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Java replaceAll but the specified regex

Can't get my head around this for quite some time already. I have this piece of code:
getStringFromDom(doc).replaceAll("contract=\"\\d*\"|name=\"\\p{L}*\"", "");
Basically I need it to work literally the opposite way - to replace everything BUT the specified regex. I've been trying to do it with the negative lookahead to no avail.

For your particular task, I think
getStringFromDom(doc).replaceAll(".*?(contract=\"\\d*\"|name=\"\\p{L}*\").*", "$1");
should do what you need.

You want to remove everything that does not match the pattern. This is the same as simply filtering the pattern matches. Use the regex to find matches for that pattern, then collect the matches in a stringbuilder.
Matcher m = Pattern.compile(your pattern).matcher(your input);
StringBuilder sb = new StringBuilder();
while (m.find()) sb.append (m.group()).append('\n');
String result = sb.toString();

I also think that removing what your are not looking for is a double negative. Concentrate on what you are looking for and use a pattern matching for that. This example searches your document for any name attributes:
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class Test {
public static void main(String[] args) {
String input = "<AnotherDoc accNum=\"1111\" docDate=\"2017-09-26\" docNum=\"2222\" name=\"foo\"> <anotherTag>some date</anotherTag>";
Pattern pattern = Pattern.compile("name=\"[^\\\"]*\""); // value are all characters but "
Matcher matcher = pattern.matcher(input);
while (matcher.find())
System.out.println(matcher.group());
}
}
This prints:
name="foo"

RegExpr output incorrect

I am trying to get all of the output from a string that I want to match a pattern using matcher, however, I am not sure that either the string or my pattern isn't correct. I am trying to get (Server: switch) as the first pattern and so on and so forth after the newline, however, I am only getting the last three pattern as my output shows. My output is the following with the code following
found_m: Message: Mess
found_m: Token: null
found_m: Response: OK
Here is my code:
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class RegexMatches {
public static void main( String args[] ) {
// String to be scanned to find the pattern.
String line = "Server: Switch\nMessage: Mess\nToken: null\nResponse: OK";
String pattern = "([\\w]+): ([^\\n]+)";
// Create a Pattern object
Pattern r = Pattern.compile(pattern);
// Now create matcher object.
Matcher m = r.matcher(line);
if (m.find( )) {
while(m.find()) {
System.out.println("found_m: " + m.group());
}
}else {
System.out.println("NO MATCH");
}
}
}
Is my string line incorrect or my string pattern that I am not doing regexpr wrong?
Thanks in advance.

Your regex is almost correct.
Problem is that you're calling find twice: 1st time in if condition and then again in while.
You can use do-while loop instead:
if (m.find( )) {
do {
System.out.println("found_m: " + m.group());
} while(m.find());
} else {
System.out.println("NO MATCH");
}
For regex part you can use this with minor correction:
final String pattern = "(\\w+): ([^\\n]+)";
or if you don't need 2 capturing groups then use:
final String pattern = "\\w+: [^\\n]+";
As there is no need to use character class around \\w+

I'm not familiar with Java, but this regex pattern should work to capture every group and match.
([\w]+): (\w+)(?:(?:[\\][n])|$)
It basically states capture the word followed by the colon and space, then capture the next word before either the \n or the end of string.
Good luck.

Java RegExp can't get the result ater evaluating pattern

Hi I have been trying to learn RegExpresions using Java I am still at the begining and I wanted to start a little program that is given a string and outputs the syllabels split.This is what I got so far:
String mama = "mama";
Pattern vcv = Pattern.compile("([aeiou][bcdfghjklmnpqrstvwxyz][aeiou])");
Matcher matcher = vcv.matcher(mama);
if(matcher){
// the result of this should be ma - ma
}
What I am trying to do is create a pattern that checks the letters of the given word and if it finds a pattern that contains a vocale/consonant/vocale it will add a "-" like this v-cv .How can I achive this.

In the following example i matched the first vowel and used positive lookahead for the next consonant-vowel group. This is so i can split again if i have a vcvcv group.
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class Test {
public static void main(String[] args) {
new Test().run();
}
private void run() {
String mama = "mama";
Pattern vcv =
Pattern.compile("([aeiou])(?=[bcdfghjklmnpqrstvwxyz][aeiou])");
Matcher matcher = vcv.matcher(mama);
System.out.println(matcher.replaceAll("$1-"));
String mamama = "mamama";
matcher = vcv.matcher(mamama);
System.out.println(matcher.replaceAll("$1-"));
}
}
Output:
ma-ma
ma-ma-ma

try
mama.replaceAll('([aeiou])([....][aeiou])', '\1-\2');
replaceAll is a regular expression method

Your pattern only matches if the String starts with a vocal. If you want to find a substring, ignoring the beginning, use
Pattern vcv = Pattern.compile (".*([aeiou][bcdfghjklmnpqrstvwxyz][aeiou])");
If you like to ignore the end too:
Pattern vcv = Pattern.compile (".*([aeiou][bcdfghjklmnpqrstvwxyz][aeiou]).*");

How do I process following PHP regex in Java?

How do I process following PHP regex in Java:
if(preg_match("/\r\n(.*?)\$/",$req,$match)){ $data=$match[1]; }
This line is part of following function, by the way:
function getheaders($req){
$r=$h=$o=null;
if(preg_match("/GET (.*) HTTP/" ,$req,$match)){ $r=$match[1]; }
if(preg_match("/Host: (.*)\r\n/" ,$req,$match)){ $h=$match[1]; }
if(preg_match("/Origin: (.*)\r\n/",$req,$match)){ $o=$match[1]; }
if(preg_match("/Sec-WebSocket-Key2: (.*)\r\n/",$req,$match)){ $key2=$match[1]; }
if(preg_match("/Sec-WebSocket-Key1: (.*)\r\n/",$req,$match)){ $key1=$match[1]; }
if(preg_match("/\r\n(.*?)\$/",$req,$match)){ $data=$match[1]; }
return array($r,$h,$o,$key1,$key2,$data);
}
Thanks in advance!
So far I have:
Matcher matcher = Pattern.compile("\r\n(.*?)\\$").matcher(req);
while(matcher.find()){
data = matcher.group(1);
}
I am sure, however, that this is wrong.
Ok guys, thanks for your answers, but they did not help yet. May I ask you to tell me, however, what this regex means:
if(preg_match("/\r\n(.*?)\$/",$req,$match)){ $data=$match[1]; }
I know, that if it does find a match with /\r\n(.*?)\$/ in the string $req, it will save the different kinds of mathces into the array $match. BUT: what is being matched here? And what's the difference between $match[0] and $match[1]? Maybe, if I understand this, I will be able to reconstruct the way to produce equal results in Java.
Thanks Jaroslav, but:
The string I am trying to process however (the last line of the handshake sent to me by Google Chrome, is:
Cookie: 34ad04df964553fb6017b93d35dccd5f=%7C34%7C36%7C37%7C40%7C41%7C42%7C43%7C44%7C45%7C46%7C47%7C48%7C49%7C50%7C52%7C53%7C54%7C55%7C56%7C57%7C58%7C59%7C60%7C61%7C62%7C63%7C64%7C65%7C66%7C67%7C68%7C69%7C70%7C71%7C72%7C73%7C74%7C75%7C76%7C77%7C78%7C79%7C80%7C81%7C82%7C83%7C84%7C85%7C86%7C87%7C88%7C89%7C90%7C91%7C92%7C93%7C94%7C95%7C96%7C97%7C98%7C99%7C100%7C101%7C102%7C103%7C104%7C105%7C106%7C107%7C108%7C109%7C110%7C111%7C112%7C113%7C114%7C115%7C116%7C117%7C118%7C119%7C120%7C121%7C122%7C123%7C124%7C125%7C126%7C127%7C128%7C129%7C130%7C131%7C132%7C133%7C134%7C135%7C136%7C137%7C138%7C139%7C%3B%7C%3B%7C%3B%7C%3B1%3B2%3B3%3B4%3B5%3B6%3B7%3B8%3B9%3B10%3B11%3B14%3B15%3B18%3B23%3B24%3B25%3B26%3B28%3B29%3B30%3B31%3B32%3B33%3B%7C
Hey guys, I just now realize what I have been asking was irrelevant :( But one answer has been right.

Use java.util.regex.Pattern look here for instructions for this class. Here is Regular Expressions Tutorial. And here is the example:
String p = "Host: (.*)\\r\\n";
String input = "Host: example.com\r\n";
Pattern pattern = Pattern.compile(p);
Matcher matcher = pattern.matcher(input);
if(matcher.matches()) {
String output = matcher.group(1);
System.out.println(output);
} else {
System.out.println("not found");
}
Note: Matcher.find matches subsequences, Matcher.matches matches entire region.
IMHO in your example \\$ at the end may cause a problem when your input is multiline and you parse it at once.
In Java there are more convenient methods for accessing headers. At client side this is HttpURLConnection.getHeaderField. At the server side there is HttpServletRequest.getHeader.

import java.util.regex.Pattern;
import java.util.regex.Matcher;
public class SplitDemo2 {
private static final String REGEX = "/\\r\\n(.*?)\\$/";
private static final String INPUT = "/GET (.*) HTTP/";
public static void main(String[] args) {
Pattern p = Pattern.compile(REGEX);
Matcher m = p.matcher(INPUT); // get a matcher object
int count = 0;
while(m.find()) {
count++;
System.out.println("Match number "+count);
System.out.println("start(): "+m.start());
System.out.println("end(): "+m.end());
}
}
}
More info on regex http://download.oracle.com/javase/tutorial/essential/regex/matcher.html
Explanation of your regex
\r\n(.*?)\$
\r Carriage return character.
\n Line break character.
(.*?) A numbered capture group
\$ Matches a $ character.

getting a string with quotes

I have a string "Hello" hello (including the quotes) and i just want to get the Hello that has the quotes but without the quotes
i tried using regular expression but it never finds the quotes im guessing
String s = new String("string");
Pattern p = Pattern.compile("\"([^\"])\"");
Matcher m = p.matcher(n);
while (m.find()) {
s = m.group(1);
}
the while loop never gets executed, suggestions?

-- Moved the star inside the parenthesis for proper grouping ---
"\"([^\"]*)\""
Tested successfully with the code
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class Test {
public static void main(String[] args) {
String s = new String("\"Hello\" hello");
Pattern p = Pattern.compile("\"([^\"]*)\"");
Matcher m = p.matcher(s);
while (m.find()) {
System.out.println(m.group(1));
}
}
}
which produced the expected output
Hello
-- Original post follows --
You don't match anything because your regex is written to only match quoted one character strings.
"\"([^\"])*\""
is closer to what you need. Note the star, it means zero or more of the preceeding expression. In this case the preceeding expression is "anything that lacks a double quote".

I suggest you try a String which has quotes in it if you want to find any. ;)
Try
String s = "start \"string\" end";
or
String s = "\"Hello\" hello";

You can simply use indexOf("\"") in this case.