Develop Reference

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.

Check special arrangement of specific signs in a string in Java

I need to check a string whether it includes a specific arrangements of letters and numbers.
Valid arrangements are for example:
X
X-Y
A-H-K-L-J-Y
A-H-J-Y
123
12?
12*
12-17
Invalid are for example:
-X-Y
-XY
*12
?12
I have written this method in java to solve this problem (but i don´t have some experiences with regular expressions):
public boolean checkPatternMatching(String sourceToScan, String searchPattern) {
boolean patternFounded;
if (sourceToScan == null) {
patternFounded = false;
} else {
Pattern pattern = Pattern.compile(Pattern.quote(searchPattern),
Pattern.CASE_INSENSITIVE);
Matcher matcher = pattern.matcher(sourceToScan);
patternFounded = matcher.find();
}
return patternFounded;
}
How can i implemented this requirement with regular expressions?
By the way: It is a good solution to check a string, whether it includes numeric content by using the method isNumeric from the java class StringUtils?
//EDIT
The link, which was edited by the admins includes not specific arrangements of characters but only an appearance of characters with regular expressions in general !

After a good while trying to help, answering to constantly changing questions, just found out that the same was asked yesterday, and that the OP doesn't accept answers to his questions...all I have left to say is good night sir, good luck
n-th answer follows:
First pattern: [a-z](-[a-z])* : a letter, possibly followed by more letters, separated by -.
Second pattern: \d+(-\d+)*[?*]* : a number, possibly followed by more numbers, separated by -, and possibly ending with ? or *.
So join them together: ^([a-z](-[a-z])*)|(\d+(-\d+)*[?*]*)$. ^ and $ mark the beginning and the end of the string.
Few more comments on the code: you don't need to use Pattern.quote, and you should use matches() instead of find(), because find() returns true if any part of the string matches the pattern, and you want the whole string:
public static boolean checkPatternMatching(String sourceToScan, String searchPattern) {
boolean patternFounded;
if (sourceToScan == null) {
patternFounded = false;
} else {
Pattern pattern = Pattern.compile(searchPattern, Pattern.CASE_INSENSITIVE);
Matcher matcher = pattern.matcher(sourceToScan);
patternFounded = matcher.matches();
}
return patternFounded;
}
Called like this: checkPatternMatching(s, "^([a-z](-[a-z])*)|(\\d+(-\\d+)*[?*]*)$")
About the second question, this is the current implementation of StringUtils.isNumeric:
public static boolean isNumeric(final CharSequence cs) {
if (isEmpty(cs)) {
return false;
}
final int sz = cs.length();
for (int i = 0; i < sz; i++) {
if (Character.isDigit(cs.charAt(i)) == false) {
return false;
}
}
return true;
}
So no, there is nothing wrong about it, that is as simple as it gets. But you need to include an external JAR in your program, which I find unnecessary if you just want to use such a simple method.

I believe that you should first remove the Pattern.quote() method because that would turn the inputting patterns into string literals; and those are not really useful in your context.
To match the valid arrangements with letters, something like this should work:
^[a-z](?:-[a-z])*$
For the numbers (if I understood the rules correctly):
^\\d+(?:[?*]|-\\d+)*$
And if you want to combine them:
^(?:[a-z](?:-[a-z])*|\\d+(?:[?*]|-\\d+)*)$
I'm not familiar with Java itself, nor the isNumeric method, sorry.
As per your comment, if you want to accept *12 or 1?2 or 12*456, you can use:
^\\*?\\d+(?:[?*]\\d*|-\\d+)*$
Then add it to the previous regex like so:
^(?:[a-z](?:-[a-z])*|\\*?\\d+(?:[?*]\\d*|-\\d+)*)$

How to find substring of a string with whitespaces in Java?

I want to check whether string contains particular sub string and using CONTAINS() for it.
But here problem is with space.
Ex- str1= "c not in(5,6)"
I want to check whether str contains NOT IN so I am using str.contains("not in")..
But problem is that here space between NOT and IN is not decided i.e. there can be 5 spaces also..
How to solve that I can find sub string like not in with any no of spaces in between...

Use a regular expression (Pattern) to get a Matcher to match your string.
The regexp should be "not\\s+in" ("not", followed by a number of space-characters, followed by "in"):
public static void main(String[] args) {
Matcher m = Pattern.compile("not\\s+in").matcher("c not in(5,6)");
if (m.find())
System.out.println("matches");
}
Note that there is a String method called matches(String regexp). You can use regular expression ".*not\\s+in.*" to get the match but it's not really a good way to perform the pattern matching.

You should use a regex: "not\\s+in"
String s = "c not in(5,6)";
Matcher matcher = Pattern.compile("not\\s+in").matcher(s);
System.out.println(matcher.find());
Explanation: The \\s+ means any kind of white space [tab is also acceptable], and must repeat at least one [any number >=1 will be accepted].
If you want only spaces, without tabs change your regex to "not +in"

Use the String.matches() method, which checks if the string matches a regular expression (docs).
In your case:
String str1 = "c not in(5,6)";
if (str1.matches(".*not\\s+in.*")) {
// do something
// the string contains "not in"
}

Case insensitive: (?i)
Considering newline as dot . too: (?s)
str1.matches("(?is).*not\\s+in.*")

Please try following,
int result = str1.indexOf ( "not in" );
if ( result != -1 )
{
// It contains "not in"
}
else if ( result == -1 )
{
// It does not contain "not in"
}

In general you can just do this:
if (string.indexOf("substring") > -1)... //It's there

Check String whether it contains only Latin characters?

Greetings,
I am developing GWT application where user can enter his details in Japanese.
But the 'userid' and 'password' should only contain English characters(Latin Alphabet).
How to validate Strings for this?

You can use String#matches() with a bit regex for this. Latin characters are covered by \w.
So this should do:
boolean valid = input.matches("\\w+");
This by the way also covers numbers and the underscore _. Not sure if that harms. Else you can just use [A-Za-z]+ instead.
If you want to cover diacritical characters as well (ä, é, ò, and so on, those are per definition also Latin characters), then you need to normalize them first and get rid of the diacritical marks before matching, simply because there's no (documented) regex which covers diacriticals.
String clean = Normalizer.normalize(input, Form.NFD).replaceAll("\\p{InCombiningDiacriticalMarks}+", "");
boolean valid = clean.matches("\\w+");
Update: there's an undocumented regex in Java which covers diacriticals as well, the \p{L}.
boolean valid = input.matches("\\p{L}+");
Above works at Java 1.6.

public static boolean isValidISOLatin1 (String s) {
return Charset.forName("US-ASCII").newEncoder().canEncode(s);
} // or "ISO-8859-1" for ISO Latin 1
For reference, see the documentation on Charset.

There is my solution and it is working excellent
public static boolean isStringContainsLatinCharactersOnly(final String iStringToCheck)
{
return iStringToCheck.matches("^[a-zA-Z0-9.]+$");
}

There might be a better approach, but you could load a collection with whatever you deem to be acceptable characters, and then check each character in the username/password field against that collection.
Pseudo:
foreach (character in username)
{
if !allowedCharacters.contains(character)
{
throw exception
}
}

For something this simple, I'd use a regular expression.
private static final Pattern p = Pattern.compile("\\p{Alpha}+");
static boolean isValid(String input) {
Matcher m = p.matcher(input);
return m.matches();
}
There are other pre-defined classes like \w that might work better.

I successfully used a combination of the answers of user232624, Joachim Sauer and Tvaroh:
static CharsetEncoder asciiEncoder = Charset.forName("US-ASCII"); // or "ISO-8859-1" for ISO Latin 1
boolean isValid(String input) {
return Character.isLetter(ch) && asciiEncoder.canEncode(username);
}

How to validate string using regex in java

I want to validate a string which donot have numeric characters.
If my string is "javaABC" then it must be validated
If my string is "java1" then it must not be validated
I want to restrict all the integers.

Try this:
String Text = ...;
boolean HasNoNumber = Text.matches("^[^0-9]*$");
'^[^0-9]*$' = From Start(^) to end ($), there are ([...]) only non(^) number(0-9). You can use '\D' as other suggest too ... but this is easy to understand.
See more info here.

You can use this:
\D
"\D" matches non-digit characters.

Here is one way that you can search for a digit in a String:
public boolean isValid(String stringToValidate) {
if(Pattern.compile("[0-9]").matcher(stringToValidate).find()) {
// The string is not valid.
return false;
}
// The string is valid.
return true;
}

More detail is here:
http://java.sun.com/javase/6/docs/api/java/util/regex/Pattern.html

The easiest to understand is probably matching for a single digit and if found fail, instead of creating a regexp that makes sure that all characters in the string are non-digits.