[ACCEPTED]-Regex to get first two words of unknown length from a string-regex
If you have only spaces between words, split 3 by \\s+. When you split, the array would be 2 the words themselves. First two would be 1 in arr[0] and arr[1] if you do:
String[] arr = origString.split("\\s+");
With regular expressions you can do something 3 like this:
public static ArrayList<String> split2(String line, int n){
line+=" ";
Pattern pattern = Pattern.compile("\\w*\\s");
Matcher matcher = pattern.matcher(line);
ArrayList<String> list = new ArrayList<String>();
int i = 0;
while (matcher.find()){
if(i!=n)
list.add(matcher.group());
else
break;
i++;
}
return list;
}
if you want the first n words, or 2 simply this:
public static String split3(String line){
line+=" ";
Pattern pattern = Pattern.compile("\\w*\\s\\w*\\s");
Matcher matcher = pattern.matcher(line);
matcher.find();
return matcher.group();
}
if you want only the first and 1 second words.
If you want to split it on exactly the space 5 character:
String[] parts = args[i].split(" ");
If you want to split it on any 4 whitespace character (space, tab, newline, cr):
String[] parts = args[i].split("\\s");
To 3 treat multiple adjacent spaces as one separator:
String[] parts = args[i].split(" +");
Same 2 for whitespace:
String[] parts = args[i].split("\\s+");
The first two words would 1 be parts[0] and parts[1]
Assuming your "words" consist of alphanumeric 2 characters, the following regex will match 1 the first 2 words:
\w+\s+\w+
Use below method.
public static String firstLettersOfWord(String word) {
StringBuilder result = new StringBuilder();
String[] myName = word.split("\\s+");
for (String s : myName) {
result.append(s.charAt(0));
}
return result.toString();
}}
0
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