I am attempting to split a word from its punctuation:
So for example if the word is "Hello?". I want to store "Hello" in one variable and the "?" in another variable.
I tried using .split method but deletes the delimiter (the punctuation) , which means you wouldn't conserve the punctuation character.
String inWord = "hello?";
String word;
String punctuation = null;
if (inWord.contains(","+"?"+"."+"!"+";")) {
String parts[] = inWord.split("\\," + "\\?" + "\\." + "\\!" + "\\;");
word = parts[0];
punctuation = parts[1];
} else {
word = inWord;
}
System.out.println(word);
System.out.println(punctuation);
I am stuck I cant see another method of doing it.
Thanks in advance
You could use a positive lookahead to split so you don't actually use the punctuation to split, but the position right before it:
inWord.split("(?=[,?.!;])");
Further to the other suggestions, you can also use the 'word boundary' matcher '\\b'. This may not always match what you are looking for, it detects the boundary between a word and a non-word, as documented: http://docs.oracle.com/javase/7/docs/api/java/util/regex/Pattern.html
In your example, it works, though the first element in the array will be a blank string.
Here is some working code:
String inWord = "hello?";
String word;
String punctuation = null;
if (inWord.matches(".*[,?.!;].*")) {
String parts[] = inWord.split("\\b");
word = parts[1];
punctuation = parts[2];
System.out.println(parts.length);
} else {
word = inWord;
}
System.out.println(word);
System.out.println(punctuation);
You can see it running here: http://ideone.com/3GmgqD
I've also fixed your .contains
to use .matches
instead.
I think you can use the below regex. But not tried. Give it a try.
input.split("[\\p{P}]")
You could use substring here. Something like this:
String inWord = "hello?";
String word = inWord.substring (0, 5);
String punctuation = inWord.substring (5, inWord.length ());
System.out.println (word);
System.out.println (punctuation);
The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.