Using Apache Commons Lang:
!StringUtils.isAlphanumeric(String)
Alternativly iterate over String's characters and check with:
!Character.isLetterOrDigit(char)
You've still one problem left:
Your example string "abcdefà" is alphanumeric, since à is a letter. But I think you want it to be considered non-alphanumeric, right?!
So you may want to use regular expression instead:
String s = "abcdefà";
Pattern p = Pattern.compile("[^a-zA-Z0-9]");
boolean hasSpecialChar = p.matcher(s).find();
Answer from Fabian Barney on Stack OverflowUsing Apache Commons Lang:
!StringUtils.isAlphanumeric(String)
Alternativly iterate over String's characters and check with:
!Character.isLetterOrDigit(char)
You've still one problem left:
Your example string "abcdefà" is alphanumeric, since à is a letter. But I think you want it to be considered non-alphanumeric, right?!
So you may want to use regular expression instead:
String s = "abcdefà";
Pattern p = Pattern.compile("[^a-zA-Z0-9]");
boolean hasSpecialChar = p.matcher(s).find();
One approach is to do that using the String class itself. Let's say that your string is something like that:
String s = "some text";
boolean hasNonAlpha = s.matches("^.*[^a-zA-Z0-9 ].*$");
one other is to use an external library, such as Apache commons:
String s = "some text";
boolean hasNonAlpha = !StringUtils.isAlphanumeric(s);
performance - Fastest way to check a string is alphanumeric in Java - Stack Overflow
java - Regex for checking if a string is strictly alphanumeric - Stack Overflow
How to list all alphanumeric UTF-8 characters?
How do you test if there is alphanumeric characters in a string?
Use String.matches(), like:
String myString = "qwerty123456";
System.out.println(myString.matches("[A-Za-z0-9]+"));
That may not be the absolute "fastest" possible approach. But in general there's not much point in trying to compete with the people who write the language's "standard library" in terms of performance.
I've written the tests that compare using regular expressions (as per other answers) against not using regular expressions. Tests done on a quad core OSX10.8 machine running Java 1.6
Interestingly using regular expressions turns out to be about 5-10 times slower than manually iterating over a string. Furthermore the isAlphanumeric2() function is marginally faster than isAlphanumeric(). One supports the case where extended Unicode numbers are allowed, and the other is for when only standard ASCII numbers are allowed.
public class QuickTest extends TestCase {
private final int reps = 1000000;
public void testRegexp() {
for(int i = 0; i < reps; i++)
("ab4r3rgf"+i).matches("[a-zA-Z0-9]");
}
public void testIsAlphanumeric() {
for(int i = 0; i < reps; i++)
isAlphanumeric("ab4r3rgf"+i);
}
public void testIsAlphanumeric2() {
for(int i = 0; i < reps; i++)
isAlphanumeric2("ab4r3rgf"+i);
}
public boolean isAlphanumeric(String str) {
for (int i=0; i<str.length(); i++) {
char c = str.charAt(i);
if (!Character.isLetterOrDigit(c))
return false;
}
return true;
}
public boolean isAlphanumeric2(String str) {
for (int i=0; i<str.length(); i++) {
char c = str.charAt(i);
if (c < 0x30 || (c >= 0x3a && c <= 0x40) || (c > 0x5a && c <= 0x60) || c > 0x7a)
return false;
}
return true;
}
}
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