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Replace characters and keep only one of these characters

Can someone help me here? I dont understand where's the problem...
I need check if a String have more than 1 char like 'a', if so i need replace all 'a' for a empty space, but i still want only one 'a'.
String text = "aaaasomethingsomethingaaaa";
for (char c: text.toCharArray()) {
if (c == 'a') {
count_A++;//8
if (count_A > 1) {//yes
//app crash at this point
do {
text.replace("a", "");
} while (count_A != 1);
}
}
}
the application stops working when it enters the while loop. Any suggestion? Thank you very much!

If you want to replace every a in the string except for the last one then you may try the following regex option:
String text = "aaaasomethingsomethingaaaa";
text = text.replaceAll("a(?=.*a)", " ");
somethingsomething a
Demo
Edit:
If you really want to remove every a except for the last one, then use this:
String text = "aaaasomethingsomethingaaaa";
text = text.replaceAll("a(?=.*a)", "");

You can also do it like
String str = new String ("asomethingsomethingaaaa");
int firstIndex = str.indexOf("a");
firstIndex++;
String firstPart = str.substring(0, firstIndex);
String secondPart = str.substring(firstIndex);
System.out.println(firstPart + secondPart.replace("a", ""));

Maybe I'm wrong here but I have a feeling your talking about runs of any single character within a string. If this is the case then you can just use a little method like this:
public String removeCharacterRuns(String inputString) {
return inputString.replaceAll("([a-zA-Z])\\1{2,}", "$1");
}
To use this method:
String text = "aaaasomethingsomethingaaaa";
System.out.println(removeCharacterRuns(text));
The console output is:
asomethingsomethinga
Or perhaps even:
String text = "FFFFFFFourrrrrrrrrrrty TTTTTwwwwwwooo --> is the answer to: "
+ "The Meeeeeaniiiing of liiiiife, The UUUniveeeerse and "
+ "Evvvvverything.";
System.out.println(removeCharacterRuns(text));
The console output is........
Fourty Two --> is the answer to: The Meaning of life, The Universe and Everything.
The Regular Expression used within the provided removeCharacterRuns() method was actually borrowed from the answers provided within this SO Post.
Regular Expression Explanation:

How to check if all characters in a String are all letters?

I'm able to separate the words in the sentence but I do not know how to check if a word contains a character other than a letter. You don't have to post an answer just some material I could read to help me.
public static void main(String args [])
{
String sentance;
String word;
int index = 1;
System.out.println("Enter sentance please");
sentance = EasyIn.getString();
String[] words = sentance.split(" ");
for ( String ss : words )
{
System.out.println("Word " + index + " is " + ss);
index++;
}
}

What I would do is use String#matches and use the regex [a-zA-Z]+.
String hello = "Hello!";
String hello1 = "Hello";
System.out.println(hello.matches("[a-zA-Z]+")); // false
System.out.println(hello1.matches("[a-zA-Z]+")); // true
Another solution is if (Character.isLetter(str.charAt(i)) inside a loop.
Another solution is something like this
String set = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ";
String word = "Hello!";
boolean notLetterFound;
for (char c : word.toCharArray()){ // loop through string as character array
if (!set.contains(c)){ // if a character is not found in the set
notLetterfound = true; // make notLetterFound true and break the loop
break;
}
}
if (notLetterFound){ // notLetterFound is true, do something
// do something
}
I prefer the first answer though, using String#matches

For more reference goto-> How to determine if a String has non-alphanumeric characters?
Make the following changes in pattern "[^a-zA-Z^]"

Not sure if I understand your question, but there is the
Character.isAlpha(c);
You would iterate over all characters in your string and check whether they are alphabetic (there are other "isXxxxx" methods in the Character class).

You could loop through the characters in the word calling Character.isLetter(), or maybe check if it matches a regular expression e.g. [\w]* (this would match the word only if its contents are all characters).

you can use charector array to do this like..
char[] a=ss.toCharArray();
not you can get the charector at the perticulor index.
with "word "+index+" is "+a[index];

Java - Check if last characters in a string are numeric

I'd like to know what I can use to check if the last characters in a string are numeric. For example for the ID: "S000123" .
I used the following to check if it starts with E, S or X.
if(! id.startsWith("E") || ! id.startsWith("S") || ! id.startsWith("X")) {
alertinputError("Incorrect format in the ID field", lineNum);
return;
}
Thanks.

You could simply use id.matches(".\\d+"). To check if id starts with E, S or X followed by a string of digits, however, you could use
id.matches("[ESX]\\d+")
Relevant Documentation
matches

I know this is a very old post but for future readers, I thought I'll add a different approach.
I personally found regex to be very unreadable (do you know someone that can write regex without using google?)
So I prefer this approach:
char lastChar = string.charAt(string.length() - 1)
return Character.isDigit();
If you want to do it for the last N letters and not just 1:
string.substring(string.length() - n).chars()
.allMatch(Character::isDigit);

A regex might work well here.
String foo = "bar123";
Pattern p = Pattern.compile(".*[0-9]+$");
Matcher m = p.matcher(foo);
if(m.matches)
System.out.println("yeah!");
So if the string is composed of any bunch of characters followed by one or more numbers, this will print "yeah!".

try this:
id.matches("^.+?\\d$")

Try it:
String id = "S000123";
Pattern pattern = Pattern.compile("[ESX][0-9].*");
Matcher matcher = pattern.matcher(id);
System.out.println(matcher.matches());

All right then you can use following Code:
boolean flag = true;
for (int i = id.length() - 1 ; i >= id.length() - 6 ; i--)
{
if (!Character.isDigit(id.charAt(i)))
{
flag = false;
break;
}
}
if (flag)
System.out.println("ID is valid");
else
System.out.println("ID is invalid");

Java String replace with string in same position

I'm having problems with strings and I need a solution I'm trying to replace characters found at a certain position with a character found in also the same position for example
private String wordNormalize(String enteredWord,String dictionary){
String normalizedWord = null;
// remove empty spaces at beginning and at the end of the word and change to lower case
normalizedWord = enteredWord.trim().toLowerCase();
//normalize term, removing all punctuation marks
normalizedWord = normalizedWord.replaceAll("["+punctuationMarks2+"]", "[b,v]");
//normalize word removing to character if dictionary has english lang
normalizedWord = normalizedWord.replaceFirst("to ", " ");
//normalizeWord if dictionary has german
if(normalizedWord.length() > 0){
normalizedWord.replace("a,b,c","t,u,v");
/*for(int i = 0;i<normalizedWord.length();i++){
char currentChar = normalizedWord.charAt(i); // currently typed character
String s1= Character.toString(currentChar);
for(int j = 0;j<specialCharacters.length;j++){
s1.replaceAll("[ "+specialCharacters[i]+" ]",""+replaceCharactersDe[i]+"");
}
= str.replace("a,b,c","t,u,v");
}*/
}
//normalize term removing special characters and replacing them
/*for(int i = 0; i > specialCharacters.length;i++){
if(normalizedWord.equals(specialCharacters[i])){
normalizedWord = replaceCharactersDe[i];
}
}*/
return normalizedWord;
}
So if a user enters a its replaced with t and if a user enters b its replaced with u and if the user enters c it will be replaced with v and only in that order is this possible and if it is show me the right way its supposed to be done

It is not clear to me what you are trying to approach with
normalizedWord = normalizedWord.replaceAll("["+punctuationMarks2+"]", "[b,v]");
It does not seem right, but i don't know how to fix it because I don't know what it's trying to do. I guess what you are looking for is
normalizedWord = normalizedWord.replaceAll("\\p{Punct}", "");
On the other part you are doing nothing, because Strings are immutable. You want to do something like
normalizedWord = normalizedWord.replace("a,b,c","t,u,v");
but that would replace all occurrences of the substring "a,b,c" with the string "t,u,v"-
What you want is:
normalizedWord = normalizedWord.replace('a', 't');
normalizedWord = normalizedWord.replace('b', 'u');
normalizedWord = normalizedWord.replace('c', 'v');
We could work on a more general solution, but you have to show us how the dictionary , which is a String, is formatted.

Using "Predefined character classes" on a substring

I want to check the last character of the String for characters that are not a non-word character using '\W' and allow certain symbols like ". , ! etc" from the top of my head I thought of using a code similar to this.
Boolean notCompleted = true;
int deduct = 1;
while(notCompleted){
if(string.charAt(string.length() -deduct) == '\W'){ // '\W' <-- doesn't work since it accepts anything other than "escape sequences".
if(string.charAt(string.length() -deduct) == '.'||string.charAt(string.length() -deduct) == ','||string.charAt(string.length() -deduct) == '!'){
//Do nothing and move on to the while loop
}else{
//Replace the non word character with ' '.
}
}
deduct++;
if(deduct >= html.length()){
notCompleted = false;
}
}
The reason why this doesn't work is because using string.charAt only accepts "Escapes sequence".
My question is there another way to pull this off rather than doing.
string.replaceAll("\W", "");
All suggestions is greatly appreciated. Thank you.
Thanks to the tip npinti gave me I built this code. However I am getting an error line
Desired Result of fakeNewString as requested should be "! asdsdefwef.,a,,sda.sd";
fakeNewString = sb.toString(); // NullPointerException
public static void test5(){
Boolean notCompleted = true;
String fakeNewString = "!##$%^&*( asdsdefwef.,a,,sda.sd";
int start = 0, end = 1;
StringBuilder sb = null;
try{
while(notCompleted){
start++;
String tempString = fakeNewString.substring(start, end);
if(Pattern.matches("\\W$", tempString)){
if(Pattern.matches("!", tempString)||Pattern.matches(".", tempString)||Pattern.matches(",", tempString)||Pattern.matches("\"", tempString)){
//do nothing
sb.append(tempString);
}else{
//Change it to spaces.
tempString = " ";
sb.append(tempString);
}
}
end++;
if(end >= fakeNewString.length()){
notCompleted = false;
fakeNewString = sb.toString();
System.out.println(fakeNewString);
}
}
}catch (Exception e) {
// TODO: handle exception
e.printStackTrace();
}
}

You can do something like so:
Pattern pattern = Pattern.compile("\\W$");
Matcher matcher = pattern.match(string);
if (matcher.find())
{
//do something when the string ends with a non word character
}
Take a look at this tutorial for more information on regular expressions.

You can use String.replaceAll in a slightly different way to do this. It achieves the same effect as the code you're trying to write, which seems like a complex solution for a simple problem. Try this code:
string.replaceAll("[^\\w!,.]", " ");
All the invalid characters are now replaced by a space, and multiple sequential occurrences of them are replaced by multiple spaces.

Lets try to break down the question (desire) and answer it:
I want to check the last character of the String for characters that are not a non-word character using '\W' and allow certain symbols like ". , ! etc"
First we have:
I want to check the last character of the String
Expression for character X at end of string:
X$
Then:
for characters that are not a non-word character
Expression:
[^\W] i.e. \w
And also:
allow certain symbols like ". , ! etc"
Added to the expression above:
[\w.,!]
And the combined final result is:
[\w.,!]$
Ta-da! (Altho I'm guessing OP is looking for something else, I did it for teh lulz.)