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Why is this regex not matching URLs?

I have the following regex:
^(?=\w+)(-\w+)(?!\.)
Which I'm attempting to match against the following text:
www-test1.examples.com
The regex should match only the -test1 part of the string and only if it is before the first .and after the start of the expression. www can be any string but it should not be matched.
My pattern is not matching the -test1 part. What am I missing?

Java is one of the only languages that support non-fixed-length look-behinds (which basically means you can use quantifiers), so you can technically use the following:
(?<=^\w+)(-\w+)
This will match for -test without capturing the preceding stuff. However, it's generally not advisable to use non-fixed-length look-behinds, as they are not perfect, nor are they very efficient, nor are they portable across other languages. Having said that.. this is a simple pattern, so if you don't care about portability, sure, go for it.
The better solution though is to group what you want to capture, and reference the captured group (in this case, group 1):
^\w+(-\w+)
p.s. - \w will not match a dot, so no need to look ahead for it.
p.p.s. - to answer your question about why your original pattern ^(?=\w+)(-\w+)(?!\.) doesn't match. There are 2 reasons:
1) you start out with a start of string assertion, and then use a lookahead to see if what follows is one or more word chars. But lookaheads are zero-width assertions, meaning no characters are actually consumed in the match, so the pointer doesn't move forward to the next chars after the match. So it sees that "www" matches it, and moves on to the next part of the pattern, but the actual pointer hasn't moved past the start of string. So, it next tries to match your (-\w+) part. Well your string doesn't start with "-" so the pattern fails.
2) (?!\.) is a negative lookahead. Well your example string shows a dot as the very next thing after your "-test" part. So even if #1 didn't fail it, this would fail it.

The problem you're having is the lookahead. In this case, it's inappropriate if you want to capture what's between the - and the first .. The pattern you want is something like this:
(-\w+)(?=\.)
In this case, the contents of capture group 1 will contain the text you want.
Demo on Regex101

Try this:
(?<=www)\-\w+(?=\.)
Demo: https://regex101.com/r/xEpno7/1

regex with lookbehind weird behavior

I have been trying to resolve this for the past 2 days...
Please help me in understanding why this is happening. My intention is to just select the <HDR> that has a <DTL1 val="92">.....</HDR>
This is my regular expression
(?<=<HDR>).*?<DTL1\sval="3".*?</HDR>
And the input string is:
<HDR>abc<DTL1 val="1"><DTL2 val="2"></HDR><HDR><DTL1 val="92"><DTL2 val="55"></HDR><HDR><DTL1 val="3"><DTL2 val="4"></HDR>
But this regular expression selects
abc<DTL1 val="1"><DTL2 val="2"></HDR><HDR><DTL1 val="92"><DTL2 val="55"></HDR>
Can anyone please help me?

A regex engine will give you always the leftmost match in a string (even if you use a non-greedy quantifier). This is exactly what you obtain.
So, a solution is to forbid the presence of another <HDR> in the parts described by .*? that is too permissive.
You have two technics to do that, you can replace the .*? with:
(?>[^<]+|<(?!/HDR))*
or with:
(?:(?!</HDR).)*+
Most of the time, the first technic is more performant, but if your string contains an high density of <, the second way can give good results too.
The use of a possessive quantifier or an atomic group can reduce the number of steps to obtain a result in particular when the subpattern fails.
Example:
With the first way:
(?<=<HDR>)(?>[^<]+|<(?!/HDR))*<DTL1\sval="3"(?>[^<]+|<(?!/HDR))*</HDR>
or this variant:
(?<=<HDR>)(?:[^<]+|<(?!/HDR|DTL1))*+<DTL1\sval="3"(?:[^<]+|<(?!/HDR))*+</HDR>
With the second way:
(?<=<HDR>)(?:(?!</HDR).)*<DTL1\sval="3"(?:(?!</HDR).)*+</HDR>
or this variant:
(?<=<HDR>)(?:(?!</HDR|DTL1).)*+<DTL1\sval="3"(?:(?!</HDR).)*+</HDR>

Casimir et Hippolyte already gave you a couple of good solutions. I want to elaborate on a few things.
First, why your regex fails to do what you want: (?<=<HDR>).*? tells it to match any number of characters starting with the first character preceded by <HDR>, until it encounters what follows the non-greedy quantifier (<DTL1...). Well, the first character that's preceded by <HDR> is the first a, so it matches everything starting from there until the fixed string <DTL1\sval="3" is encountered.
Casimir et Hippolyte's solutions are for the generalized case, where the contents of the <HDR> tags can be anything other than nested <HDR>'s. You could also do it with a positive look-ahead:
(?<=<HDR>)(.(?!</HDR>))*<DTL1\sval="3".*?</HDR>
However, if the string is guaranteed to be in the structure shown, where the <HDR> tags only contain one or more <DTL1 val="##"> tags, so you know there won't be any closing tags within, you could do it more efficiently by replacing the first .*? with [^/]*:
(?<=<HDR>)[^/]*<DTL1\sval="3".*?</HDR>
A negated character class is more efficient than a zero-width assertion, and if you're using a negated character class, a greedy quantifier becomes more efficient than a lazy one.
Note also that by using a lookbehind to match the opening <HDR>, you're excluding it from the match, but you're including the closing </HDR>. Are you sure that's what you want? You're matching this...
<DTL1 val="3"><DTL2 val="4"></HDR>
...when presumably you want this...
<HDR><DTL1 val="3"><DTL2 val="4"></HDR>
...or this...
<DTL1 val="3"><DTL2 val="4">
So, in the fist case, don't use a lookbehind for the opening tag:
<HDR>(.(?!</HDR>))*<DTL1\sval="3".*?</HDR>
<HDR>[^/]*<DTL1\sval="3".*?</HDR>
In the second case, use a look-ahead for the closing tag:
(?<=<HDR>)(.(?!</HDR>))*<DTL1\sval="3".*?(?=</HDR>)
(?<=<HDR>)[^/]*<DTL1\sval="3".*?(?=</HDR>)

Regex index 0 how it exactly works

By compiling the following:
System.out.println(Pattern.matches(".?(\\d)$","3"));
It returns true because before 3 there is nothing and ? check for a one or zero.
However 3 is already the first character of the input which starts at 0 and end at 1. How can the jvm recognize that there is nothing before 3.
For example the following.
System.out.println(Pattern.matches(".*","hello");
It returns true as well but only the very last character gets matched with "nothing".
There should not be a "nothing" character at the beginning of a string, only at the end of it right?

This is not really about the JVM. This is about Java regular expressions.
The regular expression ".*" means "match 0 or more characters". It's easy to satisfy this, since a blank string has 0 characters, and therefore satisfies this. Whether Java regular expressions will choose to be lazy and match an empty string, or to be greedy and match the entire string depends on the implementation of Java regular expressions. If you read this excellent writeup (http://docs.oracle.com/javase/tutorial/essential/regex/quant.html) you can see that patterns like ".*" in Java are considered "reluctant" quantifiers and will prefer to take as little as possible.
Based on the information in that writeup, you can see that a pattern like ".{0,}" is a greedy version of the same expression. Perhaps you'd like to use that instead if this is truly a problem for you.

You are not interpreting your regex correctly. There is no such thing as a "nothing character" . Rather, your pattern reads: any charachter followed by a digit at the end of the string OR a digit at the end of the string.
And surely, "3" fits the second description very well.

matches method tries to match the input exactly.
so there's no need to use ^,$..

validating input string "RX-EZ12345678912345B" using regex

I need to validate input string which should be in the below format:
<2_upper_case_letters><"-"><2_upper_case_letters><14-digit number><1_uppercase_letter>
Ex: RX-EZ12345678912345B
I tried something like this ^[IN]-?[A-Z]{0,2}?\\d{0,14}[A-Z]{0,1} but its not giving the expected result.
Any help will be appreciated.
Thanks

Your biggest problem is the [IN] at the beginning, which matches only one letter, and only if it's I or N. If you want to match two of any letters, use [A-Z]{2}.
Once you fix that, your regex will still only match RX-E. That's because [A-Z]{0,2}? starts out trying to consume nothing, thanks to the reluctant quantifier, {0,2}?. Then \d{0,14} matches zero digits, and [A-Z]{0,1} greedily consumes the E.
If you want to match exactly 2 letters and 14 digits, use [A-Z]{2} and \d{14}. And since you're validating the string, you should end the regex with the end anchor, $. Result:
^[A-Z]{2}-[A-Z]{2}\d{14}[A-Z]$
...or, as a Java string literal:
"^[A-Z]{2}-[A-Z]{2}\\d{14}[A-Z]$"
As #nhahtdh observed, you don't really have to use the anchors if you're using Java's matches() method to apply the regex, but I recommend doing so anyway. It communicates your intent better, and it makes the regex portable, in case you have to use it in a different flavor/context.
EDIT: If the first two characters should be exactly IN, it would be
^IN-[A-Z]{2}\d{14}[A-Z]$

Simply translating your requirements into a java regex:
"^[A-Z]{2}-[A-Z]{2}\\d{14}[A-Z]$"
This will allow you to use:
if (!input.matches("^[A-Z]{2}-[A-Z]{2}\\d{14}[A-Z]$")) {
// do something because input is invalid
}

Not sure what you are trying to do at the beginning of your current regex.
"^[A-Z]{2}-[A-Z]{2}\\d{14}[A-Z]$"
The regex above will strictly match the input string as you specified. If you use matches function, ^ and $ may be omitted.
Since you want exact number of repetitions, you should specify it as {<number>} only. {<number>,<number>} is used for variable number of repetitions. And ? specify that the token before may or may not appear - if it must be there, then specifying ? is incorrect.

^[A-Z]{2}-[A-Z]{2}\\d{14}[A-Z]$
This should solve your purpose. You can confirm it from here

This should solve your problem. Check out the validity here
^[A-Z]{2}-[A-Z]{2}[0-9]{14}[A-Z]$

^([A-Z]{2,2}[-]{1,1}[A-Z]{2,2}[0-9]{14,14}[A-Z]{1,1}){1,1}$

Why doesn't this Java regular expression work?

I need to create a regular expression that allows a string to contain any number of:
alphanumeric characters
spaces
(
)
&
.
No other characters are permitted. I used RegexBuddy to construct the following regex, which works correctly when I test it within RegexBuddy:
\w* *\(*\)*&*\.*
Then I used RegexBuddy's "Use" feature to convert this into Java code, but it doesn't appear to work correctly using a simple test program:
public class RegexTest
{
public static void main(String[] args)
{
String test = "(AT) & (T)."; // Should be valid
System.out.println("Test string matches: "
+ test.matches("\\w* *\\(*\\)*&*\\.*")); // Outputs false
}
}
I must admit that I have a bit of a blind spot when it comes to regular expressions. Can anyone explain why it doesn't work please?

That regular expression tests for any amount of whitespace, followed by any amount of alphanumeric characters, followed by any amount of open parens, followed by any amount of close parens, followed by any amount of ampersands, followed by any amount of periods.
What you want is...
test.matches("[\\w \\(\\)&\\.]*")
As mentioned by mmyers, this allows the empty string. If you do not want to allow the empty string...
test.matches("[\\w \\(\\)&\\.]+")
Though that will also allow a string that is only spaces, or only periods, etc.. If you want to ensure at least one alpha-numeric character...
test.matches("[\\w \\(\\)&\\.]*\\w+[\\w \\(\\)&\\.]*")
So you understand what the regular expression is saying... anything within the square brackets ("[]") indicates a set of characters. So, where "a*" means 0 or more a's, [abc]* means 0 or more characters, all of which being a's, b's, or c's.

Maybe I'm misunderstanding your description, but aren't you essentially defining a class of characters without an order rather than a specific sequence? Shouldn't your regexp have a structure of [xxxx]+, where xxxx are the actual characters you want ?

The difference between your Java code snippet and the Test tab in RegexBuddy is that the matches() method in Java requires the regular expression to match the whole string, while the Test tab in RegexBuddy allows partial matches. If you use your original regex in RegexBuddy, you'll see multiple blocks of yellow and blue highlighting. That indicates RegexBuddy found multiple partial matches in your string. To get a regex that works as intended with matches(), you need to edit it until the whole test subject is highlighted in yellow, or if you turn off highlighting, until the Find First button selects the whole text.
Alternatively, you can use the anchors \A and \Z at the start and the end of your regex to force it to match the whole string. When you do that, your regex always behaves in the same way, whether you test it in RegexBuddy, or whether you use matches() or another method in Java. Only matches() requires a full string match. All other Matcher methods in Java allow partial matches.

the regex
\w* *\(*\)*&*\.*
will give you the items you described, but only in the order you described, and each one can be as many as wanted. So "skjhsklasdkjgsh((((())))))&&&&&....." works, but not mixing the characters.
You want a regex like this:
\[\w\(\)\&\.]+\
which will allow a mix of all characters.
edit: my regex knowledge is limited, so the above syntax may not be perfect.