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Regex for Alternating Numbers

I am trying to write a regex pattern for phone numbers consisting of 9 fixed digits.
I want to identify numbers that have two numbers alternating for four times such as 5XYXYXYXY
I used the below sample
number = 561616161
I tried the below pattern but it is not accurate
^5(\d)(?=\d\1).+
can someone point out what i am doing wrong?

I would use:
^(?=\d{9}$)\d*(\d)(\d)(?:\1\2){3}\d*$
Demo
Here is an explanation of the pattern:
^ from the start of the number
(?=\d{9}$) assert exactly 9 digits
\d* match optional leading digits
(\d) capture a digit in \1
(\d) capture another digit in \2
(?:\1\2){3} match the XY combination 3 more times
\d* more optional digits
$ end of the number

If you want to repeat 4 times 2 of the same pairs and matching 9 digits in total:
^(?=\d{9}$)\d*(\d\d)\1{3}\d*$
Explanation
^ Start of string
(?=\d{9}$) Positive lookahead, assert 9 digits till the end of the string
\d* Match optional digits
(\d\d)\1{3} Capture group 1, match 2 digits and then repeat what is captured in group 1 3 times
\d* Match optional digits
$ End of string
Regex demo
If you want to match a pattern repeating 4 times 2 digits where the 2 digits are not the same:
^(?=\d{9}$)\d*((\d)(?!\2)\d)\1{3}\d*$
Explanation
^ Start of string
(?=\d{9}$) Positive lookahead, assert 9 digits till the end of the string
\d* Match optional digits
( Capture group 1
(\d) Capture group 2, match a single digit
(?!\2)\d Negative lookahead, assert not the same char as captured in group 2. If that is the case, then match a single digit
) Close group 1
\1{3} Repeat the captured value of capture group 1 3 times
\d* Match optional digits
$ End of string
Regex demo

My first guess from OP's self tried regex ^5(\d)(?=\d\1).+ without any own additions was a regex is needed to verify numbers starting with 5 and followed by 4 pairs of same two digits.
^5(\d\d)\1{3}$
Demo at regex101
The same idea with the "added guess" to disallow all same digits like e.g. 511111111
^5((\d)(?!\2)\d)\1{3}$
Demo at regex101
Guessing further that 5 is a variable value and assuming if one variable at start/end with the idea of taking out invalid values early - already having seen the other nice provided answers.
^(?=\d?(\d\d)\1{3})\d{9}$
Demo at regex101
Solution 3 with solution 2's assumption of two different digits in first pairing.
^(?=\d?((\d)(?!\2)\d)\1{3})\d{9}$
Demo at regex101
Solutions 3 and 4 are most obvious playings with #4thBird's nice answer in changed order.

How to match a particular URL pattern in Python regex

I am having trouble matching a pattern of this format: p#.g.com where # is not a 1 or a 2. For instance if the pattern is p1.g.com, I don't need to match. If it it p2.g.com, I don't need to match.
But if it is any other number, such as p3.g.com or p29.g.com, then I need to match.
My current pattern is r"(?P<url>p([^1,2])\.g\.com)", but this fails if the pattern is p##.g.com, basically any two digit number it fails on. There is no upper limit on the #, so it could be a 3 or 999 or anything in between.
I also tried r"(?P<url>p([^1,2])\d+\.g\.com)" but that does not match any number beginning with a 1 or a 2. For instance 11 or 23 are not matched, which I do want matched.

Try this regex:
p(?:[03-9]|\d{2,})\.g\.com
Demo
Explanation:
Matches character p
Start of non-capturing group
Match one of:
The digits 0 or 3-9
Any double digit number like 10 or higher
Matches character .g.com

Making Regex combination of multiple chars and numbers combined

How to write a regex that will combine numbers and chars in a string in any order?
For example, If I want to read some kind of invoice number, I have example like this:
example 1: 32ah3af
example 2: 32ahPP-A2ah3af
example 3: 3A63af-3HHx3B-APe5y5-9OPiis
example 4: 3A63af 3HHx3B APe5y5 9OPiis
So each 'block' have length between 3 and 7 chars (letters or numbers) that can be in any order (letters can be lowercase or uppercase). Each. 'block' can start with letter or with number.
It can have one "block" or max 4 blocks that are separated with ' ' or -.
I know that I can make separators like: \s or \-, but I have no idea how to make these kind of blocks that have (or do not have) separator.
I tried with something like this:
([0-9]?[A-z]?){3,7}
But it does not work

You could use
^[A-Za-z0-9]{3,7}(?:[ -][A-Za-z0-9]{3,7}){0,3}\b
The pattern matches:
^ Start of string
[A-Za-z0-9]{3,7} Match 3-7 times either a lower or uppercase char a-z or number 0-9
(?: Non capture group
[ -][A-Za-z0-9]{3,7} Match either a space or - and 3-7 times either a lower or uppercase char a-z or number 0-9
){0,3} Close the non capture group and repeat 0-3 times to have a maximum or 4 occurrences
\b A word boundary to prevent a partial match
Regex demo
Note that [A-z] matches more than [A-Za-z0-9]

As long as you want to only capture / search for the invoice ids, the suggestion from Hao Wu is valid:
r'\w{3,7}'
for regex (check here).
If you can drop the remaining part, then this should be enough.
You can more precisely capture the whole string with example 1:
r'example (\d+): ((\w{3,7}[\- ]?)+)'
See here how it works. Please note how capturing groups are represented.

Regex to extract first 5 digit+character from last hyphen

I am trying to extract first 5 character+digit from last hyphen.
Here is the example
String -- X008-TGa19-ER751QF7
Output -- X008-TGa19-ER751
String -- X002-KF13-ER782cPU80
Output -- X002-KF13-ER782
My attempt -- I could manage to take element from the last -- (\w+)[^-.]*$
But now how to take first 5, then return my the entire value as the output as shown in the example.

You can optionally repeat a - and 1+ word chars from the start of the string. Then match the last - and match 5 word chars.
^\w+(?:-\w+)*-\w{5}
^ Start of string
\w+ Math 1+ word chars
(?:-\w+)* Optionally repeat - and 1+ word chars
-\w{5} Match - and 5 word chars
Regex demo
import re
regex = r"^\w+(?:-\w+)*-\w{5}"
s = ("X008-TGa19-ER751QF7\n"
"X002-KF13-ER782cPU80")
print(re.findall(regex, s, re.MULTILINE))
Output
['X008-TGa19-ER751', 'X002-KF13-ER782']
Note that \w can also match _.
If there can also be other character in the string, to get the first 5 digits or characters except _ after the last hyphen, you can match word characters without an underscore using a negated character class [^\W_]{5}
Repeat that 5 times while asserting no more underscore at the right.
^.*-[^\W_]{5}(?=[^-]*$)
Regex demo

(\w+-\w+-\w{5}) seems to capture what you're asking for.
Example:
https://regex101.com/r/PcPSim/1

If you are open for non-regex solution, you can use this which is based on splitting, slicing and joining the strings:
>>> my_str = "X008-TGa19-ER751QF7"
>>> '-'.join(s[:5] for s in my_str.split('-'))
'X008-TGa19-ER751'
Here I am splitting the string based on hyphen -, slicing the string to get at max five chars per sub-string, and joining it back using str.join() to get the string in your desired format.

^(.*-[^-]{5})[^-]*$
Capture group 1 is what you need
https://regex101.com/r/SYz9i5/1
Explanation
^(.*-[^-]{5})[^-]*$
^ Start of line
( Capture group 1 start
.* Any number of any character
- hyphen
[^-]{5} 5 non-hyphen character
) Capture group 1 end
[^-]* Any number of non-hyphen character
$ End of line
Another simpler one is
^(.*-.{5}).*$
This should be quite straight-forward.
This is making use of behaviour greedy match of first .*, which will try to match as much as possible, so the - will be the last one with at least 5 character following it.
https://regex101.com/r/CFqgeF/1/

regex to find a pair of adjacent digits with different digits around them

I'm a beginner to regex and I am trying to make an expression to find if there are two of the same digits next to each other, and the digit behind and in front of the pair is different.
For example,
123456678 should match as there is a double 6,
1234566678 should not match as there is no double with different surrounding numbers.
12334566 should match because there are two 3s.
So far i have this which works only with 1, and as long as the double is not at the start or end of the string, however I can deal with that by adding a letter at the start and end.
^.*([^1]11[^1]).*$
I know i can use [0-9] instead of the 1s but the problem is having them all be the same digit.
Thank you!

I have divided my answer into four sections.
The first section contains my solution to the problem. Readers interested in nothing else may skip the other sections.
The remaining three sections are concerned with identifying the pairs of equal digits that are preceded by a different digit and are followed by a different digit. The first of the three sections matches them; the other two capture them in a group.
I've included the last section because I wanted to share The Greatest Regex Trick Ever with those unfamiliar with it, because I find it so very cool and clever, yet simple. It is documented here. Be forewarned that, to build suspense, the author at that link has included a lengthy preamble before the drum-roll reveal.
Determine if a string contains two consecutive equal digits that are preceded by a different digit and are followed by a different digit
You can test the string as follows:
import re
r = r'(\d)(?!\1)(\d)\2(?!\2)\d'
arr = ["123456678", "1123455a666788"]
for s in arr:
print(s, bool(re.search(r, s)) )
displays
123456678 True
1123455a666788 False
Run Python code | Start your engine!1
The regex engine performs the following operations.
(\d) : match a digit and save to capture group 1 (preceding digit)
(?!\1) : next character cannot equal content of capture group 1
(\d) : match a digit in capture group 2 (first digit of pair)
\2 : match content of capture group 2 (second digit of pair)
(?!\2) : next character cannot equal content of capture group 2
\d : match a digit
(?!\1) and (?!\2) are negative lookaheads.
Use Python's regex module to match pairs of consecutive digits that have the desired property
You can use the following regular expression with Python’s regex module to obtain the matching pairs of digits.
r'(\d)(?!\1)\K(\d)\2(?=\d)(?!\2)'
Regex Engine
The regex engine performs the following operations.
(\d) : match a digit and save to capture group 1 (preceding digit)
(?!\1) : next character cannot equal content of capture group 1
\K : forget everything matched so far and reset start of match
(\d) : match a digit in capture group 2 (first digit of pair)
\2 : match content of capture group 2 (second digit of pair)
(?=\d) : next character must be a digit
(?!\2) : next character cannot equal content of capture group 2
(?=\d) is a positive lookahead. (?=\d)(?!\2) could be replaced with (?!\2|$|\D).
Save pairs of consecutive digits that have the desired property to a capture group
Another way to obtain the matching pairs of digits, which does not require the regex module, is to extract the contents of capture group 2 from matches of the following regular expression.
r'(\d)(?!\1)((\d)\3)(?!\3)(?=\d)'
Re engine
The following operations are performed.
(\d) : match a digit in capture group 1
(?!\1) : next character does not equal last character
( : begin capture group 2
(\d) : match a digit in capture group 3
\3 : match the content of capture group 3
) : end capture group 2
(?!\3) : next character does not equal last character
(?=\d) : next character is a digit
Use The Greatest Regex Trick Ever to identify pairs of consecutive digits that have the desired property
We use the following regular expression to match the string.
r'(\d)(?=\1)|\d(?=(\d)(?!\2))|\d(?=\d(\d)\3)|\d(?=(\d{2})\d)'
When there is a match, we pay no attention to which character was matched, but examine the content of capture group 4 ((\d{2})), as I will explain below.
The Trick in action
The first three components of the alternation correspond to the ways that a string of four digits can fail to have the property that the second and third digits are equal, the first and second are unequal and the third and fourth are equal. They are:
(\d)(?=\1) : assert first and second digits are equal
\d(?=(\d)(?!\2)) : assert second and third digits are not equal
\d(?=\d(\d)\3) : assert third and fourth digits are equal
It follows that if there is a match of a digit and the first three parts of the alternation fail the last part (\d(?=(\d{2})\d)) must succeed, and the capture group it contains (#4) must contain the two equal digits that have the required properties. (The final \d is needed to assert that the pair of digits of interest is followed by a digit.)
If there is a match how do we determine if the last part of the alternation is the one that is matched?
When this regex matches a digit we have no interest in what digit that was. Instead, we look to capture group 4 ((\d{2})). If that group is empty we conclude that one of the first three components of the alternation matched the digit, meaning that the two digits following the matched digit do not have the properties that they are equal and are unequal to the digits that precede and follow them.
If, however, capture group 4 is not empty, it means that none of the first three parts of the alternation matched the digit, so the last part of the alternation must have matched and the two digits following the matched digit, which are held in capture group 4, have the desired properties.
1. Move the cursor around for detailed explanations.

With regex, it is much more convenient to use a PyPi regex module with the (*SKIP)(*FAIL) based pattern:
import regex
rx = r'(\d)\1{2,}(*SKIP)(*F)|(\d)\2'
l = ["123456678", "1234566678"]
for s in l:
print(s, bool(regex.search(rx, s)) )
See the Python demo. Output:
123456678 True
1234566678 False
Regex details
(\d)\1{2,}(*SKIP)(*F) - a digit and then two or more occurrences of the same digit
| - or
(\d)\2 - a digit and then the same digit.
The point is to match all chunks of identical 3 or more digits and skip them, and then match a chunk of two identical digits.
See the regex demo.

Inspired by the answer or Wiktor Stribiżew, another variation of using an alternation with re is to check for the existence of the capturing group which contains a positive match for 2 of the same digits not surrounded by the same digit.
In this case, check for group 3.
((\d)\2{2,})|\d(\d)\3(?!\3)\d
Regex demo | Python demo
( Capture group 1
(\d)\2{2,} Capture group 2, match 1 digit and repeat that same digit 2+ times
) Close group
| Or
\d(\d) Match a digit, capture a digit in group 3
\3(?!\3)\d Match the same digit as in group 3. Match the 4th digit, but is should not be the same as the group 3 digit
For example
import re
pattern = r"((\d)\2{2,})|\d(\d)\3(?!\3)\d"
strings = ["123456678", "12334566", "12345654554888", "1221", "1234566678", "1222", "2221", "66", "122", "221", "111"]
for s in strings:
match = re.search(pattern, s)
if match and match.group(3):
print ("Match: " + match.string)
else:
print ("No match: " + s)
Output
Match: 123456678
Match: 12334566
Match: 12345654554888
Match: 1221
No match: 1234566678
No match: 1222
No match: 2221
No match: 66
No match: 122
No match: 221
No match: 111
If for example 2 or 3 digits only is also ok to match, you could check for group 2
(\d)\1{2,}|(\d)\2
Python demo

You can also use a simple way .
import re
l=["123456678",
"1234566678",
"12334566 "]
for i in l:
matches = re.findall(r"((.)\2+)", i)
if any(len(x[0])!=2 for x in matches):
print "{}-->{}".format(i, False)
else:
print "{}-->{}".format(i, True)
You can customize this based on you rules.
Output:
123456678-->True
1234566678-->False
12334566 -->True