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Java Regex : Extract a specific pattern from a string "I_INSERT_TO_TOPIC_345674_123456_4.json"

I want to extract only "_123456_4" from this string using java Regex.
I_INSERT_TO_TOPIC_345674_123456_4.json
I have tried
Pattern.compile("(_([^_]*_[^_]))") and Pattern.compile("_" + "([^[0-9]]*)" + "_[0-9]") but these do not work.

If you want to get 2 group of digits just before .json then you can use regex group to find the required match. You can modify the pattern as per your requirement.
Pattern p = Pattern.compile("(_\\d+_\\d+)\\.json");
Matcher matcher = p.matcher(s);
if (matcher.find()) {
String group = matcher.group(1);
}

【\_[0-9]\*\_[0-9]\*(?=\\.)】
You can try to see if this works

Need to extract the values that follow a pattern in the string in Java

I have a string as below:
String a= "member;range=12001-*: CN=marimar,OU=Employees,OU=Cisco
Users,DC=cisco,DC=com, CN=cadautel,OU=Employees,OU=Cisco
Users,DC=cisco,DC=com CN=rajaki,OU=Employees,OU=Cisco
Users,DC=cisco,DC=com";
I need to get the values of the CN attribute like 'marimar','cadautel,'rajaki' .
I have to use Java 7 to do that and hence I cannot use String.split() Can anybody help me out to come up with the logic.
Thanks!

String#split isn't the best tool for this job. Use a pattern matcher instead:
String a = "member;range=12001-*: CN=marimar,OU=Employees,OU=Cisco
Users,DC=cisco,DC=com, CN=cadautel,OU=Employees,OU=Cisco
Users,DC=cisco,DC=com CN=rajaki,OU=Employees,OU=Cisco
Users,DC=cisco,DC=com";
String pattern = "CN=([^,]+)";
Pattern r = Pattern.compile(pattern);
Matcher m = r.matcher(a);
while (m.find()) {
System.out.println("CN attribute: " + m.group(1) );
}
Demo

extract a set of a characters between some characters

I have a string email = John.Mcgee.r2d2#hitachi.com
How can I write a java code using regex to bring just the r2d2?
I used this but got an error on eclipse
String email = John.Mcgee.r2d2#hitachi.com
Pattern pattern = Pattern.compile(".(.*)\#");
Matcher matcher = patter.matcher
for (Strimatcher.find()){
System.out.println(matcher.group(1));
}

To match after the last dot in a potential sequence of multiple dots request that the sequence that you capture does not contain a dot:
(?<=[.])([^.]*)(?=#)
(?<=[.]) means "preceded by a single dot"
(?=#) means "followed by # sign"
Note that since dot . is a metacharacter, it needs to be escaped either with \ (doubled for Java string literal) or with square brackets around it.
Demo.

Not sure if your posting the right code. I'll rewrite it based on what it should look like though:
String email = John.Mcgee.r2d2#hitachi.com
Pattern pattern = Pattern.compile(".(.*)\#");
Matcher matcher = pattern.matcher(email);
int count = 0;
while(matcher.find()) {
count++;
System.out.println(matcher.group(count));
}
but I think you just want something like this:
String email = John.Mcgee.r2d2#hitachi.com
Pattern pattern = Pattern.compile(".(.*)\#");
Matcher matcher = pattern.matcher(email);
if(matcher.find()){
System.out.println(matcher.group(1));
}

No need to Pattern you just need replaceAll with this regex .*\.([^\.]+)#.* which mean get the group ([^\.]+) (match one or more character except a dot) which is between dot \. and #
email = email.replaceAll(".*\\.([^\\.]+)#.*", "$1");
Output
r2d2
regex demo
If you want to go with Pattern then you have to use this regex \\.([^\\.]+)# :
String email = "John.Mcgee.r2d2#hitachi.com";
Pattern pattern = Pattern.compile("\\.([^\\.]+)#");
Matcher matcher = pattern.matcher(email);
if (matcher.find()) {
System.out.println(matcher.group(1));// Output : r2d2
}
Another solution you can use split :
String[] split = email.replaceAll("#.*", "").split("\\.");
email = split[split.length - 1];// Output : r2d2
Note :
Strings in java should be between double quotes "John.Mcgee.r2d2#hitachi.com"
You don't need to escape # in Java, but you have to escape the dot with double slash \\.
There are no syntax for a for loop like you do for (Strimatcher.find()){, maybe you mean while

First and second tocen regex

How could I get the first and the second text in "" from the string?
I could do it with indexOf but this is really boring ((
For example I have a String for parse like: "aaa":"bbbbb"perhapsSomeOtherText
And I d like to get aaa and bbbbb with the help of Regex pattern - this will help me to use it in switch statement and will greatly simplify my app/

If all that you have is colon delimited string just split it:
String str = ...; // colon delimited
String[] parts = str.split(":");
Note, that split() receives regex and compilies it every time. To improve performance of your code you can use Pattern as following:
private static Pattern pColonSplitter = Pattern.compile(":");
// now somewhere in your code:
String[] parts = pColonSplitter.split(str);
If however you want to use pattern for matching and extraction of string fragments in more complicated cases, do it like following:
Pattert p = Patter.compile("(\\w+):(\\w+):");
Matcher m = p.matcher(str);
if (m.find()) {
String a = m.group(1);
String b = m.group(2);
}
Pay attention on brackets that define captured group.

Something like this?
Pattern pattern = Pattern.compile("\"([^\"]*)\"");
Matcher matcher = pattern.matcher("\"aaa\":\"bbbbb\"perhapsSomeOtherText");
while (matcher.find()) {
System.out.println(matcher.group(1));
}
Output
aaa
bbbbb

String str = "\"aaa\":\"bbbbb\"perhapsSomeOtherText";
Pattern p = Pattern.compile("\"\\w+\""); // word between ""
Matcher m = p.matcher(str);
while(m.find()){
System.out.println(m.group().replace("\"", ""));
}
output:
aaa
bbbbb

there are several ways to do this
Use StringTokenizer or Scanner with UseDelimiter method

How split a string using regex pattern

How split a [0] like words from string using regex pattern.0 can replace any integer number.
I used regex pattern,
private static final String REGEX = "[\\d]";
But it returns string with [.
Spliting Code
Pattern p=Pattern.compile(REGEX);
String items[] = p.split(lure_value_save[0]);

You have to escape the brackets:
String REGEX = "\\[\\d+\\]";

Java doesn't offer an elegant solution to extract the numbers. This is the way to go:
Pattern p = Pattern.compile(REGEX);
String test = "[0],[1],[2]";
Matcher m = p.matcher(test);
List<String> matches = new ArrayList<String>();
while (m.find()) {
matches.add(m.group());
}