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Java Pattern / Matcher not finding word break

I am having trouble with Java Pattern and Matcher. I've included a very simplified example of what I'm trying to do.
I had expected the pattern ".\b" to find the last character of the first word (or "4" in the example), but as I step through the code, m.find() always returns false. What am I missing here?
Why does the following Java code always print out "Not Found"?
Pattern p = Pattern.compile(".\b");
Matcher m = p.matcher("102939384 is a word");
int ixEndWord = 0;
if (m.find()) {
ixEndWord = m.end();
System.out.println("Found: " + ixEndWord);
} else {
System.out.println("Not Found");
}

You need to escape special characters in the regex: ".\\b"
Basically, in a String the backslash has to be escaped. So "\\" becomes the character '\'.
So the String ".\\b" becomes the litteral String ".\b", which will be used by the Pattern.

To expand upton AntonH's comment, whenever you want the "\" character to appear in a regex expression, you have to escape it so that it first appears in the string you are passing in.
As is, ".\b" is the string of a dot . followed by the special backspace character represented by \b, compared to ".\\b", which is the regex .\b.

Java Regex to Validate Full Name allow only Spaces and Letters

I want regex to validate for only letters and spaces. Basically this is to validate full name. Ex: Mr Steve Collins or Steve Collins I tried this regex. "[a-zA-Z]+\.?" But didnt work. Can someone assist me please
p.s. I use Java.
public static boolean validateLetters(String txt) {
String regx = "[a-zA-Z]+\\.?";
Pattern pattern = Pattern.compile(regx,Pattern.CASE_INSENSITIVE);
Matcher matcher = pattern.matcher(txt);
return matcher.find();
}

What about:
Peter Müller
François Hollande
Patrick O'Brian
Silvana Koch-Mehrin
Validating names is a difficult issue, because valid names are not only consisting of the letters A-Z.
At least you should use the Unicode property for letters and add more special characters. A first approach could be e.g.:
String regx = "^[\\p{L} .'-]+$";
\\p{L} is a Unicode Character Property that matches any kind of letter from any language

try this regex (allowing Alphabets, Dots, Spaces):
"^[A-Za-z\s]{1,}[\.]{0,1}[A-Za-z\s]{0,}$" //regular
"^\pL+[\pL\pZ\pP]{0,}$" //unicode
This will also ensure DOT never comes at the start of the name.

For those who use java/android and struggle with this matter try:
"^\\p{L}+[\\p{L}\\p{Z}\\p{P}]{0,}"
This works with names like
José Brasão

You could even try this expression ^[a-zA-Z\\s]*$ for checking a string with only letters and spaces (nothing else).
For me it worked. Hope it works for you as well.
Or go through this piece of code once:
CharSequence inputStr = expression;
Pattern pattern = Pattern.compile(new String ("^[a-zA-Z\\s]*$"));
Matcher matcher = pattern.matcher(inputStr);
if(matcher.matches())
{
//if pattern matches
}
else
{
//if pattern does not matches
}

please try this regex (allow only Alphabets and space)
"[a-zA-Z][a-zA-Z ]*"
if you want it for IOS then,
NSString *yourstring = #"hello";
NSString *Regex = #"[a-zA-Z][a-zA-Z ]*";
NSPredicate *TestResult = [NSPredicate predicateWithFormat:#"SELF MATCHES %#",Regex];
if ([TestResult evaluateWithObject:yourstring] == true)
{
// validation passed
}
else
{
// invalid name
}

Regex pattern for matching only alphabets and white spaces:
String regexUserName = "^[A-Za-z\\s]+$";

Accept only character with space :-
if (!(Pattern.matches("^[\\p{L} .'-]+$", name.getText()))) {
JOptionPane.showMessageDialog(null, "Please enter a valid character", "Error", JOptionPane.ERROR_MESSAGE);
name.setFocusable(true);
}

My personal choice is:
^\p{L}+[\p{L}\p{Pd}\p{Zs}']*\p{L}+$|^\p{L}+$, Where:
^\p{L}+ - It should start with 1 or more letters.
[\p{Pd}\p{Zs}'\p{L}]* - It can have letters, space character (including invisible), dash or hyphen characters and ' in any order 0 or more times.
\p{L}+$ - It should finish with 1 or more letters.
|^\p{L}+$ - Or it just should contain 1 or more letters (It is done to support single letter names).
Support for dots (full stops) was dropped, as in British English it can be dropped in Mr or Mrs, for example.

To validate for only letters and spaces, try this
String name1_exp = "^[a-zA-Z]+[\-'\s]?[a-zA-Z ]+$";

Validates such values as:
"", "FIR", "FIR ", "FIR LAST"
/^[A-z]*$|^[A-z]+\s[A-z]*$/

check this out.
String name validation only accept alphabets and spaces
public static boolean validateLetters(String txt) {
String regx = "^[a-zA-Z\\s]+$";
Pattern pattern = Pattern.compile(regx,Pattern.CASE_INSENSITIVE);
Matcher matcher = pattern.matcher(txt);
return matcher.find();
}

To support language like Hindi which can contain /p{Mark} as well in between language characters.
My solution is ^[\p{L}\p{M}]+([\p{L}\p{Pd}\p{Zs}'.]*[\p{L}\p{M}])+$|^[\p{L}\p{M}]+$
You can find all the test cases for this here
https://regex101.com/r/3XPOea/1/tests

#amal. This code will match your requirement. Only letter and space in between will be allow, no number. The text begin with any letter and could have space in between only. "^" denotes the beginning of the line and "$" denotes end of the line.
public static boolean validateLetters(String txt) {
String regx = "^[a-zA-Z ]+$";
Pattern pattern = Pattern.compile(regx,Pattern.CASE_INSENSITIVE);
Matcher matcher = pattern.matcher(txt);
return matcher.find();
}

Try with this:
public static boolean userNameValidation(String name){
return name.matches("(?i)(^[a-z])((?![? .,'-]$)[ .]?[a-z]){3,24}$");
}

For Java, you can use below for Name validation which uses Alpha (Letters) + Spaces (Blanks or tabs)
"[^\\\p{Alpha}\\\p{Blank}]"
Can get a reference from Wikipedia for ASCII values also.

Regex for checking if a string is strictly alphanumeric

How can I check if a string contains only numbers and alphabets ie. is alphanumeric?

Considering you want to check for ASCII Alphanumeric characters, Try this:
"^[a-zA-Z0-9]*$". Use this RegEx in String.matches(Regex), it will return true if the string is alphanumeric, else it will return false.
public boolean isAlphaNumeric(String s){
String pattern= "^[a-zA-Z0-9]*$";
return s.matches(pattern);
}
If it will help, read this for more details about regex: http://www.vogella.com/articles/JavaRegularExpressions/article.html

In order to be unicode compatible:
^[\pL\pN]+$
where
\pL stands for any letter
\pN stands for any number

It's 2016 or later and things have progressed. This matches Unicode alphanumeric strings:
^[\\p{IsAlphabetic}\\p{IsDigit}]+$
See the reference (section "Classes for Unicode scripts, blocks, categories and binary properties"). There's also this answer that I found helpful.

See the documentation of Pattern.
Assuming US-ASCII alphabet (a-z, A-Z), you could use \p{Alnum}.
A regex to check that a line contains only such characters is "^[\\p{Alnum}]*$".
That also matches empty string. To exclude empty string: "^[\\p{Alnum}]+$".

Use character classes:
^[[:alnum:]]*$

Pattern pattern = Pattern.compile("^[a-zA-Z0-9]*$");
Matcher matcher = pattern.matcher("Teststring123");
if(matcher.matches()) {
// yay! alphanumeric!
}

try this [0-9a-zA-Z]+ for only alpha and num with one char at-least..
may need modification so test on it
http://www.regexplanet.com/advanced/java/index.html
Pattern pattern = Pattern.compile("^[0-9a-zA-Z]+$");
Matcher matcher = pattern.matcher(phoneNumber);
if (matcher.matches()) {
}

To consider all Unicode letters and digits, Character.isLetterOrDigit can be used. In Java 8, this can be combined with String#codePoints and IntStream#allMatch.
boolean alphanumeric = str.codePoints().allMatch(Character::isLetterOrDigit);

To include [a-zA-Z0-9_], you can use \w.
So myString.matches("\\w*"). (.matches must match the entire string so ^\\w*$ is not needed. .find can match a substring)
https://docs.oracle.com/javase/6/docs/api/java/util/regex/Pattern.html

If you want to include foreign language letters as well, you can try:
String string = "hippopotamus";
if (string.matches("^[\\p{L}0-9']+$")){
string is alphanumeric do something here...
}
Or if you wanted to allow a specific special character, but not any others. For example for # or space, you can try:
String string = "#somehashtag";
if(string.matches("^[\\p{L}0-9'#]+$")){
string is alphanumeric plus #, do something here...
}

100% alphanumeric RegEx (it contains only alphanumeric, not even integers & characters, only alphanumeric)
For example:
special char (not allowed)
123 (not allowed)
asdf (not allowed)
1235asdf (allowed)
String name="^[^<a-zA-Z>]\\d*[a-zA-Z][a-zA-Z\\d]*$";

To check if a String is alphanumeric, you can use a method that goes through every character in the string and checks if it is alphanumeric.
public static boolean isAlphaNumeric(String s){
for(int i = 0; i < s.length(); i++){
char c = s.charAt(i);
if(!Character.isDigit(c) && !Character.isLetter(c))
return false;
}
return true;
}

Java Regex match digital, character, comma, and quote...

I'm new to regular expressions...
I have a problem about the regular expression that will match a string only contains:
0-9, a-z, A-Z, space, comma, and single quote?
If the string contain any char that doesn't belong the above expression, it is invalid.
Is that something like:
Pattern p = Pattern.compile("\\s[a-zA-Z0-9,']");
Matcher m = p.matcher("to be or not");
boolean b = m.lookingAt();
Thank you!

Fix your expression adding bounds:
Pattern p = Pattern.compile("^\\s[a-zA-Z0-9,']+$");
Now your can say m.find() and be sure that this returns true only if your string contains the enumerated symbols only.
BTW is it mistake that you put \\s in the beginning? This means that the string must start from single white space. If this is not the requirement just remove this.

You need to include the space inside the character class and allow more than one character:
Pattern p = Pattern.compile("[\\sa-zA-Z0-9,']*");
Matcher m = p.matcher("to be or not");
boolean b = m.matches();

Note that \s will match any whitespace character (including newlines, tabs, carriage returns, etc.) and not only the space character.
You probably want something like this:
"^[a-zA-Z0-9,' ]+$"

How to check if only chosen characters are in a string?

What's the best and easiest way to check if a string only contains the following characters:
abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789_
I want like an example like this pseudo-code:
//If String contains other characters
else
//if string contains only those letters
Please and thanks :)

if (string.matches("^[a-zA-Z0-9_]+$")) {
// contains only listed chars
} else {
// contains other chars
}

For that particular class of String use the regular expression "\w+".
Pattern p = Pattern.compile("\\w+");
Matcher m = Pattern.matcher(str);
if(m.matches()) {}
else {};
Note that I use the Pattern object to compile the regex once so that it never has to be compiled again which may be nice if you are doing this check in a-lot or in a loop. As per the java docs...
If a pattern is to be used multiple
times, compiling it once and reusing
it will be more efficient than
invoking this method each time.

My turn:
static final Pattern bad = Pattern.compile("\\W|^$");
//...
if (bad.matcher(suspect).find()) {
// String contains other characters
} else {
// string contains only those letters
}
Above searches for single not matching or empty string.
And according to JavaDoc for Pattern:
\w A word character: [a-zA-Z_0-9]
\W A non-word character: [^\w]