How can a phone number (or any number) be regex'ed in Oracle to be exactly the correct length followed by a non-digit and then potentially digits again?
e.g.
SELECT 1 FROM DUAL WHERE
REGEXP_LIKE('555-5555x123', '^[0-9]{3,4}[^[:digit:]][0-9]{4}.*$')
Where the number 555-5555 would be ok, 555-5555x123 would be ok, but 555-5555123 would not.
What can happen is someone with fat fingers is typing a phone number and makes a mistake by adding extra numbers (please don't say the input format should be restricted, it's not my data) and this should be flagged as a problem. The example is then more like 555-55545x123.
Test Cases for Oracle REGEXP_LIKE
Value Result
555-5555 ok
555-5555x123 ok
555-55551x123 fail
555-55551 fail
555-5555555 fail
Just remove the .* at the end of your expression it is responsible for matching the additional stuff.
SELECT 1 FROM DUAL WHERE
REGEXP_LIKE('555-5555x123', '^[0-9]{3,4}[^[:digit:]][0-9]{4}$')
That way it does match 3 or 4 digits, a non digit and 4 more digits.
The {3,4} and {4} are the quantifiers that define the amount of digits you want to allow. Just change them to the values you need. E.g. {4,} would match 4 or more.
^ anchors the regex to the start of the string and $ to the end.
Update
To ensure that there is a non digit after the 4 digits at the end you can use an alternation
SELECT 1 FROM DUAL WHERE
REGEXP_LIKE('555-5555x123', '^[0-9]{3,4}[^[:digit:]][0-9]{4}($|[^0-9].*$)')
Now, after your 4 digits there must be either the end of the row OR a non digit ([^0-9] is a negated character class), then anything (but newlines) till the end of the row.
I don't know if it is important in your case, but [^0-9] would also match a newline character, if you want to avoid this use [^0-9\r\n]
SELECT regextestcol FROM regexptest WHERE REGEXP_LIKE(address,'^[0-9]{3}-[0-9]{4}(\w\d{3})?$');
Description:
^ start of your search pattern
[0-9]{3}-[0-9]{4} matches three and four digit numbers separated by hypen
(\w\d{3})? matches a word , a three digit number which both can be optional together with ? sign
$ end of your search pattern
Related
I have to use a regular expression to parse values out of a swift message and there are some situations where the behaviour is not what I want.
Lets say I am after something with a particular pattern - in this case a BIC (6 letters, followed by 2 letters or digits followed by optional XXX or 3 digits)
([A-Z]{6}[A-Z0-9]{2}[XXX0-9]{0,3})
this is fine but now I want to look for these bank codes in particular fields. In swift a field is denoted with : and has some numbers and sometimes a letter.
so I want to match a BIC value in field 52A
I can do the following
(52A:[A-Z]{6}[A-Z0-9]{2}[XXX0-9]{0,3})
which would match 52A:AAAAAAAAXXX
my problem is you can have things before and after this value - and the value itself might not exist in the field you want
so I can wildcard the reg ex to allow for things before it for example
(52A:.*?[A-Z]{6}[A-Z0-9]{2}[XXX0-9]{0,3})
matches 52A:somerubbishAAAAAAAAXXX
but if there isnt something within this field - the reg ex continues to search for the pattern and this is where i have a problem.
for example the above reg ex matches this 52A:somerubbish:57D:AAAAAAAAXXX
Question
I need the reg ex to stop on the first field that is after it (it might not always be 57D but it will always follow the format [0-9]{2}[A-Z]{0,1})
so the above example shouldnt return a match as the pattern I am after is not contained in the 52A section
Does anyone know how I can do this?
Change .*? to [^:]*?:
(52A:[^:]*?[A-Z]{6}[A-Z0-9]{2}[XXX0-9]{0,3})
[^:] means "any character except :", which ensures the match doesn't run into the next field.
See live demo.
Also, unless your situation requires you to match your target as group 1, you don't need the outer brackets: the entire match (ie group 0) will be your target.
I suspect instead of [XXX0-9]{0,3} you want (XXX|\d{3})? (XXX or 3 digits, but optionally) or perhaps (XXX|\d{1,3})? (XXX or up to 3 digits, but optionally)
Using [XXX0-9]{0,3} (which is the same as [X0-9]{0,3}) is a character class notation, repeating 0-3 times an X char or a digit.
If the value itself can also contain a colon, you can match any character as "rubbish" as long as what is directly to the right is not the field format.
52A:(?:(?![0-9]{2}[A-Z]?:).)*[A-Z]{6}[A-Z0-9]{2}(?:[0-9]{3}|XXX)?
The pattern matches:
52A: Match literally
(?:(?![0-9]{2}[A-Z]?:).)* Match any character asserting not 2 digits, optional char A-Z and : directly to the right
[A-Z]{6}[A-Z0-9]{2} Match 6 chars A-Z and 2 chars A-Z or 0-9
(?:[0-9]{3}|XXX)? Optionally match 3 digits or XXX
See a regex demo.
First off, this has sort of been asked before. However I haven't been able to modify this to fit my requirement.
In short: I want a regex that matches an expression if and only if it only contains digits, and there are 5 (or more) increasing consecutive digits somewhere in the expression.
I understand the logic of
^(?=\d{5}$)1*2*3*4*5*6*7*8*9*0*$
however, this limits the expression to 5 digits. I want there to be able to be digits before and after the expression. So 1111345671111 should match, while 11111 shouldn't.
I thought this might work:
^[0-9]*(?=\d{5}0*1*2*3*4*5*6*7*8*9*)[0-9]*$
which I interpret as:
^$: The entire expression must only contain what's between these 2 symbols
[0-9]*: Any digits between 0-9, 0 or more times followed by:
(?=\d{5}0*1*2*3*4*5*6*7*8*9*): A part where at least 5 increasing digits are found followed by:
[0-9]*: Any digits between 0-9, 0 or more times.
However this regex is incorrect, as for example 11111 matches. How can I solve this problem using a regex? So examples of expressions to match:
00001459000
12345
This shouldn't match:
abc12345
9871234444
While this problem can be solved using pure regular expressions (the set of strictly ascending five-digit strings is finite, so you could just enumerate all of them), it's not a good fit for regexes.
That said, here's how I'd do it if I had to:
^\d*(?=\d{5}(\d*)$)0?1?2?3?4?5?6?7?8?9?\1$
Core idea: 0?1?2?3?4?5?6?7?8?9? matches an ascending numeric substring, but it doesn't restrict its length. Every single part is optional, so it can match anything from "" (empty string) to the full "0123456789".
We can force it to match exactly 5 characters by combining a look-ahead of five digits and an arbitrary suffix (which we capture) and a backreference \1 (which must exactly the suffix matched by the look-ahead, ensuring we've now walked ahead 5 characters in the string).
Live demo: https://regex101.com/r/03rJET/3
(By the way, your explanation of (?=\d{5}0*1*2*3*4*5*6*7*8*9*) is incorrect: It looks ahead to match exactly 5 digits, followed by 0 or more occurrences of 0, followed by 0 or more occurrences of 1, etc.)
Because the starting position of the increasing digits isn't known in advance, and the consecutive increasing digits don't end at the end of the string, the linked answer's concise pattern won't work here. I don't think this is possible without being repetitive; alternate between all possibilities of increasing digits. A 0 must be followed by [1-9]. (0(?=[1-9])) A 1 must be followed by [2-9]. A 2 must be followed by [3-9], and so on. Alternate between these possibilities in a group, and repeat that group four times, and then match any digit after that (the lookahead in the last repeated digit in the previous group will ensure that this 5th digit is in sequence as well).
First lookahead for digits followed by the end of the string, then match the alternations described above, followed by one or more digits:
^(?=\d+$)\d*?(?:0(?=[1-9])|1(?=[2-9])|2(?=[3-9])|3(?=[4-9])|4(?=[5-9])|5(?=[6-9])|6(?=[7-9])|7(?=[89])|8(?=9)){4}\d+
Separated out for better readability:
^(?=\d+$)\d*?
(?:
0(?=[1-9])|
1(?=[2-9])|
2(?=[3-9])|
3(?=[4-9])|
4(?=[5-9])|
5(?=[6-9])|
6(?=[7-9])|
7(?=[89])|
8(?=9)
){4}
\d+
The lazy quantifier in the first line there \d*? isn't necessary, but it makes the pattern a bit more efficient (otherwise it initially greedily matches the whole string, requiring lots of failing alternations and backtracking until at least 5 characters before the end of the string)
https://regex101.com/r/03rJET/2
It's ugly, but it works.
I am trying to write a regex to max a sequence of numbers that is 5 digits long or over, but I ignore any spaces, dashes, parens, or hashes when doing that analysis. Here's what I have so far.
(\d|\(|\)|\s|#|-){5,}
The problem with this is that this will match any sequence of 5 characters including those characters I want to ignore, so something like "#123 " would match. While I do want to ignore the # and space character, I still need the number itself to be 5 digits or more in order to qualify at a match.
To be clear, these would match:
1-2-3-4-5
123 45
2(134) 5
Bonus points if the matching begins and ends with a number rather than with one of those "special characters" I am excluding.
Any tips for doing this kind of matching?
If I understood requirements right you can use:
^\d(?:[()\s#-]*\d){4,}$
RegEx Demo
It always matches a digit at start. Then it is followed by 4 or more of a non-capturing group i.e. (?:[()\s#-]*\d) which means 0 or more of any listed special character followed by a digit.
So just repeat a digit, followed by any other sequence of allowed characters 5 or more times:
^(\d[()\s#-]*){5,}$
You can ensure it ends on a digit if you subtract one of the repetitions and add an explicit digit at the end:
^(\d[()\s#-]*){4,}\d$
You can suggest non-digits with \D so et would be something like:
(\d\D*){5,}
Here is a guide.
I have a barcode of the format 123456########. That is, the first 6 digits are always the same followed by 8 digits.
How would I check that a variable matches that format?
You haven't specified a language, but regexp. syntax is relatively uniform across implementations, so something like the following should work: 123456\d{8}
\d Indicates numeric characters and is typically equivalent to the set [0-9].
{8} indicates repetition of the preceding character set precisely eight times.
Depending on how the input is coming in, you may want to anchor the regexp. thusly:
^123456\d{8}$
Where ^ matches the beginning of the line or string and $ matches the end. Alternatively, you may wish to use word boundaries, to ensure that your bar-code strings are properly separated:
\b123456\d{8}\b
Where \b matches the empty string but only at the edges of a word (normally defined as a sequence consisting exclusively of alphanumeric characters plus the underscore, but this can be locale-dependent).
123456\d{8}
123456 # Literals
\d # Match a digit
{8} # 8 times
You can change the {8} to any number of digits depending on how many are after your static ones.
Regexr will let you try out the regex.
123456\d{8}
should do it. This breaks down to:
123456 - the fixed bit, obviously substitute this for what you're fixed bit is, remember to escape and regex special characters in here, although with just numbers you should be fine
\d - a digit
{8} - the number of times the previous element must be repeated, 8 in this case.
the {8} can take 2 digits if you have a minimum or maximum number in the range so you could do {6,8} if the previous element had to be repeated between 6 and 8 times.
The way you describe it, it's just
^123456[0-9]{8}$
...where you'd replace 123456 with your 6 known digits. I'm using [0-9] instead of \d because I don't know what flavor of regex you're using, and \d allows non-Arabic numerals in some flavors (if that concerns you).
Can I use
\d\d\d\d[^\d]
to check for four consecutive numbers?
For example,
411112 OK
455553 OK
1200003 OK
f44443 OK
g55553 OK
3333 OK
f4442 No
45553 No
f4444g4444 No
f44444444 No
If you want to find any series of 4 digits in a string /\d\d\d\d/ or /\d{4}/ will do. If you want to find a series of exactly 4 digits, use /[^\d]\d{4}[^\d]/. If the string should simply contain 4 consecutive digits use /^\d{4}$/.
Edit: I think you want to find 4 of the same digits, you need a backreference for that. /(\d)\1{3}/ is probably what you're looking for.
Edit 2: /(^|(.)(?!\2))(\d)\3{3}(?!\3)/ will only match strings with exactly 4 of the same consecutive digits.
The first group matches the start of the string or any character. Then there's a negative look-ahead that uses the first group to ensure that the following characters don't match the first character, if any. The third group matches any digit, which is then repeated 3 times with a backreference to group 3. Finally there's a look-ahead that ensures that the following character doesn't match the series of consecutive digits.
This sort of stuff is difficult to do in javascript because you don't have things like forward references and look-behind.
Should the numbers be part of a string, or do you want only the four numbers. In the later case, the regexp should be ^\d{4}$. The ^ marks the beginning of the string, $ the end. That makes sure, that only four numbers are valid, and nothing before or after that.
That should match four digits (\d\d\d\d) followed by a non digit character ([^\d]). If you just want to match any four digits, you should used \d\d\d\d or \d{4}. If you want to make sure that the string contains just four consecutive digits, use ^\d{4}$. The ^ will instruct the regex engine to start matching at the beginning of the string while the $ will instruct the regex engine to stop matching at the end of the string.