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Need help in regex matching

It may be very simple, but I am extremely new to regex and have a requirement where I need to do some regex matches in a string and extract the number in it. Below is my code with sample i/p and required o/p. I tried to construct the Pattern by referring to https://www.freeformatter.com/java-regex-tester.html, but my regex match itself is returning false.
Pattern pattern = Pattern.compile(".*/(a-b|c-d|e-f)/([0-9])+(#[0-9]?)");
String str = "foo/bar/Samsung-Galaxy/a-b/1"; // need to extract 1.
String str1 = "foo/bar/Samsung-Galaxy/c-d/1#P2";// need to extract 2.
String str2 = "foo.com/Samsung-Galaxy/9090/c-d/69"; // need to extract 69
System.out.println("result " + pattern.matcher(str).matches());
System.out.println("result " + pattern.matcher(str1).matches());
System.out.println("result " + pattern.matcher(str1).matches());
All of above SOPs are returning false. I am using java 8, is there is any way by which in a single statement I can match the pattern and then extract the digit from the string.
I would be great if somebody can point me on how to debug/develop the regex.Please feel free to let me know if something is not clear in my question.

You may use
Pattern pattern = Pattern.compile(".*/(?:a-b|c-d|e-f)/[^/]*?([0-9]+)");
See the regex demo
When used with matches(), the pattern above does not require explicit anchors, ^ and $.
Details
.* - any 0+ chars other than line break chars, as many as possible
/ - the rightmost / that is followed with the subsequent subpatterns
(?:a-b|c-d|e-f) - a non-capturing group matching any of the alternatives inside: a-b, c-d or e-f
/ - a / char
[^/]*? - any chars other than /, as few as possible
([0-9]+) - Group 1: one or more digits.
Java demo:
List<String> strs = Arrays.asList("foo/bar/Samsung-Galaxy/a-b/1","foo/bar/Samsung-Galaxy/c-d/1#P2","foo.com/Samsung-Galaxy/9090/c-d/69");
Pattern pattern = Pattern.compile(".*/(?:a-b|c-d|e-f)/[^/]*?([0-9]+)");
for (String s : strs) {
Matcher m = pattern.matcher(s);
if (m.matches()) {
System.out.println(s + ": \"" + m.group(1) + "\"");
}
}
A replacing approach using the same regex with anchors added:
List<String> strs = Arrays.asList("foo/bar/Samsung-Galaxy/a-b/1","foo/bar/Samsung-Galaxy/c-d/1#P2","foo.com/Samsung-Galaxy/9090/c-d/69");
String pattern = "^.*/(?:a-b|c-d|e-f)/[^/]*?([0-9]+)$";
for (String s : strs) {
System.out.println(s + ": \"" + s.replaceFirst(pattern, "$1") + "\"");
}
See another Java demo.
Output:
foo/bar/Samsung-Galaxy/a-b/1: "1"
foo/bar/Samsung-Galaxy/c-d/1#P2: "2"
foo.com/Samsung-Galaxy/9090/c-d/69: "69"

Because you match always the last number in your regex, I would Like to just use replaceAll with this regex .*?(\d+)$ :
String regex = ".*?(\\d+)$";
String strResult1 = str.replaceAll(regex, "$1");
System.out.println(!strResult1.isEmpty() ? "result " + strResult1 : "no result");
String strResult2 = str1.replaceAll(regex, "$1");
System.out.println(!strResult2.isEmpty() ? "result " + strResult2 : "no result");
String strResult3 = str2.replaceAll(regex, "$1");
System.out.println(!strResult3.isEmpty() ? "result " + strResult3 : "no result");
If the result is empty then you don't have any number.
Outputs
result 1
result 2
result 69

Here is a one-liner using String#replaceAll:
public String getDigits(String input) {
String number = input.replaceAll(".*/(?:a-b|c-d|e-f)/[^/]*?(\\d+)$", "$1");
return number.matches("\\d+") ? number : "no match";
}
System.out.println(getDigits("foo.com/Samsung-Galaxy/9090/c-d/69"));
System.out.println(getDigits("foo/bar/Samsung-Galaxy/a-b/some other text/1"));
System.out.println(getDigits("foo/bar/Samsung-Galaxy/9090/a-b/69ace"));
69
no match
no match
This works on the sample inputs you provided. Note that I added logic which will display no match for the case where ending digits could not be matched fitting your pattern. In the case of a non-match, we would typically be left with the original input string, which would not be all digits.

Java Split String by colon on both side

Can you suggest me an approach by which I can split a String which is like:
:31C:150318
:31D:150425 IN BANGLADESH
:20:314015040086
So I tried to parse that string with
:[A-za-z]|\\d:
This kind of regular expression, but it is not working . Please suggest me a regular expression by which I can split that string with 20 , 31C , 31D etc as Keys and 150318 , 150425 IN BANGLADESH etc as Values .
If I use string.split(":") then it would not serve my purpose.
If a string is like:
:20: MY VALUES : ARE HERE
then It will split up into 3 string , and key 20 will be associated with "MY VALUES" , and "ARE HERE" will not associated with key 20 .

You may use matching mechanism instead of splitting since you need to match a specific colon in the string.
The regex to get 2 groups between the first and second colon and also capture everything after the second colon will look like
^:([^:]*):(.*)$
See demo. The ^ will assert the beginning of the string, ([^:]*) will match and capture into Group 1 zero or more characters other than :, and (.*) will match and capture into Group 2 the rest of the string. $ will assert the position at the end of a single line string (as . matches any symbol but a newline without Pattern.DOTALL modifier).
String s = ":20:AND:HERE";
Pattern pattern = Pattern.compile("^:([^:]*):(.*)$");
Matcher matcher = pattern.matcher(s);
while (matcher.find()){
System.out.println("Key: " + matcher.group(1) + ", Value: " + matcher.group(2) + "\n");
}
Result for this demo: Key: 20, Value: AND:HERE

You can use the following to split:
^[:]+([^:]+):

Try with split function of String class
String[] splited = string.split(":");
For your requirements:
String c = ":31D:150425 IN BANGLADESH:todasdsa";
c=c.substring(1);
System.out.println("C="+c);
String key= c.substring(0,c.indexOf(":"));
String value = c.substring(c.indexOf(":")+1);
System.out.println("key="+key+" value="+value);
Result:
C=31D:150425 IN BANGLADESH:todasdsa
key=31D value=150425 IN BANGLADESH:todasdsa

Certain strings that should be found by a working Regex are missed, and I need help identifying why

I have a set of strings, which I cycle through, checking those against the following set of regex, to try and separate the first small section from the rest of the string. The regex works in almost all cases, but unfortunately I have no idea why it fails occasionally. I’ve been using Pattern Matcher to print out the string, if the pattern is found.
Two example working strings:
98. SORGHUM Moench - Millets Annuals or rhizomatous perennials; inflorescence …
99. MISCANTHUS Andersson - Silver-grasses Rhizomatous perennials; inflorescence …
Two example failed strings:
100. ZEA L. - Maize Annuals; male and female inflorescences separate, the …
26. POA L. (Parodiochloa C.E. Hubb.) - Meadow-grasses Annuals or perennials with or without stolons or rhizomes; sheaths overlapping or some …
Regex’s used so far:
Pattern endOfGenus = Pattern.compile("(?<=(^\\d+\\. " + genusNames[l].toUpperCase() + "))");
Pattern endOfGenusTwo = Pattern.compile("(?<=(^\\d+" + genusNames[l].toUpperCase() + "))");
Pattern endOfGenusThree = Pattern.compile("(?<=(\\d+\\. " + genusNames[l] + "))");
Pattern endOfGenusFour = Pattern.compile("(?<=(\\d+" + genusNames[l] + "))");
Pattern endOfGenusFive = Pattern.compile("(?<=(\\. " + genusNames[l] + "))");
The first of these is the one thats producing the reliable results so far.
Example Code
Pattern endOfGenus = Pattern.compile("(?<=(^\\d+\\. " + genusNames[l].toUpperCase() + "))");
Matcher endOfGenusFinder = endOfGenus.matcher(descriptionPartBits[b]);
if (endOfGenusFinder.find()) {
System.out.print(descriptionPartBits[b] + ":- ");
System.out.print(genusNames[l] + "\n");
String[] genusNameBits = descriptionPartBits[b].split("(?<=(^\\d+\\. " + genusNames[l].toUpperCase() + "))");
}
Desired Output. This is what is produced by strings that work. Strings that don't work simply don't appear in the output:
98. SORGHUM Moench - Millets Annuals or rhizomatous perennials:- Sorghum
99. MISCANTHUS Andersson - Silver-grasses Rhizomatous perennials:- Miscanthus

From regex tutorial:
Lookahead and lookbehind, collectively called "lookaround", are
zero-length assertions just like the start and end of line, and start
and end of word anchors explained earlier in this tutorial.
Lookahead and lookbehind only return true or false.
So I changed your code example:
Pattern endOfGenus = Pattern.compile("(?<=(^\\d+\\. ZEA L))(.+)$");
// Matcher matcher = endOfGenus.matcher("98. SORGHUM Moench - Millets Annuals or rhizomatous perennials; inflorescence …");
Matcher matcher = endOfGenus.matcher("100. ZEA L. - Maize Annuals; male and female inflorescences separate, the …");
while (matcher.find()) {
String group1 = matcher.group(1);
String group2 = matcher.group(2);
System.out.println("group1=" + group1);
System.out.println("group2=" + group2);
}
Group 1 is matched by (^\\d+\\. ZEA L). Group 2 is matched by (.+).

Java pattern matching using regex

I am new to java coding and using pattern matching.I am reading this string from file. So, this will give compilation error. I have a string as follows :
String str = "find(\"128.210.16.48\",\"Hello Everyone\")" ; // no compile error
I want to extract "128.210.16.48" value and "Hello Everyone" from above string. This values are not constant.
can you please give me some suggestions?
Thanks

I suggest you to use String#split() method but still if you are looking for regex pattern then try it and get the matched group from index 1.
("[^"][\d\.]+"|"[^)]*+)
Online demo
Sample code:
String str = "find(\"128.210.16.48\",\"Hello Everyone\")";
String regex = "(\"[^\"][\\d\\.]+\"|\"[^)]*+)";
Pattern pattern = Pattern.compile(regex);
Matcher matcher = pattern.matcher(str);
while (matcher.find()) {
System.out.println(matcher.group(1));
}
output:
"128.210.16.48"
"Hello Everyone"
Pattern explanation:
( group and capture to \1:
" '"'
[^"] any character except: '"'
[\d\.]+ any character of: digits (0-9), '\.' (1
or more times (matching the most amount
possible))
" '"'
| OR
" '"'
[^)]* any character except: ')' (0 or more
times (matching the most amount
possible))
) end of \1

Try with String.split()
String str = "find(\"128.210.16.48\",\"Hello Everyone\")" ;
System.out.println(str.split(",")[0].split("\"")[1]);
System.out.println(str.split(",")[1].split("\"")[1]);
Output:
128.210.16.48
Hello Everyone
Edit:
Explanation:
For the first string split it by comma (,). From that array choose the first string as str.split(",")[0] split the string again with doublequote (") as split("\"")[1] and choose the second element from the array. Same the second string is also done.

The accepted answer is fine, but if for some reason you wanted to still use regex (or whoever finds this question) instead of String.split here's something:
String str = "find(\"128.210.16.48\",\"Hello Everyone\")" ; // no compile error
String regex1 = "\".+?\"";
Pattern pattern1 = Pattern.compile(regex1);
Matcher matcher1 = pattern1.matcher(str);
while (matcher1.find()){
System.out.println("Matcher 1 found (trimmed): " + matcher1.group().replace("\"",""));
}
Output:
Matcher 1 found (trimmed): 128.210.16.48
Matcher 1 found (trimmed): Hello Everyone
Note: this will only work if " is only used as a separator character. See Braj's demo as an example from the comments here.

How to remove spaces in between the String

I have below String
string = "Book Your Domain And Get\n \n\n \n \n \n Online Today."
string = str.replace("\\s","").trim();
which returning
str = "Book Your Domain And Get Online Today."
But what is want is
str = "Book Your Domain And Get Online Today."
I have tried Many Regular Expression and also googled but got no luck. and did't find related question, Please Help, Many Thanks in Advance

Use \\s+ instead of \\s as there are two or more consecutive whitespaces in your input.
string = str.replaceAll("\\s+"," ")

You can use replaceAll which takes a regex as parameter. And it seems like you want to replace multiple spaces with a single space. You can do it like this:
string = str.replaceAll("\\s{2,}"," ");
It will replace 2 or more consecutive whitespaces with a single whitespace.

First get rid of multiple spaces:
String after = before.trim().replaceAll(" +", " ");

If you want to just remove the white space between 2 words or characters and not at the end of string
then here is the
regex that i have used,
String s = " N OR 15 2 ";
Pattern pattern = Pattern.compile("[a-zA-Z0-9]\\s+[a-zA-Z0-9]", Pattern.CASE_INSENSITIVE);
Matcher m = pattern.matcher(s);
while(m.find()){
String replacestr = "";
int i = m.start();
while(i<m.end()){
replacestr = replacestr + s.charAt(i);
i++;
}
m = pattern.matcher(s);
}
System.out.println(s);
it will only remove the space between characters or words not spaces at the ends
and the output is
NOR152

Eg. to remove space between words in a string:
String example = "Interactive Resource";
System.out.println("Without space string: "+ example.replaceAll("\\s",""));
Output:
Without space string: InteractiveResource

If you want to print a String without space, just add the argument sep='' to the print function, since this argument's default value is " ".

//user this for removing all the whitespaces from a given string for example a =" 1 2 3 4"
//output: 1234
a.replaceAll("\\s", "")

String s2=" 1 2 3 4 5 ";
String after=s2.replace(" ", "");
this work for me

String string_a = "AAAA BBB";
String actualTooltip_3 = string_a.replaceAll("\\s{2,}"," ");
System.out.println(String actualTooltip_3);
OUTPUT will be:AAA BBB