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Java, getting portion of pattern partially matched by input

As title says, i'd like to get the portion of the pattern that is being matched partially by the input; example:
Pattern: aabb
Input string: "aa"
At this point, i'll use hitEnd() method of Matcher class to find out if the pattern is being matched partially, like shown in this answer, but i'd also like to find out that specifically "aa" of "aabb" is matched.
Is there any way to do this in java?

This may be dirty, but here We go...
Once you know that some string hitEnd, do a second processing:
Remove the last character from the string
Search with the original regex
If It matches, then you are over and you have the part of the string
If not, go to 1 and repeat the whole process until you match
If test strings can be long, performance may be a problem. So instead of positions from last to first, try searching for blocks.
For example, considering a string of 1,000 chars:
Test 1000/2 characters: 1-500. For this example, we consider it matches
Test for first 500 chars + 500/2 (1-750 positions). For this example, We consider It does not match. So we know that the position must be placed from 500 to 750
Now test 1-625 ((750+500)/2)... If it matches, the positions must exist between 625-750. If it does not match, It must be from 500 to 625
...

There is no such function in Matcher class. However you could achieve it for example in this way:
public String getPartialMatching(String pattern, String input) {
Pattern p = Pattern.compile(pattern);
Matcher m = p.matcher(input);
int end = 0;
while(m.find()){
end = m.end();
}
if (m.hitEnd()) {
return input.substring(end);
} else {
return null;
}
}
First, iterate over all matched parts of string and skip them. For example: input = "aabbaa" m.hitEnd() will return false without skipping aabb.
Second, validate if the left part of the string partially matches.

Replace text with data & matched group contents

I don't believe I saw this when searching (believe me, I spent a good amount of time searching for this) for a solution to this so here goes.
Goal:
Match regex in a string and replace it with something that contains the matched value.
Regex used currently:
\b(Connor|charries96|Foo|Bar)\b
For the record I suck at regex incase this isn't the best way to do it.
My current code (and several other methods I tried) can only replace the text with the first match it encounters if there are multiple matches.
private Pattern regexFromList(List<String> input) {
if(input.size() < 1) {
return "";
}
StringBuilder builder = new StringBuilder();
builder.append("\\b");
builder.append("(");
for(String s : input) {
builder.append(s);
if(!s.equals(input.get(input.size() - 1)))
{
builder.append("|");
}
}
builder.append(")");
builder.append("\\b");
return Pattern.compile(builder.toString(), Pattern.CASE_INSENSITIVE);
}
Example input:
charries96's name is Connor.
Example result using TEST as the data to prepend the match with
TESTcharries96's name is TESTcharries96.
Desired result using example input:
TESTcharries96's name is TESTConnor.
Here is my current code for replacing the text:
if(highlight) {
StringBuilder builder = new StringBuilder();
Matcher match = pattern.matcher(event.getMessage());
String string = event.getMessage();
if (match.find()) {
string = match.replaceAll("TEST" + match.group());
// I do realise I'm using #replaceAll but that's mainly given it gives me the same result as other methods so why not just cut to the chase.
}
builder.append(string);
return builder.toString();
}
EDIT:
Working example of desired result on RegExr

There are a few problems here:
You are taking the user input as is and build the regex:
builder.append(s);
If there are special character in the user input, it might be recognized as meta character and cause unexpected behavior.
Always use Pattern.quote if you want to match a string as it is passed in.
builder.append(Pattern.quote(s));
Matcher.replaceAll is a high level function which resets the Matcher (start the match all over again), and search for all the matches and perform the replacement. In your case, it can be as simple as:
String result = match.replaceAll("TEST$1");
The StringBuilder should be thrown away along with the if statement.
Matcher.find, Matcher.group are lower level functions for fine grain control on what you want to do with a match.
When you perform replacement, you need to build the result with Matcher.appendReplacement and Matcher.appendTail.
A while loop (instead of if statement) should be used with Matcher.find to search for and perform replacement for all matched.

How to best strip out certain strings in a file?

If I have a file with the following content:
11:17 GET this is my content #2013
11:18 GET this is my content #2014
11:19 GET this is my content #2015
How can I use a Scanner and ignore certain parts of a `String line = scanner.nextLine();?
The result that I like to have would be:
this is my content
this is my content
this is my content
So I'd like to trip everything from the start until GET, and then take everything until the # char.
How could this easily be done?

You can use the String.indexOf(String str) and String.indexOf(char ch) methods. For example:
String line = scanner.nextLine();
int start = line.indexOf("GET");
int end = line.indexOf('#');
String result = line.substring(start + 4, end);

One way might be
String strippedStart = scanner.nextLine().split(" ", 3)[2];
String result = strippedStart.substring(0, strippedStart.lastIndexOf("#")).trim();
This assumes the are always two space separated tokens at the beginning (11:22 GET or 11:33 POST, idk).

You could do something like this:-
String line ="11:17 GET this is my content #2013";
int startIndex = line.indexOf("GET ");
int endIndex = line.indexOf("#");
line = line.substring(startIndex+4, endIndex-1);
System.out.println(line);

In my opinion the best solution for your problem would be using Java regex. Using regex you can define which group or groups of text you want to retrieve and what kind of text comes where. I haven't been working with Java in a long time, so I'll try to help you out from the top of my head. I'll try to give you a point in the right direction.
First off, compile a pattern:
Pattern pattern = Pattern.compile("^\d{1,2}:\d{1,2} GET (.*?) #\d+$", Pattern.MULTILINE);
First part of the regex says that you expect one or two digits followed by a colon followed by one or two digits again. After that comes the GET (you can use GET|POST if you expect those words or \w+? if you expect any word). Then you define the group you want with the parentheses. Lastly, you put the hash and any number of digits with at least one digit. You might consider putting flags DOTALL and CASE_INSENSITIVE, although I don't think you'll be needing them.
Then you continue with the matcher:
Matcher matcher = pattern.matcher(textToParse);
while (matcher.find())
{
//extract groups here
String group = matcher.group(1);
}
In the while loop you can use matcher.group(1) to find the text in the group you selected with the parentheses (the text you'd like extracted). matcher.group(0) gives the entire find, which is not what you're currently looking for (I guess).
Sorry for any errors in the code, it has not been tested. Hope this puts you on the right track.

You can try this rather flexible solution:
Scanner s = new Scanner(new File("data"));
Pattern p = Pattern.compile("^(.+?)\\s+(.+?)\\s+(.*)\\s+(.+?)$");
Matcher m;
while (s.hasNextLine()) {
m = p.matcher(s.nextLine());
if (m.find()) {
System.out.println(m.group(3));
}
}
This piece of code ignores first, second and last words from every line before printing them.
Advantage is that it relies on whitespaces rather than specific string literals to perform the stripping.

Java Regex is including new line in match

I'm trying to match a regular expression to textbook definitions that I get from a website.
The definition always has the word with a new line followed by the definition. For example:
Zither
Definition: An instrument of music used in Austria and Germany It has from thirty to forty wires strung across a shallow sounding board which lies horizontally on a table before the performer who uses both hands in playing on it Not to be confounded with the old lute shaped cittern or cithern
In my attempts to get just the word (in this case "Zither") I keep getting the newline character.
I tried both ^(\w+)\s and ^(\S+)\s without much luck. I thought that maybe ^(\S+)$ would work, but that doesn't seem to successfully match the word at all. I've been testing with rubular, http://rubular.com/r/LPEHCnS0ri; which seems to successfully match all my attempts the way I want, despite the fact that Java doesn't.
Here's my snippet
String str = ...; //Here the string is assigned a word and definition taken from the internet like given in the example above.
Pattern rgx = Pattern.compile("^(\\S+)$");
Matcher mtch = rgx.matcher(str);
if (mtch.find()) {
String result = mtch.group();
terms.add(new SearchTerm(result, System.nanoTime()));
}
This is easily solved by triming the resulting string, but that seems like it should be unnecessary if I'm already using a regular expression.
All help is greatly appreciated. Thanks in advance!

Try using the Pattern.MULTILINE option
Pattern rgx = Pattern.compile("^(\\S+)$", Pattern.MULTILINE);
This causes the regex to recognise line delimiters in your string, otherwise ^ and $ just match the start and end of the string.
Although it makes no difference for this pattern, the Matcher.group() method returns the entire match, whereas the Matcher.group(int) method returns the match of the particular capture group (...) based on the number you specify. Your pattern specifies one capture group which is what you want captured. If you'd included \s in your Pattern as you wrote you tried, then Matcher.group() would have included that whitespace in its return value.

With regular expressions the first group is always the complete matching string. In your case you want group 1, not group 0.
So changing mtch.group() to mtch.group(1) should do the trick:
String str = ...; //Here the string is assigned a word and definition taken from the internet like given in the example above.
Pattern rgx = Pattern.compile("^(\\w+)\s");
Matcher mtch = rgx.matcher(str);
if (mtch.find()) {
String result = mtch.group(1);
terms.add(new SearchTerm(result, System.nanoTime()));
}

A late response, but if you are not using Pattern and Matcher, you can use this alternative of DOTALL in your regex string
(?s)[Your Expression]
Basically (?s) also tells dot to match all characters, including line breaks
Detailed information: http://www.vogella.com/tutorials/JavaRegularExpressions/article.html

Just replace:
String result = mtch.group();
By:
String result = mtch.group(1);
This will limit your output to the contents of the capturing group (e.g. (\\w+)) .

Try the next:
/* The regex pattern: ^(\w+)\r?\n(.*)$ */
private static final REGEX_PATTERN =
Pattern.compile("^(\\w+)\\r?\\n(.*)$");
public static void main(String[] args) {
String input = "Zither\n Definition: An instrument of music";
System.out.println(
REGEX_PATTERN.matcher(input).matches()
); // prints "true"
System.out.println(
REGEX_PATTERN.matcher(input).replaceFirst("$1 = $2")
); // prints "Zither = Definition: An instrument of music"
System.out.println(
REGEX_PATTERN.matcher(input).replaceFirst("$1")
); // prints "Zither"
}

How to find a whole word in a String in Java?

I have a String that I have to parse for different keywords.
For example, I have the String:
"I will come and meet you at the 123woods"
And my keywords are
'123woods'
'woods'
I should report whenever I have a match and where. Multiple occurrences should also be accounted for.
However, for this one, I should get a match only on '123woods', not on 'woods'. This eliminates using String.contains() method. Also, I should be able to have a list/set of keywords and check at the same time for their occurrence. In this example, if I have '123woods' and 'come', I should get two occurrences. Method execution should be somewhat fast on large texts.
My idea is to use StringTokenizer but I am unsure if it will perform well. Any suggestions?

The example below is based on your comments. It uses a List of keywords, which will be searched in a given String using word boundaries. It uses StringUtils from Apache Commons Lang to build the regular expression and print the matched groups.
String text = "I will come and meet you at the woods 123woods and all the woods";
List<String> tokens = new ArrayList<String>();
tokens.add("123woods");
tokens.add("woods");
String patternString = "\\b(" + StringUtils.join(tokens, "|") + ")\\b";
Pattern pattern = Pattern.compile(patternString);
Matcher matcher = pattern.matcher(text);
while (matcher.find()) {
System.out.println(matcher.group(1));
}
If you are looking for more performance, you could have a look at StringSearch: high-performance pattern matching algorithms in Java.

Use regex + word boundaries as others answered.
"I will come and meet you at the 123woods".matches(".*\\b123woods\\b.*");
will be true.
"I will come and meet you at the 123woods".matches(".*\\bwoods\\b.*");
will be false.

Hope this works for you:
String string = "I will come and meet you at the 123woods";
String keyword = "123woods";
Boolean found = Arrays.asList(string.split(" ")).contains(keyword);
if(found){
System.out.println("Keyword matched the string");
}
http://codigounico.blogspot.com/

How about something like Arrays.asList(String.split(" ")).contains("xx")?
See String.split() and How can I test if an array contains a certain value.

Got a way to match Exact word from String in Android:
String full = "Hello World. How are you ?";
String one = "Hell";
String two = "Hello";
String three = "are";
String four = "ar";
boolean is1 = isContainExactWord(full, one);
boolean is2 = isContainExactWord(full, two);
boolean is3 = isContainExactWord(full, three);
boolean is4 = isContainExactWord(full, four);
Log.i("Contains Result", is1+"-"+is2+"-"+is3+"-"+is4);
Result: false-true-true-false
Function for match word:
private boolean isContainExactWord(String fullString, String partWord){
String pattern = "\\b"+partWord+"\\b";
Pattern p=Pattern.compile(pattern);
Matcher m=p.matcher(fullString);
return m.find();
}
Done

public class FindTextInLine {
String match = "123woods";
String text = "I will come and meet you at the 123woods";
public void findText () {
if (text.contains(match)) {
System.out.println("Keyword matched the string" );
}
}
}

Try to match using regular expressions. Match for "\b123wood\b", \b is a word break.

The solution seems to be long accepted, but the solution could be improved, so if someone has a similar problem:
This is a classical application for multi-pattern-search-algorithms.
Java Pattern Search (with Matcher.find) is not qualified for doing that. Searching for exactly one keyword is optimized in java, searching for an or-expression uses the regex non deterministic automaton which is backtracking on mismatches. In worse case each character of the text will be processed l times (where l is the sum of the pattern lengths).
Single pattern search is better, but not qualified, too. One will have to start the whole search for every keyword pattern. In worse case each character of the text will be processed p times where p is the number of patterns.
Multi pattern search will process each character of the text exactly once. Algorithms suitable for such a search would be Aho-Corasick, Wu-Manber, or Set Backwards Oracle Matching. These could be found in libraries like Stringsearchalgorithms or byteseek.
// example with StringSearchAlgorithms
AhoCorasick stringSearch = new AhoCorasick(asList("123woods", "woods"));
CharProvider text = new StringCharProvider("I will come and meet you at the woods 123woods and all the woods", 0);
StringFinder finder = stringSearch.createFinder(text);
List<StringMatch> all = finder.findAll();

A much simpler way to do this is to use split():
String match = "123woods";
String text = "I will come and meet you at the 123woods";
String[] sentence = text.split();
for(String word: sentence)
{
if(word.equals(match))
return true;
}
return false;
This is a simpler, less elegant way to do the same thing without using tokens, etc.

You can use regular expressions.
Use Matcher and Pattern methods to get the desired output

You can also use regex matching with the \b flag (whole word boundary).

To Match "123woods" instead of "woods" , use atomic grouping in regular expresssion.
One thing to be noted is that, in a string to match "123woods" alone , it will match the first "123woods" and exits instead of searching the same string further.
\b(?>123woods|woods)\b
it searches 123woods as primary search, once it got matched it exits the search.

Looking back at the original question, we need to find some given keywords in a given sentence, count the number of occurrences and know something about where. I don't quite understand what "where" means (is it an index in the sentence?), so I'll pass that one... I'm still learning java, one step at a time, so I'll see to that one in due time :-)
It must be noticed that common sentences (as the one in the original question) can have repeated keywords, therefore the search cannot just ask if a given keyword "exists or not" and count it as 1 if it does exist. There can be more then one of the same. For example:
// Base sentence (added punctuation, to make it more interesting):
String sentence = "Say that 123 of us will come by and meet you, "
+ "say, at the woods of 123woods.";
// Split it (punctuation taken in consideration, as well):
java.util.List<String> strings =
java.util.Arrays.asList(sentence.split(" |,|\\."));
// My keywords:
java.util.ArrayList<String> keywords = new java.util.ArrayList<>();
keywords.add("123woods");
keywords.add("come");
keywords.add("you");
keywords.add("say");
By looking at it, the expected result would be 5 for "Say" + "come" + "you" + "say" + "123woods", counting "say" twice if we go lowercase. If we don't, then the count should be 4, "Say" being excluded and "say" included. Fine. My suggestion is:
// Set... ready...?
int counter = 0;
// Go!
for(String s : strings)
{
// Asking if the sentence exists in the keywords, not the other
// around, to find repeated keywords in the sentence.
Boolean found = keywords.contains(s.toLowerCase());
if(found)
{
counter ++;
System.out.println("Found: " + s);
}
}
// Statistics:
if (counter > 0)
{
System.out.println("In sentence: " + sentence + "\n"
+ "Count: " + counter);
}
And the results are:
Found: Say
Found: come
Found: you
Found: say
Found: 123woods
In sentence: Say that 123 of us will come by and meet you, say, at the woods of 123woods.
Count: 5

If you want to identify a whole word in a string and change the content of that word you can do this way. Your final string stays equals, except the word you treated. In this case "not" stays "'not'" in final string.
StringBuilder sb = new StringBuilder();
String[] splited = value.split("\\s+");
if(ArrayUtils.isNotEmpty(splited)) {
for(String valor : splited) {
sb.append(" ");
if("not".equals(valor.toLowerCase())) {
sb.append("'").append(valor).append("'");
} else {
sb.append(valor);
}
}
}
return sb.toString();