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How to match a string in this way?

I need to check if a String matches this specific pattern.
The pattern is:
(Numbers)(all characters allowed)(numbers)
and the numbers may have a comma ("." or ",")!
For instance the input could be 500+400 or 400,021+213.443.
I tried Pattern.matches("[0-9],?.?+[0-9],?.?+", theequation2), but it didn't work!
I know that I have to use the method Pattern.match(regex, String), but I am not being able to find the correct regex.

Dealing with numbers can be difficult. This approach will deal with your examples, but check carefully. I also didn't do "all characters" in the middle grouping, as "all" would include numbers, so instead I assumed that finding the next non-number would be appropriate.
This Java regex handles the requirements:
"((-?)[\\d,.]+)([^\\d-]+)((-?)[\\d,.]+)"
However, there is a potential issue in the above. Consider the following:
300 - -200. The foregoing won't match that case.
Now, based upon the examples, I think the point is that one should have a valid operator. The number of math operations is likely limited, so I would whitelist the operators in the middle. Thus, something like:
"((-?)[\\d,.]+)([\\s]*[*/+-]+[\\s]*)((-?)[\\d,.]+)"
Would, I think, be more appropriate. The [*/+-] can be expanded for the power operator ^ or whatever. Now, if one is going to start adding words (such as mod) in the equation, then the expression will need to be modified.
You can see this regular expression here

In your regex you have to escape the dot \. to match it literally and escape the \+ or else it would make the ? a possessive quantifier. To match 1+ digits you have to use a quantifier [0-9]+
For your example data, you could match 1+ digits followed by an optional part which matches either a dot or a comma at the start and at the end. If you want to match 1 time any character you could use a dot.
Instead of using a dot, you could also use for example a character class [-+*] to list some operators or list what you would allow to match. If this should be the only match, you could use anchors to assert the start ^ and the end $ of the string.
\d+(?:[.,]\d+)?.\d+(?:[.,]\d+)?
In Java:
String regex = "\\d+(?:[.,]\\d+)?.\\d+(?:[.,]\\d+)?";
Regex demo
That would match:
\d+(?:[.,]\d+)? 1+ digits followed by an optional part that matches . or , followed by 1+ digits
. Match any character (Use .+) to repeat 1+ times
Same as the first pattern

Regular expression match fails if only whitespace after the - character

I am working on a regular expression where the pattern is:
1.0.0[ - optional description]/1.0.0.0[ - optional description].txt
The [ - optional description] part is of course, optional. So some possible VALID values are
1.0.0/1.0.0.0.txt
1.0.0/1.0.0.0 - xyz.txt
1.0.0 - abc/1.0.0.0 - xyz.txt
1.0.0 - abc/1.0.0.0.txt
To be a little more robust in the pattern matching, I'd like to match zero or more spaces before and after the "-" character. So all these would be valid too.
1.0.0 - abc/1.0.0.0 - xyz.txt
1.0.0-abc/1.0.0.0-xyz.txt
1.0.0 -abc/1.0.0.0- xyz.txt
To do this matching, I have the following regular expression (Java code):
String part1 = "((\\d+.{1}\\d+.{1}\\d+)(\\s*-\\s*(.+))?)";
String part2 = "((\\d+.{1}\\d+.{1}\\d+.{1}\\d+)(\\s*-\\s*(.+))?\\.sql)";
pattern = Pattern.compile(part1+ "/" + part2);
So far this regular expression is working well. But while unit testing I found a case I can't quite figure out yet. The use case is if the string contains the "-" character is surrounded by 1 or more spaces, but there is no description after the "-" character. This would look like:
1.0.0 - /1.0.0.0.txt
1.0.0- /1.0.0.0-xyz.txt
In these cases, I want the pattern match to FAIL. But with my current regular expression the match succeeds. I think what I want is if there is a "-" character surrounded by any number of spaces like " - " then there must also be at least 1 non-space character following it. But I can't quite figure out the regex for this.
Thanks!

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

In the group that matches the optional part, you need to replace .+ with \\S+ where \S means any non-whitespace character. This enforces the optional part to include non-whitespace character in order to match the pattern:
String part1
= "((\\d+\\.\\d+\\.\\d+)(\\s*-\\s*(\\S+))?)";
String part2
= "((\\d+\\.\\d+\\.\\d+.{1}\\d+)(\\s*-\\s*(\\S+))?\\.txt)";
Also note that .{1} (which is the same as just .) matches any character. From the examples, you want to match a dot, so it should be replaced with \.

Something like
^\d+\.\d+\.\d+(?:\s*-\s*[^\/\s]+)?\/\d+\.\d+\.\d+\.\d+?(?:\s*-\s*[^.\s]+)?\.\w+$
Check it out here at regex101.

Regex Match word that include a Dot

I have a Question I have this Sentence for Example:
"HalloAnna daveca.nn dave anna ca. anna"
And I only wanna match the single Standing "ca." .
My RegEx is like that :
(?i)\b(ca\.)\b
But this doesn't work and I don't know why. Any ideas ?
//Update
I excecute it with:
testSource.replaceAll()
and with
pattern.matcher(testSource).replaceAll().
both doesn´t work.

You must escape the dot and assert a non-word following:
(?i)\bca\.(?=\W)
See live demo.

You should use it like this:
Pattern.compile("(?i)\\b(ca\\.)(?=\\W)").matcher(a).replaceAll("SOME TEXT");
Which if you omit the java escapes gives a regex: (?i)\b(ca\.)\W.
Every \ in normal regex has to be escaped in java - \\.
Also, before a word you have word boundary (\b), but it applies only to a part in String where you have a change from whitespace to a alphanumeric character or the other way around. But in your case you have a dot, which is not an alphanumeric character, so you can't use \b at the end. You can use \W which means that a non-word character is following the dot. But to use \W you need to ignore it in the capture group (so it won't be replaced) - (?=.
Another issue was that you used ., which matches any character, but you actually want to match the real dot, so to do that you have to escape it - \., which in java String becomes \\..

Regular expression for match

How to find such lines in a file
######## this_is_a_line.sh ########
I tried the below regular expression but it does not work
(#)+( )+(A-Za-z0-9_)+(.sh)( )+(#)+
But this doesn't seem to work. Can anyone let me know what is wrong?

You used (...) (a capturing group) instead of [...] (a character class). Use the character class:
#+ +[A-Za-z0-9_]+\.sh +#+
^^^^^^^^^^^^^
See regex demo (note most of the capture groups are redundant here, and I removed them. Also, . must be escaped to match a literal dot.)
The [A-Za-z0-9_]+ matches 1 or more letters, digits or _.
The (A-Za-z0-9_)+ matches 1 or more sequences of A-Za-z0-9_ (see demo).
Also, in Java, you can use \w to match [A-Za-z0-9_] and shorten your regex to
#+ +\w+\.sh +#+
Do not forget that you need to double each \ in the pattern string in Java (String pattern = "#+ +\\w+\\.sh +#+";).

Regex for "* word"

Any Regex masters out there? I need a regular expression in Java that matches:
"RANDOMSTUFF SPECIFICWORD"
Including the quotation marks.
Thus I need
to match the first quote,
RANDOMSTUFF (any number of words with spaces between preceding SPECIFICWORD)
SPECIFICWORD (a specific word which I won't specify here.)
and the ending quote.
I don't want to match things such as:
RANDOMSTUFF SPECIFICWORD
"RANDOMSTUFF NOTTHESPECIFICWORD"
"RANDOMSTUFF SPECIFICWORD MORERANDOMSTUFF"

\".*\sSPECIFICWORD\"
If you don't want to allow quotes in between, use \"[^"]*\sSPECIFICWORD\"
. matches any character
* says 0 or more of the preceding character (in this case, 0 or more of any characters)
\s matches any whitespace character
SPECIFICWORD will be treated as a string literal, assuming there are no special characters (escape them if there are)
\" matches the quote
[^"] means any character except a quote (the ^ is what makes it 'except')
Also, this link could be useful. Regex's are powerful expressions and are applicable across virtually any language, so it would be a good thing to become comfortable with using them.
EDIT:
As several other posters have pointed out, adding ^ to the beginning and $ to the end will only match if the entire line matches.
^ matches the beginning of the line
$ matches the end of the line

^.*\s+SPECIFICWORD"$
'^' matches 'from the start of the line'
.* matches anything
\s+ matches 'any amount of whitespace, but at least some'
SPECIFICWORD" is a string literal
$ means 'this is the end of the line'
Note that ^ and $ are not always 'line'-based; most languages allow you to specify a 'multiline' mode that would cause them to match 'start of the string/end of the string' instead of one line at a time.

Will this string be matched as a line by line basis or will it be found within the text? If so, you can add anchors to ensure that it matches the string.
^(\".*\sSPECIFICWPRD\")$
Saying, at the start of the line, look for a double quote followed by zero or more random characters followed by a single whitespace, followed by the specific word, followed by a double quote at the end of the string.
Optionally, there are excellent tools for designing regex patterns and seeing what they match in real time.
Here are a couple of examples:
http://gskinner.com/RegExr/
http://regex101.com/r/zC3fM1

Try:
\"[\w\s]*SPECIFICWORD\"
Works like this:
\" matches opening quote
[\w\s]* matches zero or more of the characters from the following sets:
[a-zA-Z_0-9] (\w part)
[ \t\n\x0B\f\r] (\s part)
SPECIFICWORD matches the SPECIFICWORD
\" matches closing quote