Develop Reference

Regex for extracting all heading digits from a string

I am trying to extract all heading digits from a string using Java regex without writing additional code and I could not find something to work:
"12345XYZ6789ABC" should give me "12345".
"X12345XYZ6789ABC" should give me nothing
public final class NumberExtractor {
private static final Pattern DIGITS = Pattern.compile("what should be my regex here?");
public static Optional<Long> headNumber(String token) {
var matcher = DIGITS.matcher(token);
return matcher.find() ? Optional.of(Long.valueOf(matcher.group())) : Optional.empty();
}
}

Use a word boundary \b:
\b\d+
See live demo.
If you strictly want to match only digits at the start of the input, and not from each word (same thing when the input contains only one word), use ^:
^\d+
Pattern DIGITS = Pattern.compile("\\b\\d+"); // leading digits of all words
Pattern DIGITS = Pattern.compile("^\\d+"); // leading digits of input

I'd think something like "^[0-9]*" would work. There's a \d that matches other Unicode digits if you want to include them as well.
Edit: removed errant . from the string.

Matching everything after the first comma in a string

I am using java to do a regular expression match. I am using rubular to verify the match and ideone to test my code.
I got a regex from this SO solution , and it matches the group as I want it to in rubular, but my implementation in java is not matching. When it prints 'value', it is printing the value of commaSeparatedString and not matcher.group(1) I want the captured group/output of println to be "v123_gpbpvl-testpv1,v223_gpbpvl-testpv1-iso"
String commaSeparatedString = "Vtest7,v123_gpbpvl-testpv1,v223_gpbpvl-testpv1-iso";
//match everything after first comma
String myRegex = ",(.*)";
Pattern pattern = Pattern.compile(myRegex);
Matcher matcher = pattern.matcher(commaSeparatedString);
String value = "";
if (matcher.matches())
value = matcher.group(1);
else
value = commaSeparatedString;
System.out.println(value);
(edit: I left out that commaSeparatedString will not always contain 2 commas. Rather, it will always contain 0 or more commas)

If you don't have to solve it with regex, you can try this:
int size = commaSeparatedString.length();
value = commaSeparatedString.substring(commaSeparatedString.indexOf(",")+1,size);
Namely, the code above returns the substring which starts from the first comma's index.
EDIT:
Sorry, I've omitted the simpler version. Thanks to one of the commentators, you can use this single line as well:
value = commaSeparatedString.substring( commaSeparatedString.indexOf(",") );

The definition of the regex is wrong. It should be:
String myRegex = "[^,]*,(.*)";

You are yet another victim of Java's misguided regex method naming.
.matches() automatically anchors the regex at the beginning and end (which is in total contradiction with the very definition of "regex matching"). The method you are looking for is .find().
However, for such a simple problem, it is better to go with #DelShekasteh's solution.

I would do this like
String commaSeparatedString = "Vtest7,v123_gpbpvl-testpv1,v223_gpbpvl-testpv1-iso";
System.out.println(commaSeparatedString.substring(commaSeparatedString.indexOf(",")+1));

Here is another approach with limited split
String[] spl = "Vtest7,v123_gpbpvl-testpv1,v223_gpbpvl-testpv1-iso".split(",", 2);
if (spl.length == 2)
System.out.println(spl[1]);
Byt IMHO Del's answer is best for your case.

I would use replaceFirst
String commaSeparatedString = "Vtest7,v123_gpbpvl-testpv1,v223_gpbpvl-testpv1-iso";
System.out.println(commaSeparatedString.replaceFirst(".*?,", ""));
prints
v123_gpbpvl-testpv1,v223_gpbpvl-testpv1-iso
or you could use the shorter but obtuse
System.out.println(commaSeparatedString.split(",", 2)[1]);

How to remove special characters from a string?

I want to remove special characters like:
- + ^ . : ,
from an String using Java.

That depends on what you define as special characters, but try replaceAll(...):
String result = yourString.replaceAll("[-+.^:,]","");
Note that the ^ character must not be the first one in the list, since you'd then either have to escape it or it would mean "any but these characters".
Another note: the - character needs to be the first or last one on the list, otherwise you'd have to escape it or it would define a range ( e.g. :-, would mean "all characters in the range : to ,).
So, in order to keep consistency and not depend on character positioning, you might want to escape all those characters that have a special meaning in regular expressions (the following list is not complete, so be aware of other characters like (, {, $ etc.):
String result = yourString.replaceAll("[\\-\\+\\.\\^:,]","");
If you want to get rid of all punctuation and symbols, try this regex: \p{P}\p{S} (keep in mind that in Java strings you'd have to escape back slashes: "\\p{P}\\p{S}").
A third way could be something like this, if you can exactly define what should be left in your string:
String result = yourString.replaceAll("[^\\w\\s]","");
This means: replace everything that is not a word character (a-z in any case, 0-9 or _) or whitespace.
Edit: please note that there are a couple of other patterns that might prove helpful. However, I can't explain them all, so have a look at the reference section of regular-expressions.info.
Here's less restrictive alternative to the "define allowed characters" approach, as suggested by Ray:
String result = yourString.replaceAll("[^\\p{L}\\p{Z}]","");
The regex matches everything that is not a letter in any language and not a separator (whitespace, linebreak etc.). Note that you can't use [\P{L}\P{Z}] (upper case P means not having that property), since that would mean "everything that is not a letter or not whitespace", which almost matches everything, since letters are not whitespace and vice versa.
Additional information on Unicode
Some unicode characters seem to cause problems due to different possible ways to encode them (as a single code point or a combination of code points). Please refer to regular-expressions.info for more information.

This will replace all the characters except alphanumeric
replaceAll("[^A-Za-z0-9]","");

As described here
http://developer.android.com/reference/java/util/regex/Pattern.html
Patterns are compiled regular expressions. In many cases, convenience methods such as String.matches, String.replaceAll and String.split will be preferable, but if you need to do a lot of work with the same regular expression, it may be more efficient to compile it once and reuse it. The Pattern class and its companion, Matcher, also offer more functionality than the small amount exposed by String.
public class RegularExpressionTest {
public static void main(String[] args) {
System.out.println("String is = "+getOnlyStrings("!&(*^*(^(+one(&(^()(*)(*&^%$##!#$%^&*()("));
System.out.println("Number is = "+getOnlyDigits("&(*^*(^(+91-&*9hi-639-0097(&(^("));
}
public static String getOnlyDigits(String s) {
Pattern pattern = Pattern.compile("[^0-9]");
Matcher matcher = pattern.matcher(s);
String number = matcher.replaceAll("");
return number;
}
public static String getOnlyStrings(String s) {
Pattern pattern = Pattern.compile("[^a-z A-Z]");
Matcher matcher = pattern.matcher(s);
String number = matcher.replaceAll("");
return number;
}
}
Result
String is = one
Number is = 9196390097

Try replaceAll() method of the String class.
BTW here is the method, return type and parameters.
public String replaceAll(String regex,
String replacement)
Example:
String str = "Hello +-^ my + - friends ^ ^^-- ^^^ +!";
str = str.replaceAll("[-+^]*", "");
It should remove all the {'^', '+', '-'} chars that you wanted to remove!

To Remove Special character
String t2 = "!##$%^&*()-';,./?><+abdd";
t2 = t2.replaceAll("\\W+","");
Output will be : abdd.
This works perfectly.

Use the String.replaceAll() method in Java.
replaceAll should be good enough for your problem.

You can remove single char as follows:
String str="+919595354336";
String result = str.replaceAll("\\\\+","");
System.out.println(result);
OUTPUT:
919595354336

If you just want to do a literal replace in java, use Pattern.quote(string) to escape any string to a literal.
myString.replaceAll(Pattern.quote(matchingStr), replacementStr)

Extract an ISBN with regex

I have an extremely long string that I want to parse for a numeric value that occurs after the substring "ISBN". However, this grouping of 13 digits can be arranged differently via the "-" character. Examples: (these are all valid ISBNs) 123-456-789-123-4, OR 1-2-3-4-5-67891234, OR 12-34-56-78-91-23-4. Essentially, I want to use a regex pattern matcher on the potential ISBN to see if there is a valid 13 digit ISBN. How do I 'ignore' the "-" character so I can just regex for a \d{13} pattern? My function:
public String parseISBN (String sourceCode) {
int location = sourceCode.indexOf("ISBN") + 5;
String ISBN = sourceCode.substring(location); //substring after "ISBN" occurs
int i = 0;
while ( ISBN.charAt(i) != ' ' )
i++;
ISBN = ISBN.substring(0, i); //should contain potential ISBN value
Pattern pattern = Pattern.compile("\\d{13}"); //this clearly will find 13 consecutive numbers, but I need it to ignore the "-" character
Matcher matcher = pattern.matcher(ISBN);
if (matcher.find()) return ISBN;
else return null;
}

Alternative 1:
pattern.matcher(ISBN.replace("-", ""))
Alternative 2: Something like
Pattern.compile("(\\d-?){13}")
Demo of second alternative:
String ISBN = "ISBN: 123-456-789-112-3, ISBN: 1234567891123";
Pattern pattern = Pattern.compile("(\\d-?){13}");
Matcher matcher = pattern.matcher(ISBN);
while (matcher.find())
System.out.println(matcher.group());
Output:
123-456-789-112-3
1234567891123

Try this:
Pattern.compile("\\d(-?\\d){12}")

Use this pattern:
Pattern.compile("(?:\\d-?){13}")
and strip all dashes from the found isbn number

Do it in one step with a pattern recognizing everything, and optional dashes between digits. No need to fiddle with ISBN offset + substrings.
ISBN(\d(-?\d){12})
If you want the raw number, strip dashes from the first matched subgroup afterwards.
I am not a Java guy so I won't show you code.

If you're going to be calling the method a lot, the best thing you can do is not compile the Pattern inside it. Otherwise, each time you call the method you'll spend more time creating the regex than you will actually searching for it.
But after looking at your code again, I think you have a bigger problem, performance-wise. All that business of locating "ISBN" and then creating substrings to apply the regex to is completely unnecessary. Let the regex do that stuff; it's what they're for. The following regex finds the "ISBN" sentinel and the following thirteen digits, if they're there:
static final Pattern isbnPattern = Pattern.compile(
"\\bISBN[^A-Z0-9]*+(\\d(?:-*+\\d){12})", Pattern.CASE_INSENSITIVE );
The [^A-Z0-9]*+ gobbles up whatever characters may appear between the "ISBN" and the first digit. The possessive quantifier (*+) prevents needless backtracking; if the next character is not a digit, the regex engine immediately quits that match attempt and resumes scanning for another "ISBN" instance.
I used another possessive quantifier for the optional hyphens, plus a non-capturing group ((?:...)) for the repeated portion; that gives another slight performance gain over the capturing groups most of the other responders are using. But I used a capturing group for the whole number, so it can be extracted from the overall match easily. With these changes, your method reduces to this:
public String parseISBN (String source) {
Matcher m = isbnPattern.matcher(source);
return m.find() ? m.group(1) : null;
}
...and it's much more efficient, too. Note that we haven't addressed how the strings are getting into memory. If you're doing the I/O yourself, it's possible there are significant performance gains to be achieved in that area, too.

You can strip out the dashes with string manipulation, or you could use this:
"\\b(?:\\d-?){13}\\b"
It has the added bonus of making sure the string doesn't start or end with -.

Try stripping the dashes out, and regex the new string

you can try this
"(?:[0-9]{9}[0-9X]|[0-9]{13}|[0-9][0-9-]{11}[0-9X]|[0-9][0-9-]{15}[0-9])(?![0-9-])"

Regular expression to match unescaped special characters only

I'm trying to come up with a regular expression that can match only characters not preceded by a special escape sequence in a string.
For instance, in the string Is ? stranded//? , I want to be able to replace the ? which hasn't been escaped with another string, so I can have this result : **Is Dave stranded?**
But for the life of me I have not been able to figure out a way. I have only come up with regular expressions that eat all the replaceable characters.
How do you construct a regular expression that matches only characters not preceded by an escape sequence?

Use a negative lookbehind, it's what they were designed to do!
(?<!//)[?]
To break it down:
(
?<! #The negative look behind. It will check that the following slashes do not exist.
// #The slashes you are trying to avoid.
)
[\?] #Your special charactor list.
Only if the // cannot be found, it will progress with the rest of the search.
I think in Java it will need to be escaped again as a string something like:
Pattern p = Pattern.compile("(?<!//)[\\?]");

Try this Java code:
str="Is ? stranded//?";
Pattern p = Pattern.compile("(?<!//)([?])");
m = p.matcher(str);
StringBuffer sb = new StringBuffer();
while (m.find()) {
m.appendReplacement(sb, m.group(1).replace("?", "Dave"));
}
m.appendTail(sb);
String s = sb.toString().replace("//", "");
System.out.println("Output: " + s);
OUTPUT
Output: Is Dave stranded?

I was thinking about this and have a second simplier solution, avoiding regexs. The other answers are probably better but I thought I might post it anyway.
String input = "Is ? stranded//?";
String output = input
.replace("//?", "a717efbc-84a9-46bf-b1be-8a9fb714fce8")
.replace("?", "Dave")
.replace("a717efbc-84a9-46bf-b1be-8a9fb714fce8", "?");
Just protect the "//?" by replacing it with something unique (like a guid). Then you know any remaining question marks are fair game.

Use grouping. Here's one example:
import java.util.regex.*;
class Test {
public static void main(String[] args) {
Pattern p = Pattern.compile("([^/][^/])(\\?)");
String s = "Is ? stranded//?";
Matcher m = p.matcher(s);
if (m.matches)
s = m.replaceAll("$1XXX").replace("//", "");
System.out.println(s + " -> " + s);
}
}
Output:
$ java Test
Is ? stranded//? -> Is XXX stranded?
In this example, I'm:
first replacing any non-escaped ? with "XXX",
then, removing the "//" escape sequences.
EDIT Use if (m.matches) to ensure that you handle non-matching strings properly.
This is just a quick-and-dirty example. You need to flesh it out, obviously, to make it more robust. But it gets the general idea across.

Match on a set of characters OTHER than an escape sequence, then a regex special character. You could use an inverted character class ([^/]) for the first bit. Special case an unescaped regex character at the front of the string.

String aString = "Is ? stranded//?";
String regex = "(?<!//)[^a-z^A-Z^\\s^/]";
System.out.println(aString.replaceAll(regex, "Dave"));
The part of the regular expression [^a-z^A-Z^\\s^/] matches non-alphanumeric, whitespace or non-forward slash charaters.
The (?<!//) part does a negative lookbehind - see docco here for more info
This gives the output Is Dave stranded//?

try matching:
(^|(^.)|(.[^/])|([^/].))[special characters list]

I used this one:
((?:^|[^\\])(?:\\\\)*[ESCAPABLE CHARACTERS HERE])
Demo: https://regex101.com/r/zH1zO3/4