What is the regular expression for . and .. ?
if(key.matches(".")) {
do something
}
The matches accepts String which asks for regular expression. Now i need to remove all DOT's inside my MAP.
+1
A:
[+*?.] Most special characters have no meaning inside the square brackets. This expression matches any of +, *, ? or the dot.
Sachin Shanbhag
2010-10-05 09:47:37
+6
A:
. matches any character so needs escaping i.e. \., or \\. within a Java string (because \ itself has special meaning within Java strings.)
You can then use \.\. or \.{2} to match exactly 2 dots.
mikej
2010-10-05 09:47:48
\. does not work with above code in key.matches
John
2010-10-05 10:03:04
As mentioned, `\.` is the regular expression but to construct a String for this expression in Java you will need `\\.` because \ is used within Java strings for things like \n, \t etc. so \\ is needed for a literal \ within a String. Also, note that `String.matches` requires the regexp to match the **entire** String. If you want to do a substring search you will need to use `Matcher.find`
mikej
2010-10-05 14:32:21
Also, can you clarify the requirements a bit. If you just need to check if the key contains a dot then you can use `key.contains` and don't really need a regexp at all.
mikej
2010-10-05 14:33:13
+1
A:
Use String.Replace() if you just want to replace the dots from string. Alternative would be to use Pattern-Matcher with StringBuilder, this gives you more flexibility as you can find groups that are between dots. If using the latter, i would recommend that you ignore empty entries with "\\.+".
public static int count(String str, String regex) {
int i = 0;
Pattern p = Pattern.compile(regex);
Matcher m = p.matcher(str);
while (m.find()) {
m.group();
i++;
}
return i;
}
public static void main(String[] args) {
int i = 0, j = 0, k = 0;
String str = "-.-..-...-.-.--..-k....k...k..k.k-.-";
// this will just remove dots
System.out.println(str.replaceAll("\\.", ""));
// this will just remove sequences of ".." dots
System.out.println(str.replaceAll("\\.{2}", ""));
// this will just remove sequences of dots, and gets
// multiple of dots as 1
System.out.println(str.replaceAll("\\.+", ""));
/* for this to be more obvious, consider following */
System.out.println(count(str, "\\."));
System.out.println(count(str, "\\.{2}"));
System.out.println(count(str, "\\.+"));
}
The output will be:
--------kkkkk--
-.--.-.-.---kk.kk.k-.-
--------kkkkk--
21
7
11
Margus
2010-10-05 10:35:12