How to check if a word is present in a sentence using Java? [duplicate]

Answer 1

This is a rather clumsy workaround, but should achieve the right results. Find the substring you are looking for within the string, and find the charachters before and after your substring. Check these using their ascii values (int)substring.charAt(x);to see if they are letters or not. If they are both eithr not letters, or fall outside the bounds of the string, you know you have found a word. Otherwise, you know that it is simply part of a word.

The logic will be very long- which is why I am not coding it for you, but give this a shot and let me know if you need clarification.

Answer 2

If you want to match a word in a sentence even with punctuation, you'll need a regex like this:

  static boolean matchesWord(String toMatch, String matchIn) {
     return Pattern.matches(".*([^A-Za-z]|^)"+toMatch+"([^A-Za-z]|$).*", matchIn);
  }

(You could use \W, but that doesn't count underscores as punctuation.)

Just concatenating spaces onto the beginning and the end won't match, for example, the word "programmer" in the string "I am a Java programmer" because there's no space at the end. It also won't match words directly before or after punctuation.

Answer 3

Question:

How do you define a word?

Possible answer:

Bunch of characters separated by some other characters. This second set of characters is defined by what you choose. Suppose you choose these to be . ,?;. So if you split the input string with these characters (called delimiters), you'll get a bunch of string which are words. Now to find if the input contains the word or not, loop over these strings to check if they match your query.

Code:

boolean isPresent(String query, String s) {    
    String [] deli = s.split("[.\\s,?;]+");

    for(int i=0;i<deli.length;i++)
        if(query.equals(deli[i]))
            return true;

    return false;    
}

tl;dr:

If you want a word to be defined as anything which consists of alphabets, numbers and underscore, there is a regex ready for you: \W+.

String [] deli = s.split("\\W+");

Consider reading this article if you want to learn more about Java Regex.

Answer 4

Take a look at the String.matches() method. If you construct the regular expression properly it should be able to do what you want it to. A good place to start for regular expressions would be the Java tutorials: http://docs.oracle.com/javase/tutorial/essential/regex/

Answer 5

Since you want to search a word there are three cases:

1. word at start of sentence means no space at start but space at end.
2. word between the sentence space at both ends.
3. word at end only space at end.

To cover all the three cases, one possible solution is:

String str = "I am a JAVA programmer";
String[] splited = str.split("\\b+"); //split on word boundries
Arrays.asList(splited).contains("ram"); //search array for word

Here is working Demo

Answer 6

Hlo. You can split the sentence as array and then you put into List. after that you can use contains method to check you word is present or not. Kindly try this code..

import java.util.ArrayList;
import java.util.Arrays;


 public class karthitest {
  public static void main(String[] args) {
    String sentence = "I am Karthick";
    String word = "I";

    if(isWordExist(sentence, word)){
    System.out.println("Word is exist");
    }
}

public static boolean isWordExist(String sentence, String word){
    boolean ans = Boolean.FALSE;        
    ArrayList<String> wordList = null;

    try {

        if(sentence != null && word != null){
            wordList = new ArrayList<String>(Arrays.asList(sentence.split("[^a-zA-z]+")));              
            if(wordList.contains(word)){
                ans = Boolean.TRUE;
            }
        }
    } catch (Exception e) {
        e.printStackTrace();
        // TODO: handle exception
    }
    return ans;
}

}