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Regex with prefix and optional suffix

This is maybe the 100+1 question regarding regex optional suffixes on SO, but I didn't find any, that could help me :(
I need to extract a part of string from the common pattern:
prefix/s/o/m/e/t/h/i/n/g/suffix
using a regular expression. The prefix is constant and the suffix may not appear at all, so prefix/(.+)/suffix doesn't meet my requirements. Pattern prefix/(.+)(?:/suffix)? returns s/o/m/e/t/h/i/n/g/suffix. The part (?:/suffix)? must be somehow more greedy.
I want to get s/o/m/e/t/h/i/n/g from these input strings:
prefix/s/o/m/e/t/h/i/n/g/suffix
prefix/s/o/m/e/t/h/i/n/g/
prefix/s/o/m/e/t/h/i/n/g
Thanks in advance!

Try
prefix\/(.+?)\/?(?:suffix|$)
The regex need to know when the match is done, so match either suffix or end of line ($), and make the capture non greedy.
See it here at regex101.

Try prefix(.*?)(?:/?(?:suffix|$)) if there are characters allowed before prefix of after suffix.
This requires the match to be as short as possible (reluctant quantifier) and be preceeded by one of 3 things: a single slash right before the end of the input, /suffix or the end of the input. That would match /s/o/m/e/t/h/i/n/g in the test cases you provided but would match more for input like prefix/s/o/m/e/t/h/i/n/g/suff (which is ok IMO since you don't know whether /suff is meant to be part of the match or a typo in the suffix).

Need regular expression for pattern this

I need a regular expression for below pattern
It can start with / or number
It can only contain numbers, no text
Numbers can have space in between them.
It can contain /*, at least 1 number and space or numbers and /*
Valid Strings:
3232////33 43/323//
3232////3343/323//
/3232////343/323//
Invalid Strings:
/sas/3232/////dsds/
/ /34343///// /////
///////////
My Problem is, it can have space between numbers like /3232 323/ but not / /.
How to validate it ?
I have tried so far:
(\\d[\\d ]*/+) , (/*\\d[\\d ]*/+) , (/*)(\\d*)(/*)

This regex should work for you:
^/*(?:\\d(?: \\d)*/*)+$
Live Demo: http://www.rubular.com/r/pUOYFwV8SQ

My solution is not so simple but it works
^(((\d[\d ]*\d)|\d)|/)*((\d[\d ]*\d)|\d)(((\d[\d ]*\d)|\d)|/)*$

Just use lookarounds for the last criteria.
^(?=.*?\\d)([\\d/]*(?:/ ?(?!/)|\\d ?))+$
The best would have been to use conditional regex, but I think Java doesn't support them.
Explanation:
Basically, numbers or slashes, followed by one number and a space, or one slash and a space which is not followed by another slash. Repeat that. The space is made optional because I assume there's none at the end of your string.

Try this java regex
/*(\\d[\\d ]*(?<=\\d)/+)+
It meets all your criteria.
Although you didn't specifically state it, I have assumed that a space may not appear as the first or last character for a number (ie spaces must be between numbers)

"(?![A-z])(?=.*[0-9].*)(?!.*/ /.*)[0-9/ ]{2,}(?![A-z])"
this will match what you want but keep in mind it will also match this
/3232///// from /sas/3232/////dsds/
this is because part of the invalid string is correct
if you reading line by line then match the ^ $ and if you are reading an entire block of text then search for \r\n around the regex above to match each new line

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}$

How to use regex to remove punctuations in a sentence

I am trying to take from a file all the valid words. Valid words are defined as normal characters that can appear like so:
don't won't can't
and I have to ignore commas periods and exclamation points.
I have gotten the expression to just get characters but now it won't get words like don't and can't or won't.
This is the expression I am using "[^A-Za-z]+" and I have tried "\'[^A-Za-z]+" but this breaks and allows all characters. Does anyone have any idea what I can use to get normal words including don't and won't and can't and such words.
Thank you very much

[^A-Za-z] Would mean anything NOT matching those character ranges! Try this:
[A-Za-z']
You may need to escape the single quote, in which case you'll probably need to escape the slash that escapes it:
[A-Za-z\\']

Another way (using abbreviations) is: \b[\w']+

This will match letters from any language and exclude numbers.
\b[\p{L}\!\'\?]+
Here is a very good resource for regular expressions.
http://www.regular-expressions.info/

What's wrong with this regex?

I am trying the following code on Java:
String test = "http://asda.aasd.sd.google.com/asdasdawrqwfqwfqwfqwf";
String regex = "[http://]{0,1}([a-zA-Z]*.)*\\.google\\.com/[-a-zA-Z/_.?&=]*";
System.out.println(test.matches(regex));
It does work for several minutes (after that I killed the VM) with no result.
Can anyone help me?
BTW: What will you recommend me to do to speed up weblink-testng regexes in future?

[http://] is a character class, meaning any one of those characters from the set.
Just leave those particular square brackets off if it must start with http://. If it's optional, you can use (http://)?.
One obvious problem is that you're looking for the sequence ([a-zA-Z]+.)*\\.google - this will do a lot of backtracking due to that naked . which means "any character" rather than the literal period that you wanted.
But even if you replace it with what you meant, ([a-zA-Z]+\\.)*\\.google, you still have a problem - this will then require two . characters immediately before google. You should instead try:
String regex = "(http://)?([a-zA-Z]+\\.)*google\\.com/[-a-zA-Z/_.?&=]*";
That returns immediately for me with a true match.
Keep in mind that this currently requires the / at the end of google.com. If that's a problem, it's a minor fix, but I've left it there since you had it in your original regex.

You are trying to match the scheme as a character class using square brackets. That means only zero or one of the characters from that set. You want a subpattern, with parentheses. You can also change {0,1} to just say ?.
Also, you should remove the period just before google\\.com because you're already looking for a period in the subdomain subpattern of your regex. As cherouvim points out, you forgot to escape that period as well.
String regex = "(http://)?([a-zA-Z]+\\.)*google\\.com/[-a-zA-Z/_.?&=]*";

In the ([a-zA-Z]*.) part you either need to escape the . (because right now it means "all characters") or remove it.

There are two problems with the regular expression.
The first is easy, as was mentioned by others. You need to match "http://" as a subpattern, not as a character class. Change the brackets to parentheses.
The second problem causes the very poor performance. It's causing the regex to backtrack repeatedly, trying to match the pattern.
What you're trying to do is match zero or more subdomains, which are groups of letters followed by a dot. Since you want to match the dot explicitly, escape the dot. Also remove the dot in front of "google" so you can match "http://google.com/etc" (ie, no leading dot in front of google).
So your expression becomes:
String regex = "(http://){0,1}([a-zA-Z]+\\.)*google\\.com/[-a-zA-Z/_.?&=]*";
Running this regex on your example takes just a fraction of a second.

Assuming you fix the ([a-zA-Z]*\\.) you need to change * to + so the part becomes ([a-zA-Z]+\\.). Otherwise you'll be accepting http://...google.com and this is not valid.

By grouping part before google.com I assume you are looking for part of URL host name. I think that rexep is powerful tool, but you can simply use URL Java class. There is getHost() method. Then you can check if host name ends with google.com and split it or use some simplier regexp with only host name.
URL url = new URL("http://asda.aasd.sd.google.com/asdasdawrqwfqwfqwfqwf");
String host = url.getHost();
if (host.endsWith("google.com"))
{
String [] parts = host.split("\\.");
for (String s: parts)
System.out.println(s);
}