Trying to create a regex that ignores a proper integer (1, 5, 999, etc.) and forward slashes (/), but finds a match in everything else. For example, it would find a match the following:
test
test1
test-1
but ignores
1
55
7
This is for a mod rewrite.
[^-0-9\/]+ should do the trick, I think. It'll match any string that contains a non-digit.
Edited to add minus sign which would also be allowed in an integer, and then to include the forward slashes mentioned in the question.
This is a very old question, but I noticed that the currently accepted answer ([^-0-9\/]) will not match any part of a number string that has a dash/minus (-) in the middle or at the end or purely consists of dashes.
So the regex will not find a match in strings like 12-34, 1234-, --12, -, or -----, even though these are clearly not valid integer numbers and should thus be caught.
To include these in the regex, you can use something like this:
[^-0-9]+|[0-9]+(?=-)|^-$|-{2,}
This will match
either any part of a string that contains a non-digit, non-minus character (as does the regex from the currently accepted answer)
or any number that is followed by a minus sign - (using a positive lookahead, taking care of the numbers with dashes anywhere but the beginning)
or any string that consists of one single dash
or any part of a string with two or more consecutive dashes
This will thus find a match in any string that is not a positive or negative integer, see also https://regex101.com/r/8zJwCy/1
To also include (or rather exclude) slashes, as the OP requested, they can be added to the first character group.
[^-0-9\/]+|[0-9]+(?=-)|^-$|-{2,}
Note that this will also not match pure series of slashes (/, //, ///,...) and multiple slashes between integers (1//2, 1//-2, //1, 2//) which may or may not be desired (the currently accepted answer behaves identically), see https://regex101.com/r/8zJwCy/3
To also catch these, add another two optional groups that match series of slashes, analogous to the last two ones that take care of the dashes. Or, if you want a single individual slash (/) to be valid, just add one group analogous to the last one, see https://regex101.com/r/8zJwCy/4
[^-0-9\/]+|[0-9]+(?=-)|^-$|-{2,}|\/{2,}
This would match line by line - is that what you need?
^[0-9\/]*$
This should do the job!
^[0-9\/]+$
Related
The first thing I tried to do, is get the regex matching what I DON'T want. This way, I could just flip it to NOT accept that same input. This is where I came up with the first part of this regex.
Accept all 9 digit numbers, where all 9 digits are identical (without dashes): "^(\d)\1{8}$". This expression works as expected (as seen here: (https://regex101.com/r/Ez8YC3/1)).
The second expression should do the same, with dashes formatted as follows xxx-xx-xxxx: "^(\d)\1{8}$". This expressions works as expected (as seen here: https://regex101.com/r/bodzIX/1).
Now what I want to do at this point, is combine them together to look for BOTH conditions. However when I do that it seems to break, and only match 9 digit numbers that are identical throughout WITH dashes: "^(\d)\1{2}-(\d)\1{1}-(\d)\1{3}$|^(\d)\1{8}$". This can be seen here: https://regex101.com/r/lPnksf/1.
I may be getting a little ahead of myself here, but in order to show my work as much as possible, I also tried flipping those regex separately, which also did not work as expected.
Condition #1 flipped: "^(?!(\d)\1{8})$". Can be seen here: https://regex101.com/r/ed51yk/1.
Condition #2 flipped: "^(?!(\d)\1{2}-(\d)\1{1}-(\d)\1{3})$". Can be seen here: https://regex101.com/r/UYfoMK/1.
I would expect the two expressions (when flipped) to match any 9 digit number (with or without dashes) where all numbers are not identical. How ever this does not happen at all.
This is the final regex that I came up with, which is clearly not doing what I would expect it to: "^(?!(\d)\1{2}-(\d)\1{1}-(\d)\1{3})$|^(?!(\d)\1{8})$". Can be seen here: https://regex101.com/r/9eHhF5/1
At the end of the day, I want to combine these 2 expressions, with this one (that already works as intended): "^(?!000|666|9\d\d)\d{3}-(?!00)\d\d-(?!0000)\d\d\d\d$". Can be seen here: https://regex101.com/r/AdRI8i/1.
I am still pretty new to regex, and really want to understand why I can't simply wrap the condition in (?!...) in order to match the opposite condition.
Thank you in advance
What you want to do is not flip, but reverse the regex logic.
Yes, to reverse the pattern logic, you should use a negative lookahead, but there are caveats.
First, the $ end of string anchor: if it was at the end of the "positive" regex, it must also be moved to the lookahead in the reverse pattern. So, your ^(?!(\d)\1{8})$ regex must be written as ^(?!(\d)\1{8}$). Same goes for your second regex.
Next, mind that each subsequent capturing group gets an incremented ID number, so you cannot keep the same backreferences when you "join" patterns with OR | operator. You must adjust these IDs to reflect their new values in the new regex.
So, you want to match a string that matches ^(?!000|666|9\d\d)\d{3}-(?!00)\d\d-(?!0000)\d\d\d\d$ first (let's note \d\d\d\d = \d{4}), then you can add restrictions with lookaheads:
(?!(\d)\1{8}$) - fails the match if, immediately from the current position, it matches identical 9 digits and then the string end comes
(?!(\d)\2\2-(\d)\2-(\d)\2{3}$) - (note the ID incrementing continuation) fails the match if, immediately from the current position, it matches identical to the first one 3 digits, -, identical 2 digits, -, identical 5 digits, and then the string end comes.
So, to follow your logic, you can use
^(?!(\d)\1{8}$)(?!(\d)\2\2-(\d)\2-(\d)\2{3}$)(?!000|666|9\d\d)\d{3}-(?!00)\d\d-(?!0000)\d{4}$
See the regex demo
As the lookaheads are non-consuming patterns, i.e. the regex index remains at the same position after matching their pattern sequences where it was before, the 3 lookaheads will all be tried at the start of the string (see the ^ anchor). If any of the three negative lookaheads at the start fails, the whole string match will be failed right away.
By this Regex you match what you dont want as social security number:
^(?:(\d)\1{8})|(?:(\d)\2{2}-\2{2}-\2{4})$
Demo
By this regex you match only what you want:
^(?!(?:(\d)\1{8})|(?:(\d)\2{2}-\2{2}-\2{4})).*$
Demo
Given a file name of 22-PLUMB-CLR-RECTANGULAR.0001.rfa I need a RegEx to match it. Basically it's any possible characters, then . and 4 digits and one of four possible file extensions.
I tried ^.?\.\d{4}\.(rvt|rfa|rte|rft)$ , which I would have thought would be correct, but I guess my RegEx understanding has not progressed as much as I thought/hoped. Now, .?\.\d{4}\.(rvt|rfa|rte|rft)$ does work and the ONLY difference is that I am not specifying the start of the string with ^. In a different situation where the file name is always in the form journal.####.txt I used ^journal\.\d{4}\.txt$ and it matched journal.0001.txt like a champ. Obviously when I am specifying a specific string, not any characters with .? is the difference, but I don't understand the WHY.
That never matches the mentioned string since ^.? means match beginning of input string then one optional single character. Then it looks for a sequence of dots and digits and nothing's there. Because we didn't yet pass the first character.
Why does it work without ^? Because without ^ it is allowed to go through all characters to find a match and it stops right before R and continues matching up to the end.
That's fine but with first approach it should be ^.*. Kleene star matches every thing greedily then backtracks but ? is the operator here which makes preceding pattern optional. That means one character, not many characters.
I'm looking for a way to use regex to search for obviously false phone numbers that have the same digit repeating. The numbers are all formatted and stored as follows:
(111)111-1111
I'm not able to alter the text in any way.
I've tried modifying a few of the regex lines I've seen such as:
^([0-9])\1{2}.\1{3}.\1{4}$
which was for finding repeating digits with a period in between the numbers. However, I haven't figured out how to get around the first character as a parenthesis.
Any help would be appreciated!
You misunderstand the purpose of the . Dot Operator. It is not to match a period, it matches anything. In that (quite badly) regex, it serves only to skip the - – and because it matches anything, it will also match something like 11121113111.
Use this regexp instead:
^\(?([0-9])\1{2}\)?\1{3}-?\1{4}$
This checks for parentheses around the first group, optionally so it will still work without; and specifically checks for the presence of a dash between the second and third group of digits, also optionally.
I am using a RegEx to test if a string is valid. The string must start and end with a number ([0-9]), but can contain comma's within.
I came up with this example, but it fails for strings less than 3 characters (for example 1 or 15 are as valid as 1,8). Presumably this is because I am specifically testing for a first and last character, but I don't know any other way of doing this.
How can I change this RegEx to match my requirements. Thanks.
^[0-9]+[0-9\,]+[0-9]$
Use this:
^[0-9]+(,[0-9])?$
the ,[0-9] part will be optional
visualized:
if you want allow for multiple comma-number groups... then replace the ? with *.
if you want to allow groups of numbers after the comma (which didn't seem to be the case in your example), then you should put + after that number group as well.
if both of the above mentioned are desired, your final regex could look like this:
^[0-9]+(,[0-9]+)*$
^\d+(?:,\d+)*$
should work.
Always have one or more digits at the start, optionally followed by any number of comma-separated other groups of one or more digits.
If you allow commas next to each other, then the second + should be a *, I think.
I would say the regex
\d(,?\d)*
Should satisfy for 1 or more digits that can be separated by only one comma. Note, 1,,2 fails
I've got this RegEx example: http://regexr.com?34hihsvn
I'm wondering if there's a more elegant way of writing it, or perhaps a more optimised way?
Here are the rules:
Digits and dashes only.
Must not contain more than 10 digits.
Must have two hyphens.
Must have at least one digit between each hyphen.
Last number must only be one digit.
I'm new to this so would appreciate any hints or tips.
In case the link expires, the text to search is
----------
22-22-1
22-22-22
333-333-1
333-4444-1
4444-4444-1
4444-55555-1
55555-4444-1
666666-7777777-1
88888888-88888888-1
1-1-1
88888888-88888888-22
22-333-
333-22
----------
My regex is: \b((\d{1,4}-\d{1,5})|(\d{1,5}-\d{1,4}))-\d{1}\b
I'm using this site for testing: http://gskinner.com/RegExr/
Thanks for any help,
Nick
Here is a regex I came up with:
(?=\b[\d-]{3,10}-\d\b)\b\d+-\d+-\d\b
This uses a look-ahead to validate the information before attempting the match. So it looks for between 3-10 characters in the class of [\d-] followed by a dash and a digit. And then after that you have the actual match to confirm that the format of your string is actually digit(dash)digit(dash)digit.
From your sample strings this regex matches:
22-22-1
333-333-1
333-4444-1
4444-4444-1
4444-55555-1
55555-4444-1
1-1-1
It also matches the following strings:
22-7777777-1
1-88888888-1
Your regexp only allows a first and second group of digits with a maximum length of 5. Therefore, valid strings like 1-12345678-1 or 123456-1-1 won't be matched.
This regexp works for the given requirements:
\b(?:\d\-\d{1,8}|\d{2}\-\d{1,7}|\d{3}\-\d{1,6}|\d{4}\-\d{1,5}|\d{5}\-\d{1,4}|\d{6}\-\d{1,3}|\d{7}\-\d{1,2}|\d{8}\-\d)\-\d\b
(RegExr)
You can use this with the m modifier (switch the multiline mode on):
^\d(?!.{12})\d*-\d+-\d$
or this one without the m modifier:
\b\d(?!.{12})\d*-\d+-\d\b
By design these two patterns match at least three digits separated by hyphens (so no need to put a {5,n} quantifier somewhere, it's useless).
Patterns are also build to fail faster:
I have chosen to start them with a digit \d, this way each beginning of a line or word-boundary not followed by a digit is immediately discarded. Other thing, using only one digit, I know the remaining string length.
Then I test the upper limit of the string length with a negative lookahead that test if there is one more character than the maximum length (if there are 12 characters at this position, there are 13 characters at least in the string). No need to use more descriptive that the dot meta-character here, the goal is to quickly test the length.
finally, I describe the end of string without doing something particular. That is probably the slower part of the pattern, but it doesn't matter since the overwhelming majority of unnecessary positions have already been discarded.