money when they do.
All orders are prepaid, just pick them up and deliver them to hungry diners. No waiting in line or fumbling with receipts and prepaid cards.
It's fast and easy to start working. Get started today.
Apply Now
Why choose ?
More orders than any other takeout platform
100% of our restaurants are official partners
Competitive pay: Per order fee + mileage + tips
We guarantee an hourly minimum of $10.50/hour*
Create your own schedule & work the hours you want
Word boundaries in your regular expression will grant you a bit more control.
Since word boundaries are a bit strict, we need to introduce an OR condition to address both cases which will satisfy your regex.
(\b[\d]{2}\.[\d]{2}\b)|(\b[\d]{2}\b)
Edit: Try this one,
\b[\d]{2}\b(\.[\d]{2})?
The first example has a chance to fail as it is order dependent due to the way it short-circuits. This I believe should address multiple cases properly.
I think this should work:
(?<!\d)((\d+\.\d\d)|(\d\d))(?!%|\d)
Demo (and explanation)
EDIT:
Improved version:
(?<!\d)(\d{1,2}(?:\.\d{1,2})?)(?!%|\d)
Demo (and explanation)
You can try this variant: (\d{1,}|[\d.])\b(?!%)
It uses negative lookahead (?!%) to exclude digits following by % sign.
Details at regex101
I'm in need of a regex, which takes a minimum and a maximum number to determine valid input, And I want the maximum and minimum to be dynamic.
I have been trying to get this done using this link
https://stackoverflow.com/a/13473595/1866676
But couldn't get it to work. Can someone please let me know how to do this.
Let's say I want to make a html5 input box, and I Want it to only receive numbers from 100 to 1999
What would a regex for this like this look like?
First off, while it is possible to do this, I think if there is a simpler way to choose a number range such as <input type="number" min="1" max="100">, that way would be preferred.
Having said that, here's how the kind of regex you requested works:
ones: ^[0-9]$ // just set the numbers -- matches 0 to 9
tens: ^[1-3]?[0-9]$ //set max tens and max ones -- matches 0 to 39
tens where max does not end in 9 ^[1-2]?[0-9]$|^[3][0-4]$ // 0 to 34
only tens: ^[1][5-9]$|^[2-3][0-9]$|^[4][0-5]$ // 15 to 45
Here, lets pick an arbitrary number 1234 to 2345
^[1][2][3][4-9]$|
^[1][2][4-9][0-9]$|
^[1][3-9][0-9][0-9]$|
^[2][0-2][0-9][0-9]$|
^[2][3][0-3][0-9]$|
^[2][3][4][0-5]$
https://regex101.com/r/pP8rQ7/4
Basically the ending of the middle series always needs to be a straight range that can reach 9 unless we are dealing with the ones place, and if it cant, you have to build it upwards toward the middle each time we have a value that can't start in 0 and then once we reach a value that cant end in 9 break early and set it in the next condition.
Notice the pattern, as each place solidifies. Also keep in mind that when dealing with going from lower to higher places, optional operators ? should be used.
Its a bit complex, but its nowhere near impossible to design a custom range with a bit of thought.
If you are more specific, we can craft an exact example, but this is generally how it is done:beginning-range|middle-range|end-range
You should only need beginning or end-ranges in certain cases like if the min or max does not end in 9. the ? means that the range that comes after it is optional. (so for example in the first case it lets us have both single and double numbers.
so for 100 - 1999 it's quite simple actually because you have lots of 9's and 0's
/^[1-9][0-9][0-9]$|^[1][0-9][0-9][0-9]$/
https://regex101.com/r/pP8rQ7/1
Note: Single values don't need ranges [n] I just added them for readability.
Edit: There used to be a regex range generator at: http://gamon.webfactional.com/regexnumericrangegenerator/. It appears to be offline now.
Essentially, you can't.
For every numeric range, there exists a regex that will match numbers in that range, therefore it is possible to write code that can generate a such regex. But such a regex is not a simple reformatting of the range ends.
However, such code would require colossal effort and complexity to write compared to code that simply checked the number using numeric methods.
With HTML 5 simply put a range input...
<form>
Quantity (between 100 and 1999):
<input type="number" name="quantity" min="100" max="1999">
</form>
with regex:
^([12345679])(\d)(\d)|^(1)(\d)(\d)(\d)
So if you need to create the regex dinamically it's possible but a bit tricky and complex
I am using a regular expression to match all UK bank card number formats; I have done research and managed to find/amend a regex that covers the majority of formats. However, I have a bit of an edge case where one is not matching and I do not know why, or how to resolve. This is what I am using:
(\b[4|5|6](\d){3}[\s|-]?((\d){4}[\s|-]?){2}(\d){4}\b)|(\b(\d){4}[\s|-]?(\d){6}[\s|-]?(\d){5}\b)
This is an example card number that does not work: 6759000000005
This is an example card number that does work: 675900000000555
Apologies if this is an easy question, I am fairly new to regular expression syntax. Any help to resolve would be greatly appreciated. Thanks.
See here the demo
The regex is (\b[4|5|6]\d{3}[\s-]?(\d{4}[\s-]?){2}\d{1,4}\b)|(\b\d{4}[\s-]?\d{6}[\s-]?\d{5}\b)
I'm not an expert of UK cards, so I can't tell what is the expected format as you did not gave exemples with spaces or hyphens in them...
If you can refine the requirements it's handlable.
A more generic card number validation (without separators, so you'll need to strip them before) would be
\d{6}\d{1,12}\d
As per the requirements of the norm (found nothing on the minimum length of the account identifier):
An ISO/IEC 7812 card number is most commonly 16 digits in length,[1]
and can be up to 19 digits. The structure is as follows:
a six-digit Issuer Identification Number (IIN) (previously called the
"Bank Identification Number" (BIN)) the first digit of which is the
Major Industry Identifier (MII), a variable length (up to 12 digits)
individual account identifier, a single check digit calculated using
the Luhn algorithm.[2]
I'm pretty sure this hasn't actually been answered yet on this site. For once and for all, what is the smallest regex that matches a numeric string that is in the range of a 32-bit signed integer, in the range -2147483648 to 2147483647.
I must use regex for validation - that is the only option available to me.
I have tried
\d{1,10}
but I can't figure out how to restrict it to the valid number range.
To aid developing in regex, it should match:
-2147483648
-2099999999
-999999999
-1
0
1
999999999
2099999999
2147483647
It should not match:
-2147483649
-2200000000
-11111111111
2147483648
2200000000
11111111111
I have set up an on-line live demo (on rubular) that has my attempt and the test cases above.
Note: The shortest regex that works will be accepted. Efficiency of regex will not be considered (unless there's a tie for shortest length).
I really hope it is just puzzler and no one will use regex for this problem in real world. Proper solution would be converting number from string to numeric type like BigInteger. This should allow us to check its range using proper methods or operators, like compareTo, >, <.
To make life easier you can use this page (dead link) to generate regex for ranges. So regex for range 0 - 2147483647 can look like
\b([0-9]{1,9}|1[0-9]{9}|2(0[0-9]{8}|1([0-3][0-9]{7}|4([0-6][0-9]{6}|7([0-3][0-9]{5}|4([0-7][0-9]{4}|8([0-2][0-9]{3}|3([0-5][0-9]{2}|6([0-3][0-9]|4[0-7])))))))))\b
(friendlier way)
\b(
[0-9]{1,9}|
1[0-9]{9}|
2(0[0-9]{8}|
1([0-3][0-9]{7}|
4([0-6][0-9]{6}|
7([0-3][0-9]{5}|
4([0-7][0-9]{4}|
8([0-2][0-9]{3}|
3([0-5][0-9]{2}|
6([0-3][0-9]|
4[0-7]
)))))))))\b
and range 0 - 2147483648
\b([0-9]{1,9}|1[0-9]{9}|2(0[0-9]{8}|1([0-3][0-9]{7}|4([0-6][0-9]{6}|7([0-3][0-9]{5}|4([0-7][0-9]{4}|8([0-2][0-9]{3}|3([0-5][0-9]{2}|6([0-3][0-9]|4[0-8])))))))))\b
So we can just combine these ranges and write it as
range of 0-2147483647 OR "-" range of 0-2147483648
which will give us
\b([0-9]{1,9}|1[0-9]{9}|2(0[0-9]{8}|1([0-3][0-9]{7}|4([0-6][0-9]{6}|7([0-3][0-9]{5}|4([0-7][0-9]{4}|8([0-2][0-9]{3}|3([0-5][0-9]{2}|6([0-3][0-9]|4[0-7])))))))))\b|-\b([0-9]{1,9}|1[0-9]{9}|2(0[0-9]{8}|1([0-3][0-9]{7}|4([0-6][0-9]{6}|7([0-3][0-9]{5}|4([0-7][0-9]{4}|8([0-2][0-9]{3}|3([0-5][0-9]{2}|6([0-3][0-9]|4[0-8])))))))))\b.
[edit]
Since Bohemian noticed in his comment final regex can be in form -?regex1|-2147483648 so here is little shorter version (also changed [0-9] to \d)
^-?(\d{1,9}|1\d{9}|2(0\d{8}|1([0-3]\d{7}|4([0-6]\d{6}|7([0-3]\d{5}|4([0-7]\d{4}|8([0-2]\d{3}|3([0-5]\d{2}|6([0-3]\d|4[0-7])))))))))$|^-2147483648$
If you will use it in Java String#matches(regex) method on each line you can also skip ^ and $ parts since they will be added automatically to make sure entire string matches regex.
I know this regex is very ugly, but just shows why regex is not good tool for range validation.
Edit:
This is the shortest regex you can get and the best way to do it:
We check every digit starting from the left, if it reaches it's limit and all the previous did, we put control on the next one.
for the range (-2147483647 to 2147483647) it could be a - signe or not. for -2147483648 it must be a - signe.
So finaly we get this:
^-?([0-9]{1,9}|[0-1][0-9]{9}|20[0-9]{8}|21[0-3][0-9]{7}|214[0-6][0-9]{6}|2147[0-3][0-9]{5}|21474[0-7][0-9]{4}|214748[0-2][0-9]{3}|2147483[0-5][0-9]{2}|21474836[0-3][0-9]|214748364[0-7])$|^(-2147483648)$
And this is a Live Demo
^(429496729[0-6]|42949672[0-8]\d|4294967[01]\d{2}|429496[0-6]\d{3}|42949[0-5]\d{4}|4294[0-8]\d{5}|429[0-3]\d{6}|42[0-8]\d{7}|4[01]\d{8}|[1-3]\d{9}|[1-9]\d{8}|[1-9]\d{7}|[1-9]\d{6}|[1-9]\d{5}|[1-9]\d{4}|[1-9]\d{3}|[1-9]\d{2}|[1-9]\d|\d)$
Kindly try this i tested randomly not thoroughly.
only for the numbers above zero. add '-' and adjust last number pattern for negative numbers.
(^\d{1,9}$|^1\d{9}$|^20\d{8}$|^21[0-3]\d{7}$|^214[0-6]\d{6}$|^2147[0-3]\d{5}$|^21474[0-7]\d{4}$|^214748[0-2]\d{3}$|^2147483[0-5]\d{2}$|^21474836[0-3]\d$|^214748364[0-7]$)
one should never use regex for this type of work.
I would like to know how can I construct a regex to know if a number in base 2 (binary) is multiple of 3. I had read in this thread Check if a number is divisible by 3 but they dont do it with a regex, and the graph someone drew is wrong(because it doesn't accept even numbers). I have tried with: ((1+)(0*)(1+))(0) but it doesn't works for some values. Hope you can help me.
UPDATE:
Ok, thanks all for your help, now I know how to draw the NFA, here I left the graph and the regular expresion:
In the graph, the states are the number in base 10 mod 3.
For example: to go to state 1 you have to have 1, then you can add 1 or 0, if you add 1, you would have 11(3 in base 10), and this number mod 3 is 0 then you draw the arc to the state 0.
((0*)((11)*)((1((00) *)1) *)(101 *(0|((00) *1 *) *0)1) *(1(000)+1*01)*) *
And the other regex works, but this is shorter.
Thanks a lot :)
I know this is an old question, but an efficient answer is yet to be given and this question pops up first for "binary divisible by 3 regex" on Google.
Based on the DFA proposed by the author, a ridiculously short regex can be generated by simplifying the routes a binary string can take through the DFA.
The simplest one, using only state A, is:
0*
Including state B:
0*(11)*0*
Including state C:
0*(1(01*0)*1)*0*
And include the fact that after going back to state A, the whole process can be started again.
0*((1(01*0)*1)*0*)*
Using some basic regex rules, this simplifies to
(1(01*0)*1|0)*
Have a nice day.
If I may plug my solution for this code golf question! It's a piece of JavaScript that generates regexes (probably inefficiently, but does the job) for divisibility for each base.
This is what it generates for divisibility by 3 in base 2:
/^((((0+)?1)(10*1)*0)(0(10*1)*0|1)*(0(10*1)*(1(0+)?))|(((0+)?1)(10*1)*(1(0+)?)|(0(0+)?)))$/
Edit: comparing to Asmor's, probably very inefficient :)
Edit 2: Also, this is a duplicate of this question.
For some who is learning and searching how to do this:
see this video:
https://www.youtube.com/watch?v=SmT1DXLl3f4&t=138s
write state quations and solve them with Axden's Theorem
The way I did is visible in the image-result is the same as pointed out by user #Kert Ojasoo. I hope i did it corretly because i spent 2 days to solve it...
n+2n = 3n. Thus, 2 adjacent bits set to 1 denote a multiple of 3. If there are an odd number of adjacent 1s, that would not be 3.
So I'd propose this regex:
(0*(11)?)+