Develop Reference

how to create a regular expression to validate the numbers are not same even separated by hyphen (-)

Using the following regex
^(\d)(?!\1+$)\d{3}-\d{1}$
It works for the pattern but I need to validate that all numbers are not the same even after /separated by the hyphen (-).
Example:
0000-0 not allowed (because of all are same digits)
0000-1 allowed
1111-1 not allowed (because of all are same digits)
1234-2 allowed

TheFourthBird's answer surely works that uses a negative lookahead. Here is another variant of this regex that might be slightly faster:
^(\d)(?!\1{3}-\1$)\d{3}-\d$
RegEx Demo
Explanation:
^(\d) matches and captures first digit after start in group #1
(?!\1{3}-\1$) is a negative lookahead that will fail the match if we have 3 repetitions and a hyphen and another repeat of 1st digit.

You could exclude only - or the same digit only to the right till the end of the string:
^(\d)(?!(?:\1|-)*$)\d{3}-\d$
^ Start of string
(\d) Capture group 1, match a digit
(?! Negative lookahead, assert what is to the right is not
(?:\1|-)*$ Optionally repeat either the backrefernce to what is already captured or - till the end of the string
) Close the non capture group
\d{3}-\d Match 3 digits - and a digit
$ End of string
Regex demo
If you don't want to match double -- or an - at the end of the string and match optional repetitions:
^(\d)(?!(?:\1|-)*$)\d*(?:-\d+)*$
Explanation
^ Start of string
(\d) Capture a single digits in group 1
(?!(?:\1|-)*$) Negative lookahead, assert not only - and the same digit till the end of the string
\d* Match optional digits
(?:-\d+)* Optionally repeat matching - and 1+ digits
$ End of string
Regex demo

You'll need a back reference, for example:
^(\d){4}-\1$

Regex to allow a space instead of following numbers after first two letters

I need a RegEx that allow a single space after two letters i.e. AB123 should not be allowed but AB 123 should be allowed ?

Here is the regex [a-zA-Z]{2}\s\S*
[a-zA-Z] means character from a to Z
{2} means character twice
\s means white space
\S means non white space.
* duplicate with 0 or more
https://regex101.com/r/uWYci4/1

This pattern will do the work: ^[a-zA-Z]{2} \d+$
Explanation:
^ - match beginning of a string
[a-zA-Z]{2} - match two letters (upper- or lowercase),
- match space
\d+ - match one or more digits
$ - match end of a string
Demo

Regex match numbers with spaces but not without spaces

Trying to match a string of numbers with spaces in between while ignoring other strings of numbers without spaces in between them. I'd like to match 16 characters.
eg. Would like to match 12345 67890 1234 but NOT 1234567890123456
I have tried this:
[0-9 ]{16}
But this matches both sets of strings.

I used and corrected #Wiktor Stribiżew regex, because original regex will match a space at the beginning and the end of the number.
Regex: \b(?![0-9]{16})\d[0-9 ]{14}\d\b
Details:
\b assert position at a word boundary (^\w|\w$|\W\w|\w\W)
(?!) Negative Lookahead
[] Match a single character present in the list 0-9
{n} Matches exactly n times
\d matches a digit (equal to [0-9])
RegEx demo

You can use this regex to enforcees at least one space in between numbers:
\d+(?:\h+\d+)+
RegEx Demo
\d+: Match 1+ digits
(?:\h+\d+)+: Match 1+ group of 1+ whitespace and 1+ digits

Java Regular Expression (greedy/nongreedy)

So I'm trying to separate the following two groups formatted as:
FIRST - GrouP second.group.txt
The first group can contain any character
The second group is a dot(.) delimited string.
I'm using the following regex to separate these two groups:
([A-Z].+).*?([a-z]+\.[a-z]+)
However, it gives a wrong result:
1: FIRST - GrouP second.grou
2: p.txt
I don't understand because I'm using "nongreedy" separater (.*?) instead of the greedy one (. *)
What am I doing wrong here?
Thanks

You can this regex to match both groups:
\b([A-Z].+?)\s*\b([a-z]+(?:\.[a-z]+)+)\b
RegEx Demo
Breakup:
\b # word boundary
([A-Z].+?) # match [A-Z] followed by 1 or more chars (lazy)
\s* # match 0 or more spaces
\b # word boundary
([a-z]+ # match 1 or more of [a-z] chars
(?:\.[a-z]+)+) # match a group of dot followed by 1 or more [a-z] chars
\b # word boundary
PS: (?:..) is used for non-capturing group.

This is one possible solution that should be pretty compact:
(.*?-\s*\S+)|(\S+\.?)+
https://regex101.com/r/iW8mE5/1
It is looking for anything followed by a dash, zero or more spaces, and then non-whitespace characters. And if it doesn't find that, it looks for non-whitespace followed by an optional decimal.

regex pattern to match specific number pattern , skip if there different pattern

Requirement:
If pattern 57XXXXXXX OR 57XXXXXXX-X found in a sentence , then copy this matched pattern (X- denotes 7 integer number and 57 are constant values must be there), else ignore complete sentence.
I have written a regex pattern 57[0-9]{7}|-[0-9]{1} to do match both the pattern.
If below pattern found(8 digits after 57 instead 7 , then still above regex still gets the matching pattern (actually expecting regex to not match)
for e.g. 5712345678-0 (after 57 , 8 digits in sentance) --> regex matches and gives 571234567-0
Using java to compile above pattern.

You could try this:
\b57\d{7}(?:-\d)?\b
Here's what it looks like:
In Java, that would be Pattern.compile("\\b57\\d{7}(?:-\\d)?\\b").

Not very different but allows letters and underscores around:
(?:(?<=[^0-9])|^)57[0-9]{7}(?:-[0-9])?(?:(?=[^0-9])|$)

You could use lookahead assertions in this case.
57\d{7}(?:-\d)?(?!\d)
Regular expression:
57 '57'
\d{7} digits (0-9) (7 times)
(?: group, but do not capture:
- '-'
\d digits (0-9)
)? end of grouping
(?! look ahead to see if there is not:
\d digits (0-9)
) end of look-ahead
Or:
(?:57\d{7})(?:-\d)?(?!\d)
Regular expression:
(?: group, but do not capture:
57 '57'
\d{7} digits (0-9) (7 times)
) end of grouping
(?: group, but do not capture
- '-'
\d digits (0-9)
)? end of grouping
(?! look ahead to see if there is not:
\d digits (0-9)
) end of look-ahead