Trying to find a String element inside an Array of strings

Question

I am trying to write a method that takes a String, converts it into an Array of Strings, and then searches for a specific String within that Array. Here is my code:

public static void ThereOrNaw ( String s, String s2 ){ 

    String[] words = s.split("\\s");
    for (int i = 0; i < words.length; i++) {
        words[i] = words[i].replaceAll("[^\\w]", "");
    }
    for ( int i = 0; i < words.length; i++) {
        if( s2 == words[i] ) {
            System.out.println("Found");
        }
        else if( s2 != words[i] && (i + 1) == words.length) {
            System.out.println("Not Found");
        }
    }
}

However, when I run this method it always returns "Not Found" even if the element is in the Array. Unless, the Array of Strings has ONLY the element I am looking for. Any ideas why this isn't working?

Your code (and algorithm) seem over-complicated, how is your implementation supposed to differ from `System.out.println(s.contains(s2) ? "Found" : "Not Found");`? — Elliott Frisch, Mar 11 '17 at 23:58
Elliot Frisch is correct. To add more, for case sensitive approach, `s.toLowerCase().contains(s2.toLowerCase()) ? "Found" : "Not Found"` — Pons, Mar 12 '17 at 00:11

Aleksandr Podkutin · Answer 1 · 2017-03-12T00:11:46.943

0

You should use equals() not ==.

== compare your String objects not by content, but by real address in memory(in simple words). So if you will use equals() you will compare String objects by content. Also you can use equalsIgnoreCase() method if you do not care about case.

And another one little advice, check that s2 not null before you will invoke equals(), because you don't know, which object client will send to this method.

For example:

if (s2 != null && s2.equals(words[i])) { ...

else if (s2 != null && !s2.equals(words[i]) ...

edited Mar 12 '17 at 00:11

answered Mar 11 '17 at 23:56

Aleksandr Podkutin

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Appears to work thanks!! I'm just curious for further in depth understanding since I'm very new to Java. Why does `equals` work but `==` does not? – Daniel Utvich Mar 12 '17 at 00:03
@DanielUtvich I supplemented my answer. – Aleksandr Podkutin Mar 12 '17 at 00:07

score 0 · Answer 2 · answered Mar 12 '17 at 00:02

Well as he said you don't have to use == comparing words, == compare objects

Look i made the change in your code for you:

 public static void main(String[] args) {

    thereOrNaw("hola","hola");
    thereOrNaw("hola","mundo");
}

public static void thereOrNaw ( String s, String s2 ){ 

    String[] words = s.split("\\s");
    for (int i = 0; i < words.length; i++) {
        words[i] = words[i].replaceAll("[^\\w]", "");
    }
    for ( int i = 0; i < words.length; i++) {
        if( s2.equals(words[i]) ) {
            System.out.println("Found");
        }
        else if( s2 != words[i] && (i + 1) == words.length) {
            System.out.println("Not Found");
        }
    }
}

And it gaves me:

Found Not Found

Trying to find a String element inside an Array of strings

2 Answers2