How can I check if a string contains only numbers and alphabets ie. is alphanumeric?
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2Try this: /^[a-z0-9]+$/i Check [here][1] too [1]: http://stackoverflow.com/questions/388996/regex-for-javascript-to-allow-only-alphanumeric – Vimalnath Jun 28 '12 at 09:34
12 Answers
Considering you want to check for ASCII Alphanumeric characters, Try this:
"^[a-zA-Z0-9]*$"
. 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]*$";
return s.matches(pattern);
}
If it will help, read this for more details about regex: http://www.vogella.com/articles/JavaRegularExpressions/article.html
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8-1 as this doesn't cover all alphanumberic characters. M42's response is better. – tster Jan 14 '13 at 22:41
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1Do you really need to add the boundary matchers (^,$) at the beginning and end? It seems like `String pattern="[a-zA-Z0-9]+";` would behave the same way since + is greedy. – krick Jun 29 '17 at 21:08
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Need regex to check if string not contain "test" and contain only alphabet and space. please help – shiva Sep 11 '18 at 07:03
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@krick Yes, you do need the anchors `^, $` for validation regex. Try testing this string `abc@def` at some site like https://regex101.com/ and you'll see why. You are right that `+` is greedy, but that means it will match `abc` and `def` instead of `a`, `b`, `c`, `d`, `e`, `f` which a lazy `+?` would. Note: [Java's String.matches() will automatically add the `^,$` anchors](https://stackoverflow.com/questions/1894624/is-regex-in-java-anchored-by-default-with-both-a-and-character) if you don't have them, but it's probably a good practice to include them. – wisbucky Nov 07 '18 at 00:35
In order to be unicode compatible:
^[\pL\pN]+$
where
\pL stands for any letter
\pN stands for any number

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Not working. Giving: java.util.regex.PatternSyntaxException: Incorrect Unicode property near index 5. – Abdalrahman Shatou Sep 13 '16 at 09:13
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@Erk I believe that has been around since Java 7: https://docs.oracle.com/javase/7/docs/api/java/util/regex/Pattern.html – slugmandrew Jan 05 '21 at 15:53
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@slugmandrew, good to know. I was hedging. Seems there is one as far back as java 6. But not java 5—it doesn't even seem to have a Pattern class. – Erk Jan 07 '21 at 03:52
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Works in Java May 2023. It's the most concise answer, yet still readable. (IntelliJ doesn't suggest change.) – devdanke May 05 '23 at 18:41
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.

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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}]*$"
.
That also matches empty string. To exclude empty string: "^[\\p{Alnum}]+$"
.

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

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4This is the right idea, but [POSIX character class syntax](http://en.wikipedia.org/wiki/Regular_expression#Character_classes) is [not valid in Java](http://docs.oracle.com/javase/7/docs/api/java/util/regex/Pattern.html), and the question is tagged as Java. The equivalent Java syntax for your answer is `"^[\\p{Alnum}]*$"`, as mentioned [below](http://stackoverflow.com/a/11241830/1840078). (If such tagging is not considered to be sufficiently conspicuous by site guidelines, let me know and I'll add a comment to the question itself. :)) – Mark A. Fitzgerald May 07 '14 at 15:56
Pattern pattern = Pattern.compile("^[a-zA-Z0-9]*$");
Matcher matcher = pattern.matcher("Teststring123");
if(matcher.matches()) {
// yay! alphanumeric!
}

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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()) {
}

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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);

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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

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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]*$";
If you want to include foreign language letters as well, you can try:
String string = "hippopotamus";
if (string.matches("^[\\p{L}0-9']+$")){
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...
}
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;
}

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