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Java/Regex: Negate the same character repeated twice:

If I have the following string:
--l=Richmond-Hill, NYC --m=5-day --d=hourly
I want to match the groups:
--l=Richmond-Hill, NYC
--m=5-day
--d=hourly
I came up with the following regex:
(^--[a-zA-Z]=[^-]*)
This works when the value after the equal sign doesn't have a dash. Basically, how do I negate a double dash ?

My guess is that maybe you wish to design some expression similar to:
--[a-zA-Z]=.*?(?=--|$)
Demo
Test
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class re{
public static void main(String[] args){
final String regex = "--[a-zA-Z]=.*?(?=--|$)";
final String string = "--l=Richmond-Hill, NYC --m=5-day --d=hourly\n"
+ "--l=Richmond-Hill, NYC\n"
+ "--m=5-day\n"
+ "--d=hourly";
final Pattern pattern = Pattern.compile(regex, Pattern.MULTILINE);
final Matcher matcher = pattern.matcher(string);
while (matcher.find()) {
System.out.println("Full match: " + matcher.group(0));
for (int i = 1; i <= matcher.groupCount(); i++) {
System.out.println("Group " + i + ": " + matcher.group(i));
}
}
}
}
Output
Full match: --l=Richmond-Hill, NYC
Full match: --m=5-day
Full match: --d=hourly
Full match: --l=Richmond-Hill, NYC
Full match: --m=5-day
Full match: --d=hourly
If you wish to explore/simplify/modify the expression, it's been
explained on the top right panel of
regex101.com. If you'd like, you
can also watch in this
link, how it would match
against some sample inputs.
RegEx Circuit
jex.im visualizes regular expressions:

Alternatively, you can also split your string at -- while keeping the delimiters and triming the spaces:
Pattern.compile("(?=--)")
.splitAsStream("--l=Richmond-Hill, NYC --m=5-day --d=hourly")
.map(String::trim)
.forEach(System.out::println);

How do I create a regex for this text?

I need to create a regex that checks if the text follows this format:
The first two letters will always be 'AB' than it will be a number
between 1-9 than either A or B than a dash ('-') than a bunch of
random text followed by a colon (':') and then index position that is
A letter and 2 digit number.
So like this:
AB8B-ANYLETTERS:H12
or
AB3B-ANYTHINGCANGOHERE:A77
I have done this to check the index position but cannot figure out the text before the colon.
"^.*:[A-H]\\d\\d"
So the general format is:
AB[1-9][A or B]-[ANYCHARACTERS]:[A-Z][01-99]
I am using Java.

I'm guessing that maybe this expression might validate that:
^AB[1-9][AB]-[^:]+:[A-Z][0-9]{2}$
The expression is explained on the top right panel of regex101.com, if you wish to explore/simplify/modify it, and in this link, you can watch how it would match against some sample inputs, if you like.
Test
import java.util.regex.Matcher;
import java.util.regex.Pattern;
final String regex = "^AB[1-9][AB]-[^:]+:[A-Z][0-9]{2}$";
final String string = "AB8B-ANYLETTERS:H12\n"
+ "AB3B-ANYTHINGCANGOHERE:A77";
final Pattern pattern = Pattern.compile(regex, Pattern.MULTILINE);
final Matcher matcher = pattern.matcher(string);
while (matcher.find()) {
System.out.println("Full match: " + matcher.group(0));
for (int i = 1; i <= matcher.groupCount(); i++) {
System.out.println("Group " + i + ": " + matcher.group(i));
}
}
RegEx Circuit
jex.im visualizes regular expressions:
Edit
For AC cases, we would try:
^AB[1-9][AC]-[^:]+:[A-Z][0-9]{2}$
Demo 2

Regex to retrieve any alphanumeric otp code

I am trying to get a regex to retrieve alphanumeric OTP code (length of the code maybe dynamic i.e depending on user's choice) and must contain at least one digit.
I have tried the following regex :
"[a-zA-z][0-9].[a-zA-z]"
But if a special character is there in the code it should result null instead it retrieves the characters before and after the special character which is not desired.
Some sample OTP-containing messages on which the regex is desired to work successfully:
OTP is **** for txn of INR 78.90.
**** is your one-time password.
Hi, Your OTP is ****.
Examples of Alphanumeric OTPs with at least one-digit:
78784
aZ837
987Ny
19hd35
fc82pl

It would be a bit difficult, maybe this expression might work with an i flag:
[a-z0-9]*\d[a-z0-9]*
or with word boundaries:
(?<=\b)[a-z0-9]*\d[a-z0-9]*(?=\b)
Test
import java.util.regex.Matcher;
import java.util.regex.Pattern;
final String regex = "[a-z0-9]*\\d[a-z0-9]*";
final String string = "78784\n"
+ "aZ837\n"
+ "987Ny\n"
+ "19hd35\n"
+ "fc82pl";
final Pattern pattern = Pattern.compile(regex, Pattern.MULTILINE | Pattern.CASE_INSENSITIVE);
final Matcher matcher = pattern.matcher(string);
while (matcher.find()) {
System.out.println("Full match: " + matcher.group(0));
for (int i = 1; i <= matcher.groupCount(); i++) {
System.out.println("Group " + i + ": " + matcher.group(i));
}
}
The expression is explained on the top right panel of regex101.com, if you wish to explore/simplify/modify it, and in this link, you can watch how it would match against some sample inputs, if you like.

Try this one
[0-9a-zA-Z]*[0-9]+[0-9a-zA-Z]*
This evaluates your desired result.
I have tested this in this site

Your regex is almost correct.
It should be \b[a-zA-z]*[0-9]+[a-zA-z0-9]*\b.

How can i split address in streetname and streetnumber in nifi?

I'm trying to split a given address (Muster Straße 114 a) in streetname and streetnumber. I'm working with nifi. The situation is the following: i have a FlowFile-Attribute (order_address) which has as FlowFile-content e.g Muster Straße 114 a, and i need to split it into sepereate attributes.
I tried
/\A\s*(?:?:\s*)?(\pN+[a-zA-Z]?(?:\s*[-\/\pP]\s*\pN+[a-zA-Z]?)*)\s*,?\s*(?P(?:[a-zA-Z]\s*|\pN\pL{2,}\s\pL)\S[^,#]*?(?<!\s))s*(?:(?:[,\/]|(?=\#))\s*(?!\s*\.(?P(?!\s).*?))? | ?:(?P.*?),\s*(?=.*[,\/]))??!\s*\.)(?P[^0-9#]\s*\S(?:[^,#](?!\b\pN+\s))*?(?<!\s))\s*[\/,]?\s*(?:\sNo[.:])?\s*(?P\pN+\s*-?[a-zA-Z]?(?:\s*[-\/\pP]?\s*\pN+(?:\s*[\-a-zA-Z])?)*|[IVXLCDM]+(?!.*\b\pN+\b))(?<!\s)\s*(?:(?:[,\/]|(?=\#)|\s)\s*(?!\s*No\.)\s*(?P(?!\s).*?))?)\s*\Z/xu
but it's not working for me

If we'd like to just separate our addresses into two parts, one including the digits and one without, we could find several expressions that'd cover this rule, such as:
(.*?)([\d].*)
Demo
Test
import java.util.regex.Matcher;
import java.util.regex.Pattern;
final String regex = "(.*?)([\\d].*)";
final String string = "Muster Straße 114 a";
final Pattern pattern = Pattern.compile(regex, Pattern.MULTILINE);
final Matcher matcher = pattern.matcher(string);
while (matcher.find()) {
System.out.println("Full match: " + matcher.group(0));
for (int i = 1; i <= matcher.groupCount(); i++) {
System.out.println("Group " + i + ": " + matcher.group(i));
}
}
RegEx Circuit
jex.im visualizes regular expressions:

In nifi, you can use the Nifi Expression language to manipulate FlowFile-Attributes. So i used the UpadateAttribute-Processor, to create the new FlowFile-Attributes street_name and streed_number.
I used the replaceAll method with a simple regex, to get streetnumber and streetname.
^(\D*)(?:.*)
^\D*(.*)
This two regex did it.
Here you find the screenshot of the processor:

Java Regex to remove all full stops apart from real numbers within a text

I am trying to write a simple regex to remove all . apart from the ones which occur in real numbers.
E.g.
The value was 0.19 psi. The water level has to be brought to normal. Mtl.temp is going to be high..
The below regex selects all real numbers.
((\+|-)?([0-9]+)(\.[0-9]+)?)|((\+|-)?\.?[0-9]+)
I could do the other way wherein I could select for pattern wherein it selects . preceded by a word and succeed by space. But, the input test is not written in proper grammatical manner.

you can use the regex
\.(?!\d)
regex101 demo
import java.util.regex.Matcher;
import java.util.regex.Pattern;
final String regex = "\\.(?!\\d)";
final String string = ".12 . 0.123 Hi.there I am .invalid.";
final Pattern pattern = Pattern.compile(regex);
final Matcher matcher = pattern.matcher(string);
while (matcher.find()) {
System.out.println("Full match: " + matcher.group(0));
for (int i = 1; i <= matcher.groupCount(); i++) {
System.out.println("Group " + i + ": " + matcher.group(i));
}
}