Basically what I want to do is this:
String mystr = "hello Benjamin Benny Benn hey hey";
String pattern = "Be";
Desired list = {"Benjamin","Benny","Benn"};
I know how to do this in a really simple way. What I am looking for is a fast way to do it based on regex or whatever that works for me. I want a list of all sub-strings(words) starting with a specific pattern.
Use this regex:
Be\w+
It matches all the words that start with Be
If you want word starting with anything else, simply do this:
String startsWith="Be"; // change this to match your requirements
String regexPattern=startsWith+"\\w+"; //escaped backslash
Now, you can substitute anything in startsWith, so you can match words that starts with a specific string.
Note: You'll need to escape the backslash in java. So \ becomes \\
Try,
String mystr = "hello Benjamin Benny Benn hey hey";
String pattern = "Be";
for(String str : mystr.split("\\s")){
if(str.matches(pattern+"\\w+")){
System.out.println("Matched "+str);
// Add the str to list
}
}
Here,
\\w+ indicates -
word characters (a-z, A-Z, 0-9, _) (1 or more times (matching the most amount possible))
\\s indicates -
whitespace (\n, \r, \t, \f, and " ")
I found this answer which helped me a lot:
String yourString = "hi #how are # you";
Matcher matcher = Pattern.compile("#\\s*(\\w+)").matcher(yourString);
while (matcher.find()) {
System.out.println(matcher.group(1));
}
This way even if your pre-string is separated from the main string it works fine.
I know how to do this in a really simple way.So please post your code to make your question more clear.