I Need to identify a string in a text and replace it with null string. Problem is, it is not always present as a word itself. There will be space character present between each letter or set of letters. For example:
For word "Decent", I may face the following values. D ec ent, De ce nt, De ce n t .
Is there a way to identify these strings using "Decent" word as input with any regular expression? I am very new to regular expressions. Please help!!
TIA!
\bD\s*e\s*c\s*e\s*n\s*t\s*
so you match D ec ent, De ce nt, De ce n t, decent Decent
but not blade centimeter
If you use
'D ?e ?c ?e ?n ?t ?'
it will match the word with extra spaces
\b is recommended here, to prevent the case of cutting off half the end of some word and half the beginning of another word.The expression "D\s*e\s*c\s*e\s*n\s*t" will do it. Each letter is followed by zero or more spaces. Actually \s is "whitespace characters." You could replace \s* with * (space followed by an asterisk) if you just want literal spaces.
first a bit of code:
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class WordsWithSpaces {
public static void main(String[] args) {
String test = "Descent D escent De s cent desce nd";
String word = "descent";
String pattern = "";
for(int i=0; i<word.length();i++) {
pattern = pattern+word.charAt(i)+"\\s*";
}
System.err.println("pattern is: "+pattern);
Pattern p = Pattern.compile(pattern,Pattern.CASE_INSENSITIVE);
Matcher m = p.matcher(test);
while(m.find()) {
String found = test.substring(m.start(),m.end());
System.err.println(found+" matches");
}
}
}
now for the explanation: \s is a character class for whitespace. this includes spaces and tabs and (possibly) linebreaks. in this piece of code, i take every character of the word i am looking for, and append "\s", with "*" meaning 0 or mor occurences.
also, to avoid it being case sensitive, i set the CASE_INSENSITIVE flag on the pattern.
character classes may not have the same name in your programming language of choice, but there should be one for whitespace. check your documentation.
