Develop Reference

Regex including date string, email, number

I have this regex expression:
String patt = "(\\w+?)(:|<|>)(\\w+?),";
Pattern pattern = Pattern.compile(patt);
Matcher matcher = pattern.matcher(search + ",");
I am able to match a string like
search = "firstName:Giorgio"
But I'm not able to match string like
search = "email:giorgio.rossi#libero.it"
or
search = "dataregistrazione:27/10/2016"
How I should modify the regex expression in order to match these strings?

You may use
String pat = "(\\w+)[:<>]([^,]+)"; // Add a , at the end if it is necessary
See the regex demo
Details:
(\w+) - Group 1 capturing 1 or more word chars
[:<>] - one of the chars inside the character class, :, <, or >
([^,]+) - Group 2 capturing 1 or more chars other than , (in the demo, I added \n as the demo input text contains newlines).

You can use regex like this:
public static void main(String[] args) {
String[] arr = new String[]{"firstName:Giorgio", "email:giorgio.rossi#libero.it", "dataregistrazione:27/10/2016"};
String pattern = "(\\w+[:|<|>]\\w+)|(\\w+:\\w+\\.\\w+#\\w+\\.\\w+)|(\\w+:\\d{1,2}/\\d{1,2}/\\d{4})";
for(String str : arr){
if(str.matches(pattern))
System.out.println(str);
}
}
output is:
firstName:Giorgio
email:giorgio.rossi#libero.it
dataregistrazione:27/10/2016
But you have to remember that this regex will work only for your format of data. To make up the universal regex you should use RFC documents and articles (i.e here) about email format. Also this question can be useful.
Hope it helps.

The Character class \w matches [A-Za-z0-9_]. So kindly change the regex as (\\w+?)(:|<|>)(.*), to match any character from : to ,.
Or mention all characters that you can expect i.e. (\\w+?)(:|<|>)[#.\\w\\/]*, .

Java Code to validate a name field [duplicate]

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.

How do I find multiple substrings from one string using regex in Java?

I want to find every instance of a number, followed by a comma (no space), followed by any number of characters in a string. I was able to get a regex to find all the instances of what I was looking for, but I want to print them individually rather than all together. I'm new to regex in general, so maybe my pattern is wrong?
This is my code:
String test = "1 2,A 3,B 4,23";
Pattern p = Pattern.compile("\\d+,.+");
Matcher m = p.matcher(test);
while(m.find()) {
System.out.println("found: " + m.group());
}
This is what it prints:
found: 2,A 3,B 4,23
This is what I want it to print:
found: 2,A
found: 3,B
found: 4,23
Thanks in advance!

try this regex
Pattern p = Pattern.compile("\\d+,.+?(?= |$)");

You could take an easier route and split by space, then ignore anything without a comma:
String values = test.split(' ');
for (String value : values) {
if (value.contains(",") {
System.out.println("found: " + value);
}
}

What you apparently left out of your requirements statement is where "any number of characters" is supposed to end. As it stands, it ends at the end of the string; from your sample output, it seems you want it to end at the first space.
Try this pattern: "\\d+,[^\\s]*"

Java (Regex?) split string between number/letter combination

I've been looking through pages and pages of Google results but haven't come across anything that could help me.
What I'm trying to do is split a string like Bananas22Apples496Pears3, and break it down into some kind of readable format. Since String.split() cannot do this, I was wondering if anyone could point me to a regex snippet that could accomplish this.
Expanding a bit: the above string would be split into (String[] for simplicity's sake):
{"Bananas:22", "Apples:496", "Pears:3"}

Try this
String s = "Bananas22Apples496Pears3";
String[] res = s.replaceAll("(?<=\\p{L})(?=\\d)", ":").split("(?<=\\d)(?=\\p{L})");
for (String t : res) {
System.out.println(t);
}
The first step would be to replace the empty string with a ":", when on the left is a letter with the lookbehind assertion (?<=\\p{L}) and on the right is a digit, with the lookahead assertion (?=\\d).
Then split the result, when on the left is a digit and on the right is a letter.
\\p{L} is a Unicode property that matches every letter in every language.

You need to Replace and then split the string.You can't do it with the split alone
1> Replace All the string with the following regex
(\\w+?)(\\d+)
and replace it with
$1:$2
2> Now Split it with this regex
(?<=\\d)(?=[a-zA-Z])

This should do what you want:
import java.util.regex.*;
String d = "Bananas22Apples496Pears3"
Pattern p = Pattern.compile("[A-Za-z]+|[0-9]+");
Matcher m = p.matcher(d);
while (m.find()) {
System.out.println(m.group());
}
// Bananas
// 22
// Apples
// 496
// Pears
// 3

String myText = "Bananas22Apples496Pears3";
System.out.println(myText.replaceAll("([A-Za-z]+)([0-9]+)", "$1:$2,"));

Replace \d+ by :$0 and then split at (?=[a-zA-Z]+:\d+).

Regarding String manipulation

I have a String str which can have list of values like below. I want the first letter in the string to be uppercase and if underscore appears in the string then i need to remove it and need to make the letter after it as upper case. The rest all letter i want it to be lower case.
""
"abc"
"abc_def"
"Abc_def_Ghi12_abd"
"abc__de"
"_"
Output:
""
"Abc"
"AbcDef"
"AbcDefGhi12Abd"
"AbcDe"
""

Well, without showing us that you put any effort into this problem this is going to be kinda vague.
I see two possibilities here:
Split the string at underscores, apply the answer from this question to each part and re-combine them.
Create a StringBuilder, walk through the string and keep track of whether you are
at the start of the string
after an underscore or
somewhere else
and act appropriately on the current character before appending it to the StringBuilder instance.

replace _ with space (str.replace("_", " "))
use WordUtils.capitalizeFully(str); (from commons-lang)
replace space with nothing (str.replace(" ", ""))

You can use following regexp based code:
public static String camelize(String input) {
char[] c = input.toCharArray();
Pattern pattern = Pattern.compile(".*_([a-z]).*");
Matcher m = pattern.matcher(input);
while ( m.find() ) {
int index = m.start(1);
c[index] = String.valueOf(c[index]).toUpperCase().charAt(0);
}
return String.valueOf(c).replace("_", "");
}

Use Pattern/Matcher in the java.util.regex package:
for each string that is in your array do the following:
StringBuffer output = new StringBuffer();
Matcher match = Pattern.compile("[^|_](\w)").matcher(inStr);
while(match.find()) {
match.appendReplacement(output, matcher.match(0).ToUpper());
}
match.appendTail(output);
// Will have the properly capitalized string.
String capitalized = output.ToString();
The regular expression looks for either the start of the string or an underscore "[^|_]"
Then puts the following character into a group "(\w)"
The code then goes through each of the matches in the input string capitalizing the first satisfying group.