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Java Regex to match supplied word with special charcters

I am trying to match string by using Java Pattern class.
private boolean isMatch(String searchSentence, String matchWord) {
String patternText = ".*\\b";
Pattern pattern = Pattern.compile(patternText + matchWord + "\\b.*",Pattern.CASE_INSENSITIVE);
Matcher matcher = pattern.matcher(searchSentence);
return matcher.matches();
}
This will work if i have match string, which doesn't contains any special character like ({[ etc.
I am able to find "RANCH" but not "RANCH (EAGLEFORD)".
Few more examples
Let say my input string is: "Point [-99.73586,28.38092]" then i should be able to search for "-99.73586,28.38092"
Let say my input string is: "Point [-99.73586,28.38092]" then i should be able to search for "[-99.73586,28.38092]"
Let say my input string is: "Rench RenchY" then i should be able to search for "Rench" but RenchY Should not be the part of search result.
So how can i handle these type of case.

Example of using special characters ({}) in regex
String stringToSearch = "Some lengthy string I am trying to RANCH (EAGLEFORD) and RANCH {EAGLEFORD}";
Pattern p1 = Pattern.compile("RANCH\\s[(){}\\w]+");
Matcher m = p1.matcher(stringToSearch);
while (m.find())
{
System.out.println(m.group());
}
output:
RANCH (EAGLEFORD)
RANCH {EAGLEFORD}

If you plan to match your keywords not enclosed with word chars (letters, digits, or underscores), use (?<!\w) and (?!\w) lookarounds instead of \b.
private boolean isMatch(String searchSentence, String matchWord) {
Pattern pattern = Pattern.compile("(?<!\\w)" + Pattern.quote(matchWord) + "(?!\\w)", Pattern.CASE_INSENSITIVE);
Matcher matcher = pattern.matcher(searchSentence);
return matcher.find();
}
If you only plan to find matches enclosed with whitespace/start/end of string, use (?<!\S) and (?!\S) lookarounds:
private boolean isMatch(String searchSentence, String matchWord) {
Pattern pattern = Pattern.compile("(?<!\\S)" + Pattern.quote(matchWord) + "(?!\\S)", Pattern.CASE_INSENSITIVE);
Matcher matcher = pattern.matcher(searchSentence);
return matcher.find();
}
Do not forget to Pattern.quote your literal strings.
Using Matcher#find is preferable since you do not need the initial/trailing .* and eliminates unnecessary overhead related to regex backtracking mechanism.

Regular expression 30 numbers + space + hypen + space+ 1 number

I have this code to find this pattern: 201409250200131738007947036000 - 1 ,inside the text
final String patternStr = "(\\d{30} - \\d{1})";
final Pattern p = Pattern.compile(patternStr);
final Matcher m = p.matcher(page);
if (m.matches()) {
System.out.println("SUCCESS");
}
But for any strange reasson in Java did't work, Can somebody help me where is the error please?

The reason is that the matches method checks for the entire given string to match the regex.
So i.e. if your string is 123456123412345612341234561234 - 8 it will match, if it is my number 123456123412345612341234561234 - 8 is inside other text it won't.
Use the find method to accomplish your task:
if (m.find()) {
System.out.println("SUCCESS");
}
It will search inside the given string instead of attempting to match the entire string.

From the documentation for Matcher, matches:
Attempts to match the entire region against the pattern.
As opposed to find which:
Attempts to find the next subsequence of the input sequence that matches the pattern.
So use matches to match an entire String against a pattern, use find to locate a pattern inside a String.
Try:
final String patternStr = "\\d{30}+\\s-\\s\\d";
final Pattern p = Pattern.compile(patternStr);
final Matcher m = p.matcher(page);
while (m.find()) {
System.out.printf("FOUND A MATCH: %s%n", matcher.group());
}
I edited your pattern slightly to make it more robust. This will print each match that it finds.

Find characters that match a regex's set

I have a regex w_p[a-z]
It would match input like w_pa, w_pb ... w_pz. I like to find which character exactly was matched i.e. a,b or z for the above input. Is this possible with java regex?

Yes, you need to capture:
final Pattern pattern = Pattern.compile("w_p([a-z])");
final Matcher m = pattern.matcher(input);
if (m.find())
// what is matched is in m.group(1)

Sure, use Regexpr groups. w_p([a-z]) defines a group for the character you are looking for.
Pattern p = Pattern.compile("w_p([a-z])");
Matcher matcher = p.matcher(input);
if (matcher.find()) {
String character = matcher.group(1)
}
matcher.group(0) contains all that was matched (w_pa or w_pb etc.)
matcher.group(1) contains what was found in the first () pair.
See the documentation for more information.

The REGEX will be something like this:
w_p([a-z])
So you will create a group from wich you can get the value

java regex: extract text after delimeter?

i am new to regular expressions in Java. I like to extract a string by using regular expressions.
This is my String: "Hello,World"
I like to extract the text after ",". The result would be "World". I tried this:
final Pattern pattern = Pattern.compile(",(.+?)");
final Matcher matcher = pattern.matcher("Hello,World");
matcher.find();
But what would be the next step?

You don't need Regex for this. You can simply split on comma and get the 2nd element from the array: -
System.out.println("Hello,World".split(",")[1]);
OUTPUT: -
World
But if you want to use Regex, you need to remove ? from your Regex.
? after + is used for Reluctant matching. It will only match W and stop there.
You don't need that here. You need to match until it can match.
So use greedy matching instead.
Here's the code with modified Regex: -
final Pattern pattern = Pattern.compile(",(.+)");
final Matcher matcher = pattern.matcher("Hello,World");
if (matcher.find()) {
System.out.println(matcher.group(1));
}
OUTPUT: -
World

Extending what you have, you need to remove the ? sign from your pattern to use the greedy matching and then process the matched group:
final Pattern pattern = Pattern.compile(",(.+)"); // removed your '?'
final Matcher matcher = pattern.matcher("Hello,World");
while (matcher.find()) {
String result = matcher.group(1);
// work with result
}
Other answers suggest different approaches to your problem and might offer better solution for what you need.

System.out.println( "Hello,World".replaceAll(".*,(.*)","$1") ); // output is "World"

You are using a reluctant expression and will only select a single character W, whereas you can use a greedy one and print your matched group content:
final Pattern pattern = Pattern.compile(",(.+)");
final Matcher matcher = pattern.matcher("Hello,World");
if (matcher.find()) {
System.out.println(matcher.group(1));
}
Output:
World
See Regex Pattern doc

how to use Matcher.replaceAll in java?

i have a file which contains "(*" and "*)". i want to remove everything between this two char sequences.
i used the following code but it didn't do anything with my string.
String regex = "\\(\\*.*\\*\\)";
Pattern pattern = Pattern.compile(regex);
Matcher matcher = pattern.matcher(input);
matcher.replaceAll("");
the 'input' is:
(* This program prints out a message. *)
program is
begin
write ("Hello, world!");
end;

You need to capture the return value of your matcher - it's replaceAll method returns the replaced String.
Additionally, use a regexp to match what you want to match, this time a parenthesized String. If you don't have some strange inputs, it may look like this:
String regex = "\\(\\*.*\\*\\)";
Pattern pattern = Pattern.compile(regex);
Matcher matcher = pattern.matcher(input);
String result = matcher.replaceAll("(\\*\\*)");
System.out.println(result);
This regexp in fact captures the whole region from the first comment start to the last comment end, which would usually not be what you want. To let it match non-greedy (reluctantly), use this regexp: \(\*.*?\*\) (with doubled backslashes in Java.)