i try to write regex for get only amount, from string, i do this, it's work but i want to optimise my expression, for exemple i have
125.250.230,55
this is my regew :
\d{1,3}[\,\.]{1}\d{1,3}[\,\.]{1}\d{1,3}[\,\.]{1}\d{1,3}
i want to write it with another form with a repeat group like, but it doesn't work for me
(\d{1,3}[\,\.]{1}){6}\d{1,3}
It's unclear from your example if you want to match always 4 sets of digits separated by a comma or dot, or a variable sets of digits.
If exactly 4 sets of digits use this:
(?:\d{1,3}[.,]){3}\d{1,3}
If a variable sets of digits use this:
(?:\d{1,3}[.,])+\d{1,3}
If you want to properly match sets of 3 digits, with variable number of digits at the beginning and end, such as:
1,123,123.1
12,123,123.12
123,123,123.123
1,123,123,123,123.1
Then use this:
\d{1,3}[.,](?:\d{3}[.,])*\d{1,3}
Explanation of regex:
\d{1,3} - one to three digits (0...9)
[.,] - followed by a dot or a comma
(?: ... )* - followed by a non-capturing group; the * means zero to multiple repeats
d{3}[.,] - inside non-capturing group, expect three digits, followed by a dot or comma
\d{1,3} - followed by one to three digits
You need to make the group in your exemple non capturing. See https://www.regular-expressions.info/captureall.html
Here is your regew:
(?:\d{1,3}[\,\.]){3}\d{1,3}
Related
I'm trying to figure out if my problem is solvable using regex.
I have computer name in format computer01.domain.com.
I'd like to check if number before first dot is odd or even number.
I managed to build regex to locate first character before dot ^([^.])+(?=\.) now I can't figure out how to check if it's odd or even.
If you want to know wether ther is an even or odd number, you might use 2 capture groups with an alternation.
If the first capture group exists, it is an odd number, if the second one exists then it is an even number.
If you only want to capture the single digit, you can also match the following dot instead of asserting it.
^[^.]+(?:([13579])|([02468]))\.
The pattern matches:
^ Start of string.
[^.]+ Match 1+ times any char other than a dot
(?: Non capture group
([13579]) Capture an odd number in group 1
| Or
([02468]) Capture an even number in group 2
) Close the non capture group
\. Match the dot
Regex demo
To match these examples:
1-10-1
1-7-3
10-8-5
1-7-14
11-10-12
This regex works:
^[\\d]{1,2}-[\\d]{1,2}-[\\d]{1,2}$
How could this be written in a way that just matches something like "[\d]{1,2}-?" three (n) times?
You may use:
^\d\d?(?:-\d\d?){2}$
See an online demo.
^ - Start line anchor.
\d\d? - A single digit followed by an optional one (the same as \d{1,2}).
(?:-\d\d?){2} - A non-capture group starting with an hyphen followed by the same construct as above, one or two digits. The capture group is repeated exactly two times.
$ - End string anchor.
The idea here is to avoid an optional hyphen in the attempts you made since essentially you'd start to allow whole different things like "123" and "123456". It's more appropriate to match the first element of a delimited string and then use the non-capture group to match a delimiter and the rest of the required elements exactly n-1 times.
Regex is great, but I can't for the life of me figure out how I'd express the following constraint, without spelling out the whole permutation:
2 of any digit [0-9]
3 of any other digit [0-9] excluding the above
4 of any third digit [0-9] excluding the above
I've got this monster, which is clearly not a good way of doing this, as it grows exponentially with each additional set of digits:
^(001112222|001113333|001114444|001115555|001116666|0001117777|0001118888|0001119999|0002220000|...)$
OR
^(0{2}1{3}2{4}|0{2}1{3}3{4}|0{2}1{3}4{4}|0{2}1{3}5{4}|0{2}1{3}6{4}|0{2}1{3}7{4}|0{2}1{3}8{4}|...)$
Looks like the following will work:
^((\d)\2(?!.+\2)){2}\2(\d)\3{3}$
It may look a bit tricky, using recursive patterns, but it may look more intimidating then it really is. See the online demo.
^ - Start string anchor.
( - Open 1st capture group:
(\d) - A 2nd capture group that does capture a single digit ranging from 0-9.
\2 - A backreference to what is captured in this 2nd group.
(?!.+\2) - Negative lookahead to prevent 1+ characters followed by a backreference to the 2nd group's match.
){2} - Close the 1st capture group and match this two times.
\2 - A backreference to what is most recently captured in the 2nd capture group.
(\d) - A 3rd capture group holding a single digit.
\3{3} - Exactly three backreferences to the 3rd capture group's match.
$ - End string anchor.
EDIT:
Looking at your alternations it looks like you are also allowing numbers like "002220000" as long as the digits in each sequence are different to the previous sequence of digits. If that is the case you can simplify the above to:
^((\d)\2(?!.\2)){2}\2(\d)\3{3}$
With the main difference is the "+" modifier been taken out of the pattern to allow the use of the same number further on.
See the demo
Depending on whether your target environment/framework/language supports lookaheads, you could do something like:
^(\d)\1(?!\1)(\d)\2\2(?!\1|\2)(\d)\3\3\3$
First capture group ((\d)) allows us to enforce the "two identical consecutive digits" by referencing the capture value (\1) as the next match, after which the negative lookahead ensures the next sequence doesn't start with the previous digit - then we just repeat this pattern twice
Note: If you want to exclude only the digit used in the immediately preceding sequence, change (?!\1|\2) to just (?!\2)
I am trying to block any strings that contain more than 3 numbers and prevent special characters. I have the special characters part down. I'm just missing the number part.
For example:
"Hello 1234" - Not Allowed
"Hello 123" - Allowed
I've tried the following:
/^[!?., A-Za-z0-9]+$/
/((^[!?., A-Za-z]\d)([0-9]{3}+$))/
/^((\d){2}[a-zA-Z0-9,.!? ])*$/
The last one is the closest I got as it prevents any special characters and any numbers from being entered at all.
I've looked through previous posts, but am coming up short.
Edit for clarification
Essentially I'm trying to find a way to prevent customers from entering PII on a form. No submission should be allowed that contains more than 3 numbers in a string.
Hello1234 - Not allowed
12345 - Not allowed
1111 - not allowed
No where in the comment section when the user enters the string should there be more than 3 numbers in total.
About the patterns that you tried
^[!?., A-Za-z0-9]+$ The pattern matches 1+ times any of the listed, including 1 or more digits
((^[!?., A-Za-z]\d)([0-9]{3}+$)) If {3}+ is supported, the pattern matches a single char from the character class, 1 digit followed by 3 digits
^((\d){2}[a-zA-Z0-9,.!? ])*$ The pattern repeats 0+ times matching 2 digits and 1 of the listed in the character class
You can use a negative lookahead if that is supported to assert not 4 digits in a row.
^(?!.*\d{4})[a-zA-Z0-9,.!? ]+$
regex demo
If there can not be 4 digits in total, but 0-3 occurrences:
^[a-zA-Z,.!? ]*(?:\d[a-zA-Z,.!? ]*){0,3}$
Explanation
^ Start of string
[a-zA-Z,.!? ]* Match 0+ times any of the listed (without a digit)
(?:\d[a-zA-Z,.!? ]*){0,3} Repeat 0 - 3 times matching a single digit followed by optional listed chars (Again without a digit)
$ End of string
regex demo
If you don't want to match an empty string and a lookahead is supported:
^(?!$)[a-zA-Z,.!? ]*(?:\d[a-zA-Z,.!? ]*){0,3}$
See another regex demo
Here is my two cents:
^(?!(.*\d){4})[A-Za-z ,.!?\d]+$
See the online demo
^ - Start string anchor.
(?! - Open a negative lookahead.
( - Open capture group.
.*\d - Match anything other than newline up to a digit.
){4} - Close capture group and match it 4 times.
) - Close negative lookahead.
[A-Za-z ,.!?\d]+ - 1+ Characters from specified class.
$ - End string anchor.
I think it should cover what you described.
Assuming you mean <= 3 digits, this may be a naive one but how about
[ALLOWED_CHARS]*[0-9]?[ALLOWED_CHARS]*[0-9]?[ALLOWED_CHARS]*[0-9][ALLOWED_CHARS]*?
Fill [ALLOWED_CHARS] to whatever you define is not special character and nums.
I would like to check if a phone number contains exactly 3 digits - dot - 3 digits - dot - 3 digits. (e.g. 123.456.789)
So far I have this, but it doesn't work:
^(\d{3}\){2}\d{4}$
Note that an escaped bracket \) loses its special meaning in regex and the pattern becomes invalid since the capturing group is not closed.
If you want to match a dot with a regex, you need to include it to your pattern, and if you say 3 digits must be at the end there is no point in declaring 4 digits with \d{4}.
^(\d{3}\.){2}\d{3}$
^ ^
or if we expand the first group:
^\d{3}\.\d{3}\.\d{3}$
So all the fix consists in adding a dot after the second backslash and adjusting the final limiting quantifier.
Note that for mostly "stylistics" concerns (since efficiency gain is insignificant) I'd use a non-capturing group with the first regex variant:
^(?:\d{3}\.){2}\d{3}$