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Java regex for matching #<string>vs<string>

I have a string "Waiting for match #indvspak and #indvsaus" and want to match the strings "#indvspak" and "#indvsaus" seperately.
I am using the following regex (^|)#.*vs.+?\s\b. But it matches the entire string starting from the hash sign. How can i achieve my requirement please help.

I though you want to match the string which startswith # contains vs and the whole string must be preceded by a non-space character.
"(?<!\\S)#\\S*vs\\S+"
(?<!\\S) negative look-behind asserts that the match won't be preceded by a non-space character.
Code:
String s = "Waiting for match #indvspak and #indvsaus";
Matcher m = Pattern.compile("(?<!\\S)#\\S*vs\\S+").matcher(s);
while(m.find())
{
System.out.println(m.group());
}
Output:
#indvspak
#indvsaus

You need this regex:
#[^\\s]+
it matches anything after (including) # but not spaces.
Edit:
As #AvinashRaj suggested, if you want to ensure "vs" appears in the hashtag, you should use a negative lookbehind.
I highly recommend you to go though the String API, there are many methods that can help you with your problem.

EDITED
(copied from other answer comments)
Use this:
"(?<!\\B)#\\w+vs\\o/\S#vas\\S-[]"
Easy...

Regex to match \a574322 in Java

I have long string looking like this: \c53\e59\c9\e28\c20140326\a4095\c8\c15\a546\c11 and I need to find expressions starting with \a and followed by digits. For example: \a574322
And I have no idea how to build it. I can't use:
Pattern p = Pattern.compile("\\a\\d*");
because \a is special character in regex.
When I try to group it like this:
Pattern p = Pattern.compile("(\\)(a)(\\d)*");
I get unclosed group error even though there is even number of brackets.
Can you help me with this?
Thank you all very much for solution.

You can use this regex:
\\\\a\\d+
Code Demo
Since in Java you need to double escape the \\ once for String and second time for regex engine.

You have to change your regex to:
Pattern p = Pattern.compile("(\\\\a\\d+)");
The regex is:
(\\a\d+)
The idea is to escape a backslash and then also escape the backslash for \a, and match digits too.

You need 4 \.
2 to indicate to regex that it is not a special character, but a plain \, and 2 for each to tell the Java String that these are not special characters either. So you need to represent it in code this way:
"\\\\a\\d*"
Which is actually the regex \\a\d*

\\(a)[0-9]+ this should work
you can't try your regexps on this page or some similar
http://regex101.com/

java Regex - split but ignore text inside quotes?

using only regular expression methods, the method String.replaceAll and ArrayList
how can i split a String into tokens, but ignore delimiters that exist inside quotes?
the delimiter is any character that is not alphanumeric or quoted text
for example:
The string :
hello^world'this*has two tokens'
should output:
hello
worldthis*has two tokens

I know there is a damn good and accepted answer already present but I would like to add another regex based (and may I say simpler) approach to split the given text using any non-alphanumeric delimiter which not inside the single quotes using
Regex:
/(?=(([^']+'){2})*[^']*$)[^a-zA-Z\\d]+/
Which basically means match a non-alphanumeric text if it is followed by even number of single quotes in other words match a non-alphanumeric text if it is outside single quotes.
Code:
String string = "hello^world'this*has two tokens'#2ndToken";
System.out.println(Arrays.toString(
string.split("(?=(([^']+'){2})*[^']*$)[^a-zA-Z\\d]+"))
);
Output:
[hello, world'this*has two tokens', 2ndToken]
Demo:
Here is a live working Demo of the above code.

Use a Matcher to identify the parts you want to keep, rather than the parts you want to split on:
String s = "hello^world'this*has two tokens'";
Pattern pattern = Pattern.compile("([a-zA-Z0-9]+|'[^']*')+");
Matcher matcher = pattern.matcher(s);
while (matcher.find()) {
System.out.println(matcher.group(0));
}
See it working online: ideone

You cannot in any reasonable way. You are posing a problem that regular expressions aren't good at.

Do not use a regular expression for this. It won't work. Use / write a parser instead.
You should use the right tool for the right task.

How to include symbol of the end of the line ("\n") in regexp

i have regular expression which need to cover multi lines(delete comments from pascal file)
\(\*.*?\*\)|\{.*?\}|\/\/(.*$)
this works almost fine but
\(\*.*?\*\)
and
\{.*?\}
are supposed to work for multilines, but work for single only. How to make them working right(and dont make
//(.*$)
work for multi lines)
Thanks in advance

You are looking for the Pattern.DOTALL flag. Pass it to Pattern.compile like so:
Pattern p = Pattern.compile("regex", Pattern.DOTALL);
You can also set it in the regex with (?s), like: "(?s)regex"

This will match literally everything, including newlines:
Pattern regex = Pattern.compile("[\\s\\S]*");

How to match a set of string in regexp

How do I combine ch..+ and ch..- in regexp effectively without having to scan separately?
And are we using matcher in the pattern?
My output code is like this:
ch01+
ch01-
ch02+
ch02-
...

How do I combine ch..+ and ch..- in regexp effectively without having to scan separately?
Use | (pipe) for alternation:
ch..(\+|-)
And are we using matcher in the pattern?
Depends on how you're using the regexp and the pattern. To get a concrete answer, you'll have to show some actual code, or ask a much more specific question.
N.B. If you want to restrict the two characters after ch to 0-9, you can use \d, which is a shorthand character class for [0-9]:
ch\d{2}(\+|-)

You can use a character class containing just "+" and "-" like so "[+-]".
Pattern p = Pattern.compile("ch..[+-]");
Matcher m = p.matcher("ch01+");
if (m.find()) {
// found it...