Whitespace Matching Regex - Java

Yeah, you need to grab the result of matcher.replaceAll():

String result = matcher.replaceAll(" ");
System.out.println(result);

Make Regex Match Whitespaces in Java

\s matches a white-space character and when this is used in a java string you need to escape the \ so it would be \\s. If you want to match zero-or-more white-space then use \\s*.

This will match a single domain and TLD:

([A-Za-z][A-Za-z0-9.\\-_]*)\\s*(at)\\s*([A-Za-z][A-Za-z0-9\\-_()]*\\s*(dot)\\s*[A-Za-z]+)

However, you are trying to match multiple levels of sub-domains so you need to wrap the domain part of the regular expression ([A-Za-z][A-Za-z0-9\\-_()]*\\s*(dot)\\s* in ()+ to get one-or-more of them:

([A-Za-z][A-Za-z0-9.\\-_]*)\\s*(at)\\s*(([A-Za-z][A-Za-z0-9\\-_()]*\\s*(dot)\\s*)+[A-Za-z]+)
                                       ^                                        ^^

Something like this:

public class RegexpMatch {
    static Pattern Regex = Pattern.compile(
            "([A-Za-z][A-Za-z0-9.\\-_]*)\\s*(at)\\s*(([A-Za-z][A-Za-z0-9\\-_()]*\\s*(dot)\\s*)+[A-Za-z]+)"
        );

    public static void main( final String[] args ){
        final String[] tests = {
                "abcdatcsdotuniversitydotedu",
                "abcd at cs dot university dot edu"
        };

        for ( final String test : tests )
            System.out.println( test + " - "  + ( Regex.matcher( test ).matches() ? "Match" : "No Match" ) );
    }
}

Which outputs:

abcdatcsdotuniversitydotedu - Match
abcd at cs dot university dot edu - Match

Regular expression match fails if only whitespace after the - character

Something like,

^\d+\.\d+\.\d+(?:\s*-\s*\w+)?\/\d+\.\d+\.\d+\.\d+(?:\s*-\s*\w+)?.txt$

Or you can combine the \.\d+ repetitions as

^\d+(?:\.\d+){2}(?:\s*-\s*\w+)?\/\d+(?:\.\d+){3}(?:\s*-\s*\w+)?.txt$

Regex Demo

Changes

.{1} When you want to repeat something once, no need for {}. Its implicit
(?:\s*-\s*\w+) Matches zero or more space (\s*) followed by -, another space and then \w+ a description of length greater than 1
- The ? at the end of this patterns makes this optional.
- This same pattern is repeated again at the end to match the second part.
^ Anchors the regex at the start of the string.
$ Anchors the regex at the end of the string. These two are necessary so that there is nothing other in the string.
Don't group the patterns using () unless it is necessary to capture them. This can lead to wastage of memory. Use (?:..) If you want to group patterns but not capture them

Regex match a whitespace table

You may trim the input first, and then split with 3 or more whitespaces, then check if you got the first 2 cell values and use them :

String s = "           here is a $$ cell               here  another         cells I dont care about.........";
String[] res = s.trim().split("\\s{3,}");
if (res.length > 1) {
    System.out.println(res[0]); // Item 1
    System.out.println(res[1]); // Item 2, the rest is unimportant
}

See the Java demo

RegEx to match lines consisting of whitespace only

Use string.trim().isEmpty() to check for length 0 after trimming leading and trailing whitespaces.

Java Regex match space before or after characters

You may use

^(?!\s)(?!.*\s[*?])(?!.*[*?]\s)(?:[?\s]*[a-zA-Z0-9]){2}[a-zA-Z0-9?\s]*\*?$

See the regex demo.

Usage note: if you use it with Java's .matches() method, the ^ and $ can be removed from the pattern. Remember to double escape backslashes in the string literal.

Details

^ - start of string
(?!\s) - no whitespace is allowed immediately to the right (at the start of the string)
(?!.*\s[*?]) - no whitespace is allowed after any 0+ chars, as many as possible, before * or ?
(?!.*[*?]\s) - no whitespace is allowed after any 0+ chars, as many as possible, after * or ?
(?:[?\s]*[a-zA-Z0-9]){2} - two sequences of
- [?\s]* - 0 or more ? or/and whitespaces
- [a-zA-Z0-9] - an alphanumeric char
[a-zA-Z0-9?\s]* - 0 or more letters, digits, ? or whitespaces
\*? - an optional ? char
$ - end of the string.

How to express several whitespaces in a regex?

The \s metacharacter should account for tabs. What you're missing is the quantifier. You need a + or * quantifier (depending on whether you allow no space between the two segments) to detect any number of whitespaces.

java regex find all whitespace in a string

You can use negative lookahead to check for spaces:

^(?!.* )

^ - Start matching at the beginning of the string.

(?! - Begin a negative lookahead group (the pattern inside the parentheses must not come next.

.* - Any non-newline character any number of times followed by a space.

) - Close the negative lookahead group.

Combined with the full regex pattern (also cleaned up a bit to remove redundancy):

^(?!.* )(?=.*[a-z])(?=.*[A-Z])(?=.*\\d)(?=.*[!@$%&*?])[A-Za-z\\d!@$%&*?]+

Java regex match match whitespace and non-whitespace characters 2 times

There are two problems in your code.

You need to always call, matches() or find() before invoking .group() methods on matcher object.
Second your regex is incorrectly grouped.

Currently your group will only give one/last match, so instead you need to wrap whole of your expression into group. The correct regex you need is this,

.*\n((?:\\S+\\s+){2})(.*)

Try this Java codes,

String pattern = ".*\n((?:\\S+\\s+){2})(.*)";
String str = "Filesystem     1MB-blocks   Used Available Use% Mounted on\n" +
              "/dev/sda6         72342MB 5013MB   63655MB   8% /common";
Pattern r = Pattern.compile(pattern,  Pattern.DOTALL);
Matcher m = r.matcher(str);
if (m.matches()) {
    System.out.println(m.group(1));
}

Prints,

/dev/sda6         72342MB

How to ignore whitespace in regular expression matching

Convert your regex string to

mark|anthony|joseph\s*smith|michael

i.e. replace all spaces with \s* to match zero or more whitespace characters. If you want to match ONLY space (0x20) then make it

mark|anthony|joseph *smith|michael

Whitespace Matching Regex - Java