Develop Reference

What is the Regex for decimal numbers in Java?

I am not quite sure of what is the correct regex for the period in Java. Here are some of my attempts. Sadly, they all meant any character.
String regex = "[0-9]*[.]?[0-9]*";
String regex = "[0-9]*['.']?[0-9]*";
String regex = "[0-9]*["."]?[0-9]*";
String regex = "[0-9]*[\.]?[0-9]*";
String regex = "[0-9]*[\\.]?[0-9]*";
String regex = "[0-9]*.?[0-9]*";
String regex = "[0-9]*\.?[0-9]*";
String regex = "[0-9]*\\.?[0-9]*";
But what I want is the actual "." character itself. Anyone have an idea?
What I'm trying to do actually is to write out the regex for a non-negative real number (decimals allowed). So the possibilities are: 12.2, 3.7, 2., 0.3, .89, 19
String regex = "[0-9]*['.']?[0-9]*";
Pattern pattern = Pattern.compile(regex);
String x = "5p4";
Matcher matcher = pattern.matcher(x);
System.out.println(matcher.find());
The last line is supposed to print false but prints true anyway. I think my regex is wrong though.

Update
To match non negative decimal number you need this regex:
^\d*\.\d+|\d+\.\d*$
or in java syntax : "^\\d*\\.\\d+|\\d+\\.\\d*$"
String regex = "^\\d*\\.\\d+|\\d+\\.\\d*$"
String string = "123.43253";
if(string.matches(regex))
System.out.println("true");
else
System.out.println("false");
Explanation for your original regex attempts:
[0-9]*\.?[0-9]*
with java escape it becomes :
"[0-9]*\\.?[0-9]*";
if you need to make the dot as mandatory you remove the ? mark:
[0-9]*\.[0-9]*
but this will accept just a dot without any number as well... So, if you want the validation to consider number as mandatory you use + ( which means one or more) instead of *(which means zero or more). That case it becomes:
[0-9]+\.[0-9]+

If you on Kotlin, use ktx:
fun String.findDecimalDigits() =
Pattern.compile("^[0-9]*\\.?[0-9]*").matcher(this).run { if (find()) group() else "" }!!

Your initial understanding was probably right, but you were being thrown because when using matcher.find(), your regex will find the first valid match within the string, and all of your examples would match a zero-length string.
I would suggest "^([0-9]+\\.?[0-9]*|\\.[0-9]+)$"

There are actually 2 ways to match a literal .. One is using backslash-escaping like you do there \\., and the other way is to enclose it inside a character class or the square brackets like [.]. Most of the special characters become literal characters inside the square brackets including .. So use \\. shows your intention clearer than [.] if all you want is to match a literal dot .. Use [] if you need to match multiple things which represents match this or that for example this regex [\\d.] means match a single digit or a literal dot

I have tested all the cases.
public static boolean isDecimal(String input) {
return Pattern.matches("^[-+]?\\d*[.]?\\d+|^[-+]?\\d+[.]?\\d*", input);
}

Insert character before specific character Java

I have been taking a look at the regular expressions and how to use it in Java for the problem I have to solve. I have to insert a \ before every ". This is what I have:
public class TestExpressions {
public static void main (String args[]) {
String test = "$('a:contains(\"CRUCERO\")')";
test = test.replaceAll("(\")","$1%");
System.out.println(test);
}
}
The ouput is:
$('a:contains("%CRUCERO"%)')
What I want is:
$('a:contains(\"CRUCERO\")')
I have changed % for \\ but have an error StringIndexOutofBounds don't know why. If someone can help me I would appreciate it, thank you in advance.

I have to insert a \ before every "
You can try with replace which automatically escapes all regex metacharacters and doesn't use any special characters in replacement part so you can simply use String literals you want to be put in matched part.
So lets just replace " with \" literal. You can write it as
test = test.replace("\"", "\\\"");

If you want to insert backspace before quote then use:
test = test.replaceAll("(\")","\\\\$1"); // $('a:contains(\"CRUCERO\")')
Or if you want to avoid already escaped quote then use negative lookbehind:
String test = "$('a:contains(\\\"CRUCERO\")')";
test = test.replaceAll("((?<!\\\\)\")","\\\\$1"); // $('a:contains(\"CRUCERO\")')

String result = subject.replaceAll("(?i)\"CRUCERO\"", "\\\"CRUCERO\\\"");
EXPLANATION:
Match the character string “"CRUCERO"” literally (case insensitive) «"CRUCERO"»
Ignore unescaped backslash «\»
Insert the character string “"CRUCERO” literally «"CRUCERO»
Ignore unescaped backslash «\»
Insert the character “"” literally «"»

If your goal is escape text for Java strings, then instead of regular expressions, consider using
String escaped = org.apache.commons.lang.StringEscapeUtils.
escapeJava("$('a:contains(\"CRUCERO\")')");
System.out.println(escaped);
Output:
$('a:contains(\"CRUCERO\")')
JavaDoc: http://commons.apache.org/proper/commons-lang/javadocs/api-2.6/org/apache/commons/lang/StringEscapeUtils.html#escapeJava(java.lang.String)

Regex for check string check that does not contain \ > <

I have to create a regular expression to check if the string contains only digits, alphabets and other symbols except \ > <. I am able to create for digits and alphabets. I have tried with [^\<>] to check if string doesn't contain \ > < special characters.
But it did't work. Can someone please suggest how it an be done.
Edit:
Might be a simple question, but I am just starting with regx.

\ is a special character in regex, even inside of a character class. If you want to use it as literal character, you have to escape it, so the regex would be
[^\\<>]
if you use it in Java you have to escape additionally for the string level, so it would appear as:
String regex = "[^\\\\<>]";

Matching against [^\<>] will succeed if the string contains even one character which is not a backslash or angle bracket. If you wish it to succeed only when no character in the string is one of the forbidden ones, use
^[^\\<>]+$

You can try the regular expression:
^[^\\<>]*$
e.g.
private static final Pattern REGEX_PATTERN =
Pattern.compile("^[^\\\\<>]*$");
public static void main(String[] args) {
System.out.println(
REGEX_PATTERN.matcher("Hello World!").matches()
); // prints "true"
System.out.println(
REGEX_PATTERN.matcher("<script ...></script>").matches()
); // prints "false"
}

Escaping special characters in Java Regular Expressions

Is there any method in Java or any open source library for escaping (not quoting) a special character (meta-character), in order to use it as a regular expression?
This would be very handy in dynamically building a regular expression, without having to manually escape each individual character.
For example, consider a simple regex like \d+\.\d+ that matches numbers with a decimal point like 1.2, as well as the following code:
String digit = "d";
String point = ".";
String regex1 = "\\d+\\.\\d+";
String regex2 = Pattern.quote(digit + "+" + point + digit + "+");
Pattern numbers1 = Pattern.compile(regex1);
Pattern numbers2 = Pattern.compile(regex2);
System.out.println("Regex 1: " + regex1);
if (numbers1.matcher("1.2").matches()) {
System.out.println("\tMatch");
} else {
System.out.println("\tNo match");
}
System.out.println("Regex 2: " + regex2);
if (numbers2.matcher("1.2").matches()) {
System.out.println("\tMatch");
} else {
System.out.println("\tNo match");
}
Not surprisingly, the output produced by the above code is:
Regex 1: \d+\.\d+
Match
Regex 2: \Qd+.d+\E
No match
That is, regex1 matches 1.2 but regex2 (which is "dynamically" built) does not (instead, it matches the literal string d+.d+).
So, is there a method that would automatically escape each regex meta-character?
If there were, let's say, a static escape() method in java.util.regex.Pattern, the output of
Pattern.escape('.')
would be the string "\.", but
Pattern.escape(',')
should just produce ",", since it is not a meta-character. Similarly,
Pattern.escape('d')
could produce "\d", since 'd' is used to denote digits (although escaping may not make sense in this case, as 'd' could mean literal 'd', which wouldn't be misunderstood by the regex interpeter to be something else, as would be the case with '.').

Is there any method in Java or any open source library for escaping (not quoting) a special character (meta-character), in order to use it as a regular expression?
If you are looking for a way to create constants that you can use in your regex patterns, then just prepending them with "\\" should work but there is no nice Pattern.escape('.') function to help with this.
So if you are trying to match "\\d" (the string \d instead of a decimal character) then you would do:
// this will match on \d as opposed to a decimal character
String matchBackslashD = "\\\\d";
// as opposed to
String matchDecimalDigit = "\\d";
The 4 slashes in the Java string turn into 2 slashes in the regex pattern. 2 backslashes in a regex pattern matches the backslash itself. Prepending any special character with backslash turns it into a normal character instead of a special one.
matchPeriod = "\\.";
matchPlus = "\\+";
matchParens = "\\(\\)";
...
In your post you use the Pattern.quote(string) method. This method wraps your pattern between "\\Q" and "\\E" so you can match a string even if it happens to have a special regex character in it (+, ., \\d, etc.)

I wrote this pattern:
Pattern SPECIAL_REGEX_CHARS = Pattern.compile("[{}()\\[\\].+*?^$\\\\|]");
And use it in this method:
String escapeSpecialRegexChars(String str) {
return SPECIAL_REGEX_CHARS.matcher(str).replaceAll("\\\\$0");
}
Then you can use it like this, for example:
Pattern toSafePattern(String text)
{
return Pattern.compile(".*" + escapeSpecialRegexChars(text) + ".*");
}
We needed to do that because, after escaping, we add some regex expressions. If not, you can simply use \Q and \E:
Pattern toSafePattern(String text)
{
return Pattern.compile(".*\\Q" + text + "\\E.*")
}

The only way the regex matcher knows you are looking for a digit and not the letter d is to escape the letter (\d). To type the regex escape character in java, you need to escape it (so \ becomes \\). So, there's no way around typing double backslashes for special regex chars.

The Pattern.quote(String s) sort of does what you want. However it leaves a little left to be desired; it doesn't actually escape the individual characters, just wraps the string with \Q...\E.
There is not a method that does exactly what you are looking for, but the good news is that it is actually fairly simple to escape all of the special characters in a Java regular expression:
regex.replaceAll("[\\W]", "\\\\$0")
Why does this work? Well, the documentation for Pattern specifically says that its permissible to escape non-alphabetic characters that don't necessarily have to be escaped:
It is an error to use a backslash prior to any alphabetic character that does not denote an escaped construct; these are reserved for future extensions to the regular-expression language. A backslash may be used prior to a non-alphabetic character regardless of whether that character is part of an unescaped construct.
For example, ; is not a special character in a regular expression. However, if you escape it, Pattern will still interpret \; as ;. Here are a few more examples:
> becomes \> which is equivalent to >
[ becomes \[ which is the escaped form of [
8 is still 8.
\) becomes \\\) which is the escaped forms of \ and ( concatenated.
Note: The key is is the definition of "non-alphabetic", which in the documentation really means "non-word" characters, or characters outside the character set [a-zA-Z_0-9].

Use this Utility function escapeQuotes() in order to escape strings in between Groups and Sets of a RegualrExpression.
List of Regex Literals to escape <([{\^-=$!|]})?*+.>
public class RegexUtils {
static String escapeChars = "\\.?![]{}()<>*+-=^$|";
public static String escapeQuotes(String str) {
if(str != null && str.length() > 0) {
return str.replaceAll("[\\W]", "\\\\$0"); // \W designates non-word characters
}
return "";
}
}
From the Pattern class the backslash character ('\') serves to introduce escaped constructs. The string literal "\(hello\)" is illegal and leads to a compile-time error; in order to match the string (hello) the string literal "\\(hello\\)" must be used.
Example: String to be matched (hello) and the regex with a group is (\(hello\)). Form here you only need to escape matched string as shown below. Test Regex online
public static void main(String[] args) {
String matched = "(hello)", regexExpGrup = "(" + escapeQuotes(matched) + ")";
System.out.println("Regex : "+ regexExpGrup); // (\(hello\))
}

Agree with Gray, as you may need your pattern to have both litrals (\[, \]) and meta-characters ([, ]). so with some utility you should be able to escape all character first and then you can add meta-characters you want to add on same pattern.

use
pattern.compile("\"");
String s= p.toString()+"yourcontent"+p.toString();
will give result as yourcontent as is

How to replace special characters in a string?

I have a string with lots of special characters. I want to remove all those, but keep alphabetical characters.
How can I do this?

That depends on what you mean. If you just want to get rid of them, do this:
(Update: Apparently you want to keep digits as well, use the second lines in that case)
String alphaOnly = input.replaceAll("[^a-zA-Z]+","");
String alphaAndDigits = input.replaceAll("[^a-zA-Z0-9]+","");
or the equivalent:
String alphaOnly = input.replaceAll("[^\\p{Alpha}]+","");
String alphaAndDigits = input.replaceAll("[^\\p{Alpha}\\p{Digit}]+","");
(All of these can be significantly improved by precompiling the regex pattern and storing it in a constant)
Or, with Guava:
private static final CharMatcher ALNUM =
CharMatcher.inRange('a', 'z').or(CharMatcher.inRange('A', 'Z'))
.or(CharMatcher.inRange('0', '9')).precomputed();
// ...
String alphaAndDigits = ALNUM.retainFrom(input);
But if you want to turn accented characters into something sensible that's still ascii, look at these questions:
Converting Java String to ASCII
Java change áéőűú to aeouu
ń ǹ ň ñ ṅ ņ ṇ ṋ ṉ ̈ ɲ ƞ ᶇ ɳ ȵ --> n or Remove diacritical marks from unicode chars

I am using this.
s = s.replaceAll("\\W", "");
It replace all special characters from string.
Here
\w : A word character, short for [a-zA-Z_0-9]
\W : A non-word character

You can use the following method to keep alphanumeric characters.
replaceAll("[^a-zA-Z0-9]", "");
And if you want to keep only alphabetical characters use this
replaceAll("[^a-zA-Z]", "");

Replace any special characters by
replaceAll("\\your special character","new character");
ex:to replace all the occurrence of * with white space
replaceAll("\\*","");
*this statement can only replace one type of special character at a time

Following the example of the Andrzej Doyle's answer, I think the better solution is to use org.apache.commons.lang3.StringUtils.stripAccents():
package bla.bla.utility;
import org.apache.commons.lang3.StringUtils;
public class UriUtility {
public static String normalizeUri(String s) {
String r = StringUtils.stripAccents(s);
r = r.replace(" ", "_");
r = r.replaceAll("[^\\.A-Za-z0-9_]", "");
return r;
}
}

string Output = Regex.Replace(Input, #"([ a-zA-Z0-9&, _]|^\s)", "");
Here all the special characters except space, comma, and ampersand are replaced. You can also omit space, comma and ampersand by the following regular expression.
string Output = Regex.Replace(Input, #"([ a-zA-Z0-9_]|^\s)", "");
Where Input is the string which we need to replace the characters.

Here is a function I used to remove all possible special characters from the string
let name = name.replace(/[&\/\\#,+()$~%!.„'":*‚^_¤?<>|#ª{«»§}©®™ ]/g, '').toLowerCase();

You can use basic regular expressions on strings to find all special characters or use pattern and matcher classes to search/modify/delete user defined strings. This link has some simple and easy to understand examples for regular expressions: http://www.vogella.de/articles/JavaRegularExpressions/article.html

You can get unicode for that junk character from charactermap tool in window pc and add \u e.g. \u00a9 for copyright symbol.
Now you can use that string with that particular junk caharacter, don't remove any junk character but replace with proper unicode.

For spaces use "[^a-z A-Z 0-9]" this pattern