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Regex for detecting repeating symbols

I'm looking for the regex expression that will detect repeating symbols in a String. And currently I didn't found solution that fits all my requirements.
Requirements are pretty simple:
detect any repeating symbol in a String;
to be able to setup repeating count (eg. more than twice)
Examples of required detection (of symbol 'a', more than 2 times, true if detects, false otherwise)
"Abcdefg" - false
"AbcdaBCD" - false
"abcd_ab_ab" - true (symbol 'a' used three times)
"aabbaabb" - true (symbols 'a' used four times)
Since I'm not a pro in regex and usage of them - code snippet and explanation would be appreciated!
Thanks!

I think that
(.).*\1
would work:
(.) match a single character and capture
.* match any intervening characters
\1 match the captured group again.
(You'd need to compile with the DOTALL flag, or replace . with [\s\S] or similar if the string contains characters not ordinarily matched by .)
and if you want to require that it is found at least 3 times, just change the quantifier of the second two bullets:
(.)(.*\1){2}
etc.
This is going to be pretty inefficient, though, because it's going to have to do the "search for the next matching character" between every character in the string and the end of the string, making it at least quadratic.
You might be as well off not using regular expressions, e.g.
char[] cs = str.toCharArray();
Arrays.sort(cs);
int n = numOccurrencesRequired - 1;
for (int i = n; i < cs.length; ++i) {
boolean allSame = true;
for (int j = 1; j <= n && allSame; ++j) {
allSame = cs[i] == cs[i - j];
}
if (allSame) return true;
}
return false;
This sorts all of the same characters together, allowing you just to pass over the string once looking for adjacent equal characters.
Note that this doesn't quite work for any symbol: it will split up multi-char codepoints like 🍕. You can adapt the code above to work with codepoints, rather than chars.

Try this regex: (.)(?:.*\1)
It basically matches any character (.) is followed by anything .* and itself \1. If you want to check for 2 or more repeats only add {n,} at the end with n being the number of repeats you want to check for.

Yea, such regex exists but just because the set of characters is finite.
regex: .*(a.*a|b.*b|c.*c|...|y.*y|z.*z).*
It makes no sense. Use another approach:
String string = "something";
int[] count = new int[256];
for (int i = 0; i < string.length; i++) {
int temp = int(string.charAt(i));
count[temp]++;
}
Now you have all characters counted and you can use them as you wish.

Counting comma and any text in java String

I'm trying to write a function to count specific Strings.
The Strings to count look like the following:
first any character except comma at least once -
the comma -
any chracter but at least once
example string:
test, test, test,
should count to 3
I've tried do that by doing the following:
int countSubstrings = 0;
final Pattern pattern = Pattern.compile("[^,]*,.+");
final Matcher matcher = pattern.matcher(commaString);
while (matcher.find()) {
countSubstrings++;
}
Though my solution doesn't work. It always ends up counting to one and no further.

Try this pattern instead: [^,]+
As you can see in the API, find() will give you the next subsequence that matches the pattern. So this will find your sequences of "non-commas" one after the other.

Your regex, especially the .+ part will match any char sequence of at least length 1. You want the match to be reluctant/lazy so add a ?: [^,]*,.+?
Note that .+? will still match a comma that directly follows a comma so you might want to replace .+? with [^,]+ instead (since commas can't match with this lazyness is not needed).
Besides that an easier solution might be to split the string and get the length of the array (or loop and check the elements if you don't want to allow for empty strings):
countSubstrings = commaString.split(",").length;
Edit:
Since you added an example that clarifies your expectations, you need to adjust your regex. You seem to want to count the number of strings followed by a comma so your regex can be simplified to [^,]+,. This matches any char sequence consisting of non-comma chars which is followed by a comma.
Note that this wouldn't match multiple commas or text at the end of the input, e.g. test,,test would result in a count of 1. If you have that requirement you need to adjust your regex.

So, quite good answers are already given. Very readable. Something like this should work, beware, it's not clean and probably not the fastest way to do this. But is is quite readable. :)
public int countComma(String lots_of_words) {
int count = 0;
for (int x = 0; x < lots_of_words.length(); x++) {
if (lots_of_words.charAt(x) == ',') {
count++;
}
}
return count;
}
Or even better:
public int countChar(String lots_of_words, char the_chosen_char) {
int count = 0;
for (int x = 0; x < lots_of_words.length(); x++) {
if (lots_of_words.charAt(x) == the_chosen_char) {
count++;
}
}
return count;
}

How do I remove illegal characters in a subdomain?

I'm using Java 6. Using an Amazon AWS library, I'm dynamically creating domains. However, I'm looking for a function that can strip out illegal characters from a subdomain. E.g. if my function were about to create
dave'ssite.mydomain.com
I would like to pass the string "dave'ssite" to some function, which would strip out the apostrophe, or whatever other illegal characters lurked in the subdomain.
How do I do taht? THe more specific quesiton is, how do I identify what the illegal subdomain characters are?

Subdomains are same as Domains, so most likely the allowed characters are A-Z a-z 0-9 and -. There fore you can use Regex.
...
String s = "dave's-site.mydomain.com";
//prints daves-sitemydomaincom
System.out.println(s.replaceAll("[^A-Za-z0-9\\-]",""));
...

Here is some I made for a game. Its basically the same thing except I used it for removing invalid characters from a username.
char[] validChars = {a, b, c, d, etc...};//Put all valid characters in this array, so for subdomains put all letters and numbers
public static String cleanString(String text){
StringBuilder sb = new StringBuilder("");
for(int i = 0;i < text.length();i++){
for (int j = 0; j < validCharslength; j++) {
if (validChars[j] == text.charAt(i)) {
sb.append(text.charAt(i));
break;
}
}
}
return sb.toString();
}
As said my the comment in the code, the char array contains all the valid characters and anything else will be removed. Keep in mind that his is a return method.

I stumbled on here looking for a c# solution to the same problem and well here it is. Look how elegant c# LINQ makes this ;)
if (model.UserName.All(char.IsLetterOrDigit) && !model.UserName.StartsWith("-"))
{
//oh yeah, valid subdomain
}
Not sure what the spec says about length though.

the best way for character replacement in String in java

I want to check a string for each character I replace it with other characters or keep it in the string. and also because it's a long string the time to do this task is so important. what is the best way of these, or any better idea?
for all of them I append the result to an StringBuilder.
check all of the characters with a for and charAt commands.
use switch like the previous way.
use replaceAll twice.
and if one of the first to methods is better is there any way to check a character with a group of characters, like :
if (st.charAt(i)=='a'..'z') ....
Edit:
please tell the less consuming in time way and tell the reason.I know all of these ways you said!

If you want to replace a single character (or a single sequence), use replace(), as other answers have suggested.
If you want to replace several characters (e.g., 'a', 'b', and 'c') with a single substitute character or character sequence (e.g., "X"), you should use a regular expression replace:
String result = original.replaceAll("[abc]", "X");
If you want to replace several characters, each with a different replacement (e.g., 'a' with 'A', 'b' with 'B'), then looping through the string yourself and building the result in a StringBuilder will probably be the most efficient. This is because, as you point out in your question, you will be going through the string only once.
String sb = new StringBuilder();
String targets = "abc";
String replacements = "ABC";
for (int i = 0; i < result.length; ++i) {
char c = original.charAt(i);
int loc = targets.indexOf(c);
sb.append(loc >= 0 ? replacements.charAt(loc) : c);
}
String result = sb.toString();

Check the documentation and find some good methods:
char from = 'a';
char to = 'b';
str = str.replace(from, to);

String replaceSample = "This String replace Example shows
how to replace one char from String";
String newString = replaceSample.replace('r', 't');
Output: This Stting teplace Example shows how to teplace one chat ftom Stting
Also, you could use contains:
str1.toLowerCase().contains(str2.toLowerCase())
To check if the substring str2 exists in str1
Edit.
Just read that the String come from a file. You can use Regex for this. That would be the best method.
http://docs.oracle.com/javase/tutorial/essential/regex/literals.html

This is your comment:
I want to replace all of the uppercases to lower cases and replace all
of the characters except a-z with space.
You can do it like this:
str = str.toLowerCase().replaceAll("[^a-z]", " ");
Your requirement should be part of the question, not in comment #7 under a posted answer...

You should look into regex for Java. You can match an entire set of characters. Strings have several functions: replace, replaceAll, and match, which you may find useful here.
You can match the set of alphanumeric, for instance, using [a-zA-Z], which may be what you're looking for.

java check the first and last char are uppercase

I am trying to achieve this.
I have a string of 9char (always the same). But i also know that the first and last char is always a aplhabetic, it must be. the rest in the middle are numbers. How to check for that.
I got this logic so far, syntax is my problem
string samplestring;
samplestring = a1234567B
If(samplestring.length() == 9 && samplestring.substring(0,1).uppercase && samplestring.substring(8,9) && samplestring.THE REST OF THE CHAR IN THE MIDDLE ARE DIGITS)
{
println("yes this is correct");
}
else
{
println("retype");
}
Please dont mind about the simple english just want to know the syntax but the logic is there i hope..
Also can please show me those lowercase ones how to convert to uppercase?

A regular expression would be suitable:
String s = new String("A2345678Z");
if (s.matches("[A-Z][0-9]{7}[A-Z]")))
{
}
Regular expression explained:
[A-Z] means any uppercase letter
[0-9]{7} means 7 digits

Pattern p = Pattern.compile("^[A-Za-z]\\d+[A-Za-z]$");
Matcher m = p.match("A1234567B");
if (m.matches()) {
//
}
Edit:
If there are always seven digits, you can replace the \\d+ with \\d{7}

String str="A12345678B";
char first = str.charAt(0);
char second = str.charAt(str.length()-1);
if(Character.isUpperCase(first)&& Character.isUpperCase(second)){
//do something
}