Examples:
- "1 name": Should say it has characters
- "10,000": OK
- "na123me": Should say it has characters
- "na 123, 000": Should say it has characters
views:
209answers:
4
A:
With this line you can check if your string contains only of characters given by the regex (in this case a,b,c,...z and A,B,C,...Z):
boolean doesMatch = "your string".matches( "[a-zA-Z]*" );
tangens
2010-05-20 18:43:08
that regex would match any string that has at least one alphabetic character
Kip
2010-05-20 18:46:09
I thought it would match any string that had precisely one alphabetic character, not just at least one. Doesn't it need an asterisk?
Paul
2010-05-20 18:47:42
Thanks, I added one.
tangens
2010-05-20 18:48:55
The question is unclear enough that you're both right.
WhirlWind
2010-05-20 18:49:56
This tests only if there is one character in a string
Vishal
2010-05-20 18:50:15
Doesnt * match zero to many? It should be + instead.
Shervin
2010-05-20 19:01:24
With the asterisk, it will also match an empty string.
mpez0
2010-05-20 19:02:38
This doesn't work. Java regexes are by default bounded by `^` and `$`. Test with `1name` yourself.
BalusC
2010-05-20 19:05:18
+2
A:
public static void main(String[] args)
{
Pattern p = Pattern.compile("^([^a-zA-Z]*([a-zA-Z]+)[^a-zA-Z]*)+$");
Matcher m = p.matcher("1 name");
Matcher m1 = p.matcher("10,000");
Matcher m2 = p.matcher("na123me");
Matcher m3 = p.matcher("na 123, 000");
Matcher m4 = p.matcher("13bbbb13jdfgjd43534 fkgdfkgjk34 rktekjg i54 ");
if (m.matches())
System.out.println(m.group(1));
if (m1.matches())
System.out.println(m1.group(1));
if(m2.matches())
System.out.println(m2.group(1));
if(m3.matches())
System.out.println(m3.group(1));
if (m4.matches())
System.out.println(m4.group(1));
}
The above should match any letter in both lower and upper case. If the regex returns a match, the string has a letter in it.
Result
1 name
me
na 123, 000
i54
Statements that contain no letters do not match the expression.
npinti
2010-05-20 18:43:46
Regular expressions that begin with '^.*' or end with '.*$' are poorly written. Why not just use the regular expression '[a-zA-Z]'?
David M
2010-05-20 19:25:42
@David: unlike in PHP, Perl and another languages, Java regexes are **implicitly** bounded by `^` and `$`.
BalusC
2010-05-20 19:27:01
@BalusC, incorrect! The `matches()` method uses the whole string, but the `find()` method does not.
David M
2010-05-20 19:33:42
@David: I am not that experienced, how ever just because with Java the ^ and $ are implicitly added, I do not think that they should be left out. Putting in an extra character or two won't harm, but I think it makes the code more understandable.
npinti
2010-05-20 20:07:17
+3
A:
public class HasCharacters {
public static void main( String [] args ){
if( args[0].matches(".*[a-zA-Z]+.*")){
System.out.println( "Has characters ");
} else {
System.out.println("Ok");
}
}
}
Test
$java HasCharacters "1 name"
Has characters
$java HasCharacters "10,000"
Ok
$java HasCharacters "na123me"
Has characters
$java HasCharacters "na 123, 000"
Has characters
OscarRyz
2010-05-20 18:59:21
+2
A:
The regular expression you want is [a-zA-Z], but you need to use the find() method.
This page will let you test regular expressions against input.
David M
2010-05-20 19:37:47