I need to extract the last number that is inside a string. I'm trying to do this with regex and negative lookaheads, but it's not working. This is the regex that I have:
\d+(?!\d+)
And these are some strings, just to give you an idea, and what the regex should match:
ARRAY[123] matches 123
ARRAY[123].ITEM[4] matches 4
B:1000 matches 1000
B:1000.10 matches 10
And so on. The regex matches the numbers, but all of them. I don't get why the negative lookahead is not working. Any one care to explain?
Your regex \d+(?!\d+) says
match any number if it is not immediately followed by a number.
which is incorrect. A number is last if it is not followed (following it anywhere, not just immediately) by any other number.
When translated to regex we have:
(\d+)(?!.*\d)
Rubular Link
I took it this way: you need to make sure the match is close enough to the end of the string; close enough in the sense that only non-digits may intervene. What I suggest is the following:
/(\d+)\D*\z/
\z at the end means that that is the end of the string.
\D* before that means that an arbitrary number of non-digits can intervene between the match and the end of the string.
(\d+) is the matching part. It is in parenthesis so that you can pick it up, as was pointed out by Cameron.
You can use
.*(?:\D|^)(\d+)
to get the last number; this is because the matcher will gobble up all the characters with .*, then backtrack to the first non-digit character or the start of the string, then match the final group of digits.
Your negative lookahead isn't working because on the string "1 3", for example, the 1 is matched by the \d+, then the space matches the negative lookahead (since it's not a sequence of one or more digits). The 3 is never even looked at.
Note that your example regex doesn't have any groups in it, so I'm not sure how you were extracting the number.
I still had issues with managing the capture groups
(for example, if using Inline Modifiers (?imsxXU)).
This worked for my purposes -
.(?:\D|^)\d(\D)
Related
I need some help here
Here is example of what im trying to match:
1 ScreenMail Enable friendly none Internal any 5
I need to match everything excluding the last digits (5) Meaning matching the first digit(1), spaces, letter, special characters, etc I tried using /^(\d), but after matching the first digits, it stopped. Your assistance would be appreciated.
The simplest way is probably to remove last digits with:
\d+$
\d+\s*$
See the regex demo.
You may want to use a matching regex like
^.*[^\d\s]
that matches any zero or more chars other than line break chars (.*) as many as possible and then a char other than a digit and whitespace. See this regex demo.
However, if the digits are followed with an optional whitespace, or if you allow any text after the last digits, it will fail. You can then use
^.*[^\d\s](?=\s*\d)
See this regex demo. The (?=\s*\d) positive lookahead requires zero or more whitespaces and then a digit immediately to the right of the current location.
I need to process numbers that may have optional thousand-separators, such as 1234567 and 1,234,567
I naively assumed I could achieve this with
(\d{1,3}([,]?(\d{3}))*)
This, however, matches only 123456 (not the 7) and 1,234,567 (correctly)
However, if I specify an explicit number of matches (2 in this case)
(\d{1,3}([,]?(\d{3})){2})
or a bound (such as \b)
(\d{1,3}([,]?(\d{3}))*)\b
the full match is performed.
Why does the “greedy” * quantifier stop after the first match in the first regex?
If you want to match both numbers with, and without, proper comma thousands separators, then I would use an alternation:
^(\d{1,3}(?:,\d{3})*|\d+)$
Demo
The reason is that \d{1,3} is greedy, so it matches 123 at the beginning of the number. Then the rest of the regexp will only match groups of exactly 3 digits because it uses \d{3}. A regular expression doesn't try to match the longest possible string, so it won't backtrack and shorten the match for \d{1,3} to make the rest of the regexp go further.
But if you add a word boundary \b at the end, it no longer matches with that 3-digit prefix. That causes it to backtrack until it's able to match groups of 3 digits ending with a word boundary.
I have got the following regex expression so far:
used-cars\/((?:\d+[a-z]|[a-z]+\d)[a-z\d]*)
This is sort of working, I need it to match basically ANYTHING apart from JUST numbers after used-cars/
Match:
used-cars/page-1
used-cars/1eeee
used-cars/page-1?*&_-
Not Match:
used-cars/2
used-cars/400
Can someone give me a hand? Been trying get this working for a while now!
There are few shortcomings of your regex used-cars\/((?:\d+[a-z]|[a-z]+\d)[a-z\d]*).
It's checking for used-cars/ followed by multiple digits then one character within a-z OR multiple characters within a-z then one digit.
[a-z\d]* is searching for either characters or digits which is also optional.
It's inaccurate for your pattern.
Try with following regex.
Regex: ^used-cars\/(?!\d+$)\S*$
Explanation:
used-cars\/ searches for literal used-cars/
(?!\d+$) is negative lookahead for only digits till end. If only digits are present then it won't be a match.
\S* matches zero or more characters other than whitespace.
Regex101 Demo
Using Regular Expression,
from any line of input that has at least one word repeated two or more times.
Here is how far i got.
/(\b\w+\b).*\1
but it is wrong because it only checks for single char, not one word.
input: i might be ill
output: < i might be i>ll
<> marks the matched part.
so, i try to do (\b\w+\b)(\b\w+\b)*\1
but it is not working totally.
Can someone give help?
Thanks.
this should work
(\b\w+\b).*\b\1\b
greedy algorithm will ensure longest match. If you want second instance to be a separate word you have to add the boundaries there as well. So it's the same as
\b(\w+)\b.*\b\1\b
Positive lookahead is not a must here:
/\b([A-Za-z]+)\b[\s\S]*\b\1\b/g
EXPLANATION
\b([A-Za-z]+)\b # match any word
[\s\S]* # match any character (newline included) zero or more times
\b\1\b # word repeated
REGEX 101 DEMO
To check for repeated words you can use positive lookahead like this.
Regex: (\b[A-Za-z]+\b)(?=.*\b\1\b)
Explanation:
(\b[A-Za-z]+\b) will capture any word.
(?=.*\b\1\b) will lookahead if the word captured by group is present or not. If yes then a match is found.
Note:- This will produce repeated results because the word which is matched once will again be matched when regex pointer captures it as a word.
You will have to use programming to strip off the repeated results.
Regex101 Demo
I want to match the following rules:
One dash is allowed at the start of a number.
Only values between 0 and 9 should be allowed.
I currently have the following regex pattern, I'm matching the inverse so that I can thrown an exception upon finding a match that doesn't follow the rules:
[^-0-9]
The downside to this pattern is that it works for all cases except a hyphen in the middle of the String will still pass. For example:
"-2304923" is allowed correctly but "9234-342" is also allowed and shouldn't be.
Please let me know what I can do to specify the first character as [^-0-9] and the rest as [^0-9]. Thanks!
This regex will work for you:
^-?\d+$
Explanation: start the string ^, then - but optional (?), the digit \d repeated few times (+), and string must finish here $.
You can do this:
(?:^|\s)(-?\d+)(?:["'\s]|$)
^^^^^ non capturing group for start of line or space
^^^^^ capture number
^^^^^^^^^ non capturing group for end of line, space or quote
See it work
This will capture all strings of numbers in a line with an optional hyphen in front.
-2304923" "9234-342" 1234 -1234
++++++++ captured
^^^^^^^^ NOT captured
++++ captured
+++++ captured
I don't understand how your pattern - [^-0-9] is matching those strings you are talking about. That pattern is just the opposite of what you want. You have simply negated the character class by using caret(^) at the beginning. So, this pattern would match anything except the hyphen and the digits.
Anyways, for your requirement, first you need to match one hyphen at the beginning. So, just keep it outside the character class. And then to match any number of digits later on, you can use [0-9]+ or \d+.
So, your pattern to match the required format should be:
-[0-9]+ // or -\d+
The above regex is used to find the pattern in some large string. If you want the entire string to match this pattern, then you can add anchors at the ends of the regex: -
^-[0-9]+$
For a regular expression like this, it's sometimes helpful to think of it in terms of two cases.
Is the first character messed up somehow?
If not, are any of the other characters messed up somehow?
Combine these with |
(^[^-0-9]|^.+?[^0-9])