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}$/i
Matches three letters (case insensitive flag at the end), followed by one digit that is not zero, followed by up to six more digits (which may be zero).
Related
I've written a regular expression that matches any number of letters with any number of single spaces between the letters. I would like that regular expression to also enforce a minimum and maximum number of characters, but I'm not sure how to do that (or if it's possible).
My regular expression is:
[A-Za-z](\s?[A-Za-z])+
I realized it was only matching two sets of letters surrounding a single space, so I modified it slightly to fix that. The original question is still the same though.
Is there a way to enforce a minimum of three characters and a maximum of 30?
Yes
Just like + means one or more you can use {3,30} to match between 3 and 30
For example [a-z]{3,30} matches between 3 and 30 lowercase alphabet letters
From the documentation of the Pattern class
X{n,m} X, at least n but not more than m times
In your case, matching 3-30 letters followed by spaces could be accomplished with:
([a-zA-Z]\s){3,30}
If you require trailing whitespace, if you don't you can use: (2-29 times letter+space, then letter)
([a-zA-Z]\s){2,29}[a-zA-Z]
If you'd like whitespaces to count as characters you need to divide that number by 2 to get
([a-zA-Z]\s){1,14}[a-zA-Z]
You can add \s? to that last one if the trailing whitespace is optional. These were all tested on RegexPlanet
If you'd like the entire string altogether to be between 3 and 30 characters you can use lookaheads adding (?=^.{3,30}$) 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 have a regex
/^([a-zA-Z0-9]+)$/
this just allows only alphanumerics but also if I insert only number(s) or only character(s) then also it accepts it. I want it to work like the field should accept only alphanumeric values but the value must contain at least both 1 character and 1 number.
Why not first apply the whole test, and then add individual tests for characters and numbers? Anyway, if you want to do it all in one regexp, use positive lookahead:
/^(?=.*[0-9])(?=.*[a-zA-Z])([a-zA-Z0-9]+)$/
This RE will do:
/^(?:[0-9]+[a-z]|[a-z]+[0-9])[a-z0-9]*$/i
Explanation of RE:
Match either of the following:
At least one number, then one letter or
At least one letter, then one number plus
Any remaining numbers and letters
(?:...) creates an unreferenced group
/i is the ignore-case flag, so that a-z == a-zA-Z.
I can see that other responders have given you a complete solution. Problem with regexes is that they can be difficult to maintain/understand.
An easier solution would be to retain your existing regex, then create two new regexes to test for your "at least one alphabetic" and "at least one numeric".
So, test for this :-
/^([a-zA-Z0-9]+)$/
Then this :-
/\d/
Then this :-
/[A-Z]/i
If your string passes all three regexes, you have the answer you need.
The accepted answers is not worked as it is not allow to enter special characters.
Its worked perfect for me.
^(?=.*[0-9])(?=.*[a-zA-Z])(?=\S+$).{6,20}$
one digit must
one character must (lower or upper)
every other things optional
Thank you.
While the accepted answer is correct, I find this regex a lot easier to read:
REGEX = "([A-Za-z]+[0-9]|[0-9]+[A-Za-z])[A-Za-z0-9]*"
This solution accepts at least 1 number and at least 1 character:
[^\w\d]*(([0-9]+.*[A-Za-z]+.*)|[A-Za-z]+.*([0-9]+.*))
And an idea with a negative check.
/^(?!\d*$|[a-z]*$)[a-z\d]+$/i
^(?! at start look ahead if string does not
\d*$ contain only digits | or
[a-z]*$ contain only letters
[a-z\d]+$ matches one or more letters or digits until $ end.
Have a look at this regex101 demo
(the i flag turns on caseless matching: a-z matches a-zA-Z)
Maybe a bit late, but this is my RE:
/^(\w*(\d+[a-zA-Z]|[a-zA-Z]+\d)\w*)+$/
Explanation:
\w* -> 0 or more alphanumeric digits, at the beginning
\d+[a-zA-Z]|[a-zA-Z]+\d -> a digit + a letter OR a letter + a digit
\w* -> 0 or more alphanumeric digits, again
I hope it was understandable
What about simply:
/[0-9][a-zA-Z]|[a-zA-Z][0-9]/
Worked like a charm for me...
Edit following comments:
Well, some shortsighting of my own late at night: apologies for the inconvenience...
The - incomplete - underlying idea was that only one "transition" from a digit to an alpha or from an alpha to a digit was needed somewhere to answer the question.
But next regex should do the job for a string only comprised of alphanumeric characters:
/^[0-9a-zA-Z]*([0-9][a-zA-Z]|[a-zA-Z][0-9])[0-9a-zA-Z]*$/
which in Javascript can be furthermore simplified as:
/^[0-9a-z]*([0-9][a-z]|[a-z][0-9])[0-9a-z]*$/i
In IMHO it's more straigthforward to read and understand than some other answers (no backtraking and the like).
Hope this helps.
If you need the digit to be at the end of any word, this worked for me:
/\b([a-zA-Z]+[0-9]+)\b/g
\b word boundary
[a-zA-Z] any letter
[0-9] any number
"+" unlimited search (show all results)
It's been a while that I am juggling around this. Hope you can give me
some pointers.
All I want to achieve is, the string should contain EXACTLY 4 '-' and 10 digits in any giver order.
I created this regex : ^(-\d-){10}$
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.
I'm trying to get regex for minimum requirements of a password to be minimum of 6 characters; 1 uppercase, 1 lowercase, and 1 number. Seems easy enough? I have not had any experience in regex's that "look ahead", so I would just do:
if(!pwStr.match(/[A-Z]+/) || !pwStr.match(/[a-z]+/) || !pwStr.match(/[0-9]+/) ||
pwStr.length < 6)
//was not successful
But I'd like to optimize this to one regex and level up my regex skillz in the process.
^.*(?=.{6,})(?=.*[a-zA-Z])(?=.*\d)(?=.*[!&$%&? "]).*$
^.*
Start of Regex
(?=.{6,})
Passwords will contain at least 6 characters in length
(?=.*[a-zA-Z])
Passwords will contain at least 1 upper and 1 lower case letter
(?=.*\d)
Passwords will contain at least 1 number
(?=.*[!#$%&? "])
Passwords will contain at least given special characters
.*$
End of Regex
here is the website that you can check this regex - http://rubular.com/
Assuming that a password may consist of any characters, have a minimum length of at least six characters and must contain at least one upper case letter and one lower case letter and one decimal digit, here's the one I'd recommend: (commented version using python syntax)
re_pwd_valid = re.compile("""
# Validate password 6 char min with one upper, lower and number.
^ # Anchor to start of string.
(?=[^A-Z]*[A-Z]) # Assert at least one upper case letter.
(?=[^a-z]*[a-z]) # Assert at least one lower case letter.
(?=[^0-9]*[0-9]) # Assert at least one decimal digit.
.{6,} # Match password with at least 6 chars
$ # Anchor to end of string.
""", re.VERBOSE)
Here it is in JavaScript:
re_pwd_valid = /^(?=[^A-Z]*[A-Z])(?=[^a-z]*[a-z])(?=[^0-9]*[0-9]).{6,}$/;
Additional: If you ever need to require more than one of the required chars, take a look at my answer to a similar password validation question
Edit: Changed the lazy dot star to greedy char classes. Thanks Erik Reppen - nice optimization!
My experience is if you can separate out Regexes, the better the code will read. You could combine the regexes with positive lookaheads (which I see was just done), but... why?
Edit:
Ok, ok, so if you have some configuration file where you could pass string to compile into a regex (which I've seen done and have done before) I guess it is worth the hassle. But otherwise, Even if the answers provided are corrected to match what you need, I'd still advise against it unless you intend to create such a thing. Separate regexes are just so much nicer to deal with.
I haven't tested thoroughly but here's a more efficient version of Amit's. I think his also allowed unspecified characters into the mix (which wasn't technically listed as a rule). This one won't go berserk on you if you accidentally target a large hunk of text, it will fail sooner on strings that are too long and it only allows the characters in the final class.
'.' should be used sparingly. Think of the looping it has to do to determine a match with all the characters it can represent. It's much more efficient to use negating classes.
`^(?=[^0-9]{0,9}[0-9])(?=[^a-z]{0,9}[a-z])(?=[^A-Z]{0,9}[A-Z])(?=[^##$%]{0,9}[##$%])[0-9a-zA-Z##$%]{6,10`}$
There's nothing wrong with trying to find the ideal regEx. But split it up when you need to.
RegEx tends to be explained poorly. I'll add a breakdown:
a - a single 'a' character
ab - a single 'a' character followed by a single b character
a* - 0 or more 'a' characters
a+ - one or more 'a' characters
a+b - one or any number of a characters followed by a single b character.
a{6,} - at least 6 'a' characters (would match more)
a{6,10} - 6-10 'a' characters
a{10} - exactly 10 'a' characters iirc - not very useful
^ - beginning of a string - so ^a+ would not math 'baaaa'
$ - end of a string - b$ would not find a match 'aaaba'
[] signifies a character class. You can put a variety of characters inside it and every character will be checked. By itself only whatever string character you happen to be on is matched against. It can be modified by + and * as above.
[ab]+c - one or any number of a or b characters followed by a single c character
[a-zA-Z0-9] - any letter, any number - there are a bunch of \<some key> characters representing sets like \d for 'digits' I'm guessing. \w iirc is basically [a-zA-Z_]
note: '\' is the escape key for character classes. [a\-z] for 'a' or '-' or 'z' rather than anything from a to z which is what [a-z] means
[^<stuff>] a character class with the caret in front means everything but the characters or <stuff> listed - this is critical to performance in regEx matches hitting large strings.
. - wildcard character representing most characters (exceptions are a handful of really old-school whitespace characters). Not a big deal in very small sets of characters but avoid using it.
(?=<regex stuff>) - a lookahead. Doesn't move the parser further down the string if it matches. If a lookahead fails, the whole match fails. If it succeeds, you go back to the same character before it. That's why we can string a bunch together to search if there's at least one of a given character.
So:
^ - at the beginning followed by whatever is next
(?=[^0-9]{0,9}[0-9]) - look for a digit from 0-9 preceded by up to 9 or 0 instances of anything that isn't 0-9 - next lookahead starts at the same place
etc. on the lookaheads
[0-9a-zA-Z##$%]{6,10} - 6-10 of any letter, number, or ##$% characters
No '$' is needed because I've limited everything to 10 characters anyway
I'm after a regular expression that matches a UK Currency (ie. £13.00, £9,999.99 and £12,333,333.02), but does not allow negative (-£2.17) or zero values (£0.00 or 0).
I've tried to create one myself, but I've got in a right muddle!
Any help greatfully received.
Thanks!
This'll do it (well mostly...)
/^£?[1-9]{1,3}(,\d{3})*(\.\d{2})?$/
Leverages the ^ and $ to make sure no negative or other character is in the string, and assumes that commas will be used. The pound symbol, and pence are optional.
edit: realised you said non-zero so replaced the first \d with [1-9]
Update: it's been pointed out the above won't match £0.01. The below improvement will but now there's a level of complexity where it may quite possibly be better to test /[1-9]/ first and then the above - haven't benchmarked it.
/^£?(([1-9]{1,3}(,\d{3})*(\.\d{2})?)|(0\.[1-9]\d)|(0\.0[1-9]))$/
Brief explanation:
Match beginning of string followed by optional "£"
Then match either:
a >£1 amount with potential for comma separated groupings and optional pence
OR a <£1 >=£0.10 amount
OR a <=£0.09 amount
Then match end of line
The more fractions of pence (zero in the above) you require adding to the regex the less efficient it becomes.
Under Unix/Linux, it's not always possible to type in the '£' sign in a JavaScript file, so I tend to use its hexadecimal representation, thus:
/^\xA3?\d{1,3}?([,]\d{3}|\d)*?([.]\d{1,2})?$/
This seems to take care of all combinations of UK currency amounts representation that I have come across.
/^\xA3?\d{1,}(?:\,?\d+)*(?:.\d{1,2})?$/;
Explanation:
^ Matches the beginning of the string, or the beginning of a line.
xA3 Matches a "£" character (char code 163)
? Quantifier for match between 0 and 1 of the preceding token.
\d Matches any digit character (0-9).
{1,} Match 1 or more of the preceding token.
(?: Groups multiple tokens together without creating a capture group.
\, Matches a "," character (char code 44).
{1,2} Match between 1 and 2 of the preceding token.
$ Matches the end of the string, or the end of a line if the multiline flag (
You could just make two passes:
/^£\d{1,3}(,\d{3})*(\.\d{2})?$/
to validate the format, and
/[1-9]/
to ensure that at least one digit is non-zero.
This is less efficient than doing it in one pass, of course (thanks, annakata, for the benchmark information), but for a first implementation, just "saying what you want" can significantly reduce developing time.