It's been a while that I am juggling around this. Hope you can give me
some pointers.
All I want to achieve is, the string should contain EXACTLY 4 '-' and 10 digits in any giver order.
I created this regex : ^(-\d-){10}$
It does enforce max-length of 10 on digits but I am not getting a way to implement max-length of 4 for '-'
Thanks
Ok, here's a pattern:
^(?=(?:\d*?-){4}\d*$)(?=(?:-*?\d){10}-*$).{14}$
Demo
Explanation:
The main part is ^.{14}$ which simply checks there are 14 characters in the string.
Then, there are two lookaheads at the start:
(?=(?:\d*?-){4}\d*$)
(?=(?:-*?\d){10}-*$)
The first one checks the hyphens, and the second one checks the digits and make sure the count is correct. Both match the entire input string and are very similar so let's just take a look at the first one.
(?:\d*?-){4} matches any number of digits (or none) followed by a hyphen, four times. After this match, we know there are four hyphens. (I used an ungreedy quantifier (*?) just to prevent useless backtracking, as an optimization)
\d*$ just makes sure the rest of the string is only made of digits.
Related
I've written a regular expression that matches any number of letters with any number of single spaces between the letters. I would like that regular expression to also enforce a minimum and maximum number of characters, but I'm not sure how to do that (or if it's possible).
My regular expression is:
[A-Za-z](\s?[A-Za-z])+
I realized it was only matching two sets of letters surrounding a single space, so I modified it slightly to fix that. The original question is still the same though.
Is there a way to enforce a minimum of three characters and a maximum of 30?
Yes
Just like + means one or more you can use {3,30} to match between 3 and 30
For example [a-z]{3,30} matches between 3 and 30 lowercase alphabet letters
From the documentation of the Pattern class
X{n,m} X, at least n but not more than m times
In your case, matching 3-30 letters followed by spaces could be accomplished with:
([a-zA-Z]\s){3,30}
If you require trailing whitespace, if you don't you can use: (2-29 times letter+space, then letter)
([a-zA-Z]\s){2,29}[a-zA-Z]
If you'd like whitespaces to count as characters you need to divide that number by 2 to get
([a-zA-Z]\s){1,14}[a-zA-Z]
You can add \s? to that last one if the trailing whitespace is optional. These were all tested on RegexPlanet
If you'd like the entire string altogether to be between 3 and 30 characters you can use lookaheads adding (?=^.{3,30}$) at the beginning of the RegExp and removing the other size limitations
All that said, in all honestly I'd probably just test the String's .length property. It's more readable.
This is what you are looking for
^[a-zA-Z](\s?[a-zA-Z]){2,29}$
^ is the start of string
$ is the end of string
(\s?[a-zA-Z]){2,29} would match (\s?[a-zA-Z]) 2 to 29 times..
Actually Benjamin's answer will lead to the complete solution to the OP's question.
Using lookaheads it is possible to restrict the total number of characters AND restrict the match to a set combination of letters and (optional) single spaces.
The regex that solves the entire problem would become
(?=^.{3,30}$)^([A-Za-z][\s]?)+$
This will match AAA, A A and also fail to match AA A since there are two consecutive spaces.
I tested this at http://regexpal.com/ and it does the trick.
You should use
[a-zA-Z ]{20}
[For allowed characters]{for limiting of the number of characters}
I am trying to create a regular expression (Java/JavaScript) that matches the following regex, but only when there are fewer than 13 characters total (and a minimum of 4).
(COT|MED)[ABCD]?-?[0-9]{1,4}(([JK]+[0-9]*)|(\ DDD)?) ← originally posted
(COT|MED)[ABCD]?-?[0-9]{1,4}(([JK]+[0-9]*)|(\ [A-Z]+)?)
These values should (and do) match:
MED-123
COTA-1224
MED4
COTB-892K777
MED-33 DDD
MED-234J5678
This value matches, but I don't want it to (I want to only match if there are fewer than 12 characters total):
COT-1111J11111111111111
See http://regexr.com/3bs7b http://regexr.com/3bsfv
I have tried grouping my expression and putting {4,12} at the end, but that just makes it look for 4 to 12 instances of the whole expression matching.
I feel like I am missing something simple...thanks in advance for your help!
You can use negative look-ahead:
(?!.{13,})(COT|MED)[ABCD]?-?[0-9]{1,4}(([JK]+[0-9]*)|(\ DDD)?)
Since your expression already make sure that a match starts with COT or MED and there is at least one digit after that, it already guarantees that there are at least 4 characters
I have tried grouping my expression and putting {4,12} at the end, but
that just makes it look for 4 to 12 instances of the whole expression
matching.
This looks for 4 to 12 instances of the whole expression because you didn't add a word boundary \b. Your regex works fine, just add a word boundary and your desired outcome would be achieved. Take a look at this DEMO.
Your regex seems to be very clumsy and looks a little bit hard to read. It is also very limited to certain characters example JK except if you want it to be that way. For a more general pattern, you can check this out
(COT|MED)[AB]?-?[\dJK]{1,8}(\s+D{1,3})?\b
(COT|MED): matches either COT or MED
[AB]?: matches A or B which is optional because of the presence of ?
-?: matches - which is also optional
[\dJK]{1,8}: This matches a number,or J or K with a length of at least one character and a maximum of eight characters.
(\s+D{1,3})?: matches a space or a D at least one time and a maximum of 3 times and this is optional
\b: with respect to your question this seems to be the most important and it creates a boundary for the words that have already been matched. This means that anything exceeding the matched pattern would not be captured.
See the demo here DEMO2
The answer you are looking for is
(?!\S{13})(?:COT|MED)[ABCD]?-?\d{1,4}(?:[JK]+\d*|(?: [A-Z]+)?)
See regex demo
Note that it is almost impossible to check the length of a phrase that is not a whole string or that has spaces inside since boundaries are a bit "blurred". Thus, (?!\S{13}) is a kind of a workaround that just makes sure you do not have a string without whitespace that is 13 characters long or longer.
The regex breakdown:
(?!\S{13}) - Check if the substring that follows does not consist of 13 non-whitespace characters
(?:COT|MED) - Any of the values in the alternation (COTorMED`)
[ABCD]?-? - Optional A, B, C, D and then an optional -
\d{1,4} - 1 to 4 digits
(?:[JK]+\d*|(?: [A-Z]+)?) - a group of 2 alternatives:
[JK]+\d* - J or K, 1 or more times, and then 0 or more digits
(?: [A-Z]+)? - optional space and 1 or more Latin uppercase letters
As this answer suggests, you could solve this this way:
(?=(COT|MED)[ABCD]?-?[0-9]{1,4}(([JK]+[0-9]*)|(\ DDD)?))(?={4 , 12})
so I got an assesment due and one the regular expressions is a phone digit that has 10 numbers but it also allows dashes "-". However it can only contain 2 dashes and start with 0. eg. 0x-xxxx-xxxx where x can be any digit between 0-9.
So, so far I've come up with the following regexes:
^[0-0][0-9-]{1,11}$
^[0-0][0-9-]{11}$
In the first one it works but allows any length, if I put 5 numbers it goes through as long as the first one is 0. In the second one I can put 2 dashes followed by each other or just fill it with dashes and it goes through.
Thanks for helping guys!
Match every of the three parts, separated by the dashes, like this:
^0\d-\d{4}-\d{4}$
This matches 0x-xxxx-xxxx. Demo: https://regex101.com/r/nW7wL5/1
If you also want to match the number without the dashes, use
^0\d-?\d{4}-?\d{4}$
Demo: https://regex101.com/r/gY0mC3/1
\d is the same as [0-9] but it's shorter.
I'm trying to create a regular expression in javascript for a UK bank sort code so that the user can input 6 digits, or 6 digits with a hyphen between pairs. For example "123456" or "12-34-56". Also not all of the digits can be 0.
So far I've got /(?!0{2}(-?0{2}){2})(\d{2}(-\d{2}){2})|(\d{6})/ and this jsFiddle to test.
This is my first regular expression so I'm not sure I'm doing it right. The test for 6 0-digits should fail and I thought the -? optional hyphen in the lookahead would cause it to treat it the same as 6 0-digits with hyphens, but it isn't.
I'd appreciate some help and any criticism if I'm doing it completely incorrectly!
Just to answer your question, you can validate user input with:
/^(?!(?:0{6}|00-00-00))(?:\d{6}|\d\d-\d\d-\d\d)$/.test(inputString)
It will strictly match only input in the form XX-XX-XX or XXXXXX where X are digits, and will exclude 00-00-00, 000000 along with any other cases (e.g. XX-XXXX or XXXX-XX).
However, in my opinion, as stated in other comments, I think it is still better if you force user to either always enter the hyphen, or none at all. Being extra strict when dealing with anything related to money saves (unknown) troubles later.
Since any of the digits can be zero, but not all at once, you should treat the one case where they are all zero as a single, special case.
You are checking for two digits (\d{2}), then an optional hyphen (-?), then another two digits (\d{2}) and another optional hyphen (-?), before another two digits (\d{2}).
Putting this together gives \d{2}-?\d{2}-?\d{2}, but you can simplify this further:
(\d{2}-?){2}\d{2}
You then use the following pseudocode to match the format but not 000000 or 00-00-00:
if (string.match("/(\d{2}-?){2}\d{2}/") && !string.match("/(00-?){2}00/"))
//then it's a valid code, you could also use (0{2}-?){2}0{2} to check zeros
You may wish to add the string anchors ^ (start) and $ (end) to check the entire string.
Could you please help me to provide a RegEx pattern to validate a string which satisfy:
length from 4 to 10 (strictly)
first 3 characters must be string (A-Z a-z)
the remain characters must be number without 00 as prefix, I mean ABC15 is passed but ABC0015 is not.
This problem took me so much time and I have tried so many regex patterns, but I still have no solution for it.
Thank you so much.
Use this one:
/^[a-z]{3}(?!00)\d{1,7}$/i
Explanation:
/
^ Start
[a-z]{3} Three letters.
(?!00) Must NOT be followed by two zeros.
\d{1,7} One to seven digits.
$ End.
/i ignore case flag.
Easy.
/^[a-z]{3}[1-9][0-9]{0,6}$/i
Matches three letters (case insensitive flag at the end), followed by one digit that is not zero, followed by up to six more digits (which may be zero).