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trouble with writing regex java

String always consists of two distinct alternating characters. For example, if string 's two distinct characters are x and y, then t could be xyxyx or yxyxy but not xxyy or xyyx.
But a.matches() always returns false and output becomes 0. Help me understand what's wrong here.
public static int check(String a) {
char on = a.charAt(0);
char to = a.charAt(1);
if(on != to) {
if(a.matches("["+on+"("+to+""+on+")*]|["+to+"("+on+""+to+")*]")) {
return a.length();
}
}
return 0;
}

Use regex (.)(.)(?:\1\2)*\1?.
(.) Match any character, and capture it as group 1
(.) Match any character, and capture it as group 2
\1 Match the same characters as was captured in group 1
\2 Match the same characters as was captured in group 2
(?:\1\2)* Match 0 or more pairs of group 1+2
\1? Optionally match a dangling group 1
Input must be at least two characters long. Empty string and one-character string will not match.
As java code, that would be:
if (a.matches("(.)(.)(?:\\1\\2)*\\1?")) {
See regex101.com for working examples1.
1) Note that regex101 requires use of ^ and $, which are implied by the matches() method. It also requires use of flags g and m to showcase multiple examples at the same time.
UPDATE
As pointed out by Austin Anderson:
fails on yyyyyyyyy or xxxxxx
To prevent that, we can add a zero-width negative lookahead, to ensure input doesn't start with two of the same character:
(?!(.)\1)(.)(.)(?:\2\3)*\2?
See regex101.com.
Or you can use Austin Anderson's simpler version:
(.)(?!\1)(.)(?:\1\2)*\1?

Actually your regex is almost correct but problem is that you have enclosed your regex in 2 character classes and you need to match an optional 2nd character in the end.
You just need to use this regex:
public static int check(String a) {
if (a.length() < 2)
return 0;
char on = a.charAt(0);
char to = a.charAt(1);
if(on != to) {
String re = on+"("+to+on+")*"+to+"?|"+to+"("+on+to+")*"+on+"?";
System.out.println("re: " + re);
if(a.matches(re)) {
return a.length();
}
}
return 0;
}
Code Demo

Regular Expression allows specific special characters in java

I need to know the regular expression for string that contains alphanumeric characters, #, underscore(_), full stop(.)and not any blank spaces. And also for alphanumeric characters and it allow spaces. I tried with this regex,
^[_A-Za-z0-9-\\.\\#]$ and ^[A-Za-z0-9-\\s]$
CODE:
private static final String Username_REGEX ="^[_A-Za-z0-9.#-]$";
public static boolean isUsername(EditText editText, boolean required) {
return isValid(editText, Username_REGEX,Username_MSG, required);
}
public static boolean isValid(EditText editText, String regex, String errMsg, boolean required) {
String text = editText.getText().toString().trim();
editText.setError(null);
if ( required && !hasTextemt(editText) ) return false;
if (required && !Pattern.matches(regex, text)) {
editText.setError(errMsg);
return false;
};
return true;
}
public static boolean hasTextemt(EditText editText) {
String text = editText.getText().toString().trim();
editText.setError(null);
if (text.length() == 0) {
editText.setError(emt);
return false;
}
return true;
}
Is this correct? I did not get proper result. Can anyone guide me?

Move the dash - at the end of the character class:
^[_A-Za-z0-9.#-]+$
and
^[A-Za-z0-9\\s-]+$
Between two characters it means a range.
Edit: You also need a + modifier to match one or more of the characters in the character class.

I am assuming that you are getting this input via an EditText widget. So inside the layout of the XML file you can add the following properties by which it will receive only specified characters. :
android:digits="abcdefghijklmnopqrstuvwxyz0123456789,.-#_"
note that it wont allow any capital letter.
just add any digits/keys you want your user to be able to enter. If you are not worried about the patterns and number of occurrence of any character then you don't even need any regex.
Hope it helps

Try
"[\\w#\\.]+" //for alphanumeric, #, .
"[\\w\\s]+" //for alphanumeric, spaces
Add ^ and $ if you need that matches the whole word.
PS: For testing regexp I always use RegexPlanet (not spam :P)
Hope it helps.

You are only missing a quantifier. In your expression ^[_A-Za-z0-9.#-]$, the character class [_A-Za-z0-9.#-] matches exactly one character out of the class. To allow repeated characters, you need to define a quantifier.
* short for {0,} matches 0 or more characters (==> this allows the empty string!)
+ short for {1,} matches 1 or more characters
{n,m} matches minimum n and maximum m characters.
So your regex would look like
^[_A-Za-z0-9.#-]+$
if you require 1 or more characters, or
^[_A-Za-z0-9.#-]{6,20}$
if you want at least 6 characters and at most 20.
Other things:
You can replace _A-Za-z0-9 by \w, but be aware, \w is Unicode based and contains all letters and digits from all languages.
A-Za-z is only ASCII, maybe you want to have a look at Unicode properties. With e.g. \p{L} you can match a letter of any language.

You're missing a plus sign (meaning one or more) at the end of the character class, and you can simplify considerably:
^[\\w.#]+$
Characters within a character class lose their special meanings so don't need to be escaped, except for square brackets and a couple of others.
For alphanumeric and spaces only, that is only combinations of letters, numbers and spaces:
^[a-zA-Z0-9 ]+$

Remove Special Characters For A Pattern Java

I want to remove that characters from a String:
+ - ! ( ) { } [ ] ^ ~ : \
also I want to remove them:
/*
*/
&&
||
I mean that I will not remove & or | I will remove them if the second character follows the first one (/* */ && ||)
How can I do that efficiently and fast at Java?
Example:
a:b+c1|x||c*(?)
will be:
abc1|xc*?

This can be done via a long, but actually very simple regex.
String aString = "a:b+c1|x||c*(?)";
String sanitizedString = aString.replaceAll("[+\\-!(){}\\[\\]^~:\\\\]|/\\*|\\*/|&&|\\|\\|", "");
System.out.println(sanitizedString);

I think that the java.lang.String.replaceAll(String regex, String replacement) is all you need:
http://docs.oracle.com/javase/6/docs/api/java/lang/String.html#replaceAll(java.lang.String, java.lang.String).

there is two way to do that :
1)
ArrayList<String> arrayList = new ArrayList<String>();
arrayList.add("+");
arrayList.add("-");
arrayList.add("||");
arrayList.add("&&");
arrayList.add("(");
arrayList.add(")");
arrayList.add("{");
arrayList.add("}");
arrayList.add("[");
arrayList.add("]");
arrayList.add("~");
arrayList.add("^");
arrayList.add(":");
arrayList.add("/");
arrayList.add("/*");
arrayList.add("*/");
String string = "a:b+c1|x||c*(?)";
for (int i = 0; i < arrayList.size(); i++) {
if (string.contains(arrayList.get(i)));
string=string.replace(arrayList.get(i), "");
}
System.out.println(string);
2)
String string = "a:b+c1|x||c*(?)";
string = string.replaceAll("[+\\-!(){}\\[\\]^~:\\\\]|/\\*|\\*/|&&|\\|\\|", "");
System.out.println(string);

Thomas wrote on How to remove special characters from a string?:
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]","");
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.

How can I validate white space?

I have to validate a String with only white spaces. In my String white spaces between a String is allowed but only white spaces not allowed. For example "conditions apply","conditions" etc is allowed but not " ". ie, only white space not allowed
I want a regular expression in JavaScript for this

Try this regular expression
".*\\S+.*"

Do you really need to use a regex?
if (str.trim().length() == 0)
return false;
else
return true;
As mentionned in the comments, this could be simplified to a one-liner
return str.trim().length() > 0;
or, since Java 6
return !str.trim().isEmpty();

you can do it like this:
// This does replace all whitespaces at the end of the string
String s = " ".trim();
if(s.equals(""))
System.out.println(true);
else
System.out.println(s);

what about checking if the string does not match "\\s+"?

The regular expression is ^\\s*$ is used to match a whitespace only string, you can validate against this.
^ # Match the start of the string
\\s* # Match zero of more whitespace characters
$ # Match the end of the string
Anchoring to the start and the end of the string is important here.

Regex for numeric portion of Java string

I'm trying to write a Java method that will take a string as a parameter and return another string if it matches a pattern, and null otherwise. The pattern:
Starts with a number (1+ digits); then followed by
A colon (":"); then followed by
A single whitespace (" "); then followed by
Any Java string of 1+ characters
Hence, some valid string thats match this pattern:
50: hello
1: d
10938484: 394958558
And some strings that do not match this pattern:
korfed49
: e4949
6
6:
6:sdjjd4
The general skeleton of the method is this:
public String extractNumber(String toMatch) {
// If toMatch matches the pattern, extract the first number
// (everything prior to the colon).
// Else, return null.
}
Here's my best attempt so far, but I know I'm wrong:
public String extractNumber(String toMatch) {
// If toMatch matches the pattern, extract the first number
// (everything prior to the colon).
String regex = "???";
if(toMatch.matches(regex))
return toMatch.substring(0, toMatch.indexOf(":"));
// Else, return null.
return null;
}
Thanks in advance.

Your description is spot on, now it just needs to be translated to a regex:
^ # Starts
\d+ # with a number (1+ digits); then followed by
: # A colon (":"); then followed by
# A single whitespace (" "); then followed by
\w+ # Any word character, one one more times
$ # (followed by the end of input)
Giving, in a Java string:
"^\\d+: \\w+$"
You also want to capture the numbers: put parentheses around \d+, use a Matcher, and capture group 1 if there is a match:
private static final Pattern PATTERN = Pattern.compile("^(\\d+): \\w+$");
// ...
public String extractNumber(String toMatch) {
Matcher m = PATTERN.matcher(toMatch);
return m.find() ? m.group(1) : null;
}
Note: in Java, \w only matches ASCII characters and digits (this is not the case for .NET languages for instance) and it will also match an underscore. If you don't want the underscore, you can use (Java specific syntax):
[\w&&[^_]]
instead of \w for the last part of the regex, giving:
"^(\\d+): [\\w&&[^_]]+$"

Try using the following: \d+: \w+