Manual String Breaking in JAVA

Question

Gentlemen, I am trying to do a routine that receives a certain text and a word from the user, after checking if within the text there is a word equal to what was requested. My attempt was to generate a counter that checks the character at the counter position and if it is not a blank space concatenates the character into a string. If it is a blank space, it checks whether the content of the string is equal to the text sought by the user. However my code is not returning a value, how do I correct.

Note: There may be new methods that perform the process without the need to use this process, but my intention is to learn how to work better with Strings, Characters and Loops.

Follows excerpt from the code.

        // Faz a separação manual de palavras e verifica .equals
    for(cont = 0; cont < texto.length(); cont++){
        a = texto.charAt(cont);
        if(!a.equals(" ")){
            montagem += a;
        } else {
            if(montagem.equals(palavraBusca)){
                palavras++;
            } else { 
                montagem = "";
            }
        }
    }
    lblResultado.setText(Integer.toString(palavras));

Answer 1

I believe the problem may be about method equals(); this method compares the whole object, not just the value, as is its idea (for example, "home" would equal "home").

You could overwrite the method equals(), but that’s kind of risky since another part of the code might need the original. Therefore, I believe that the most viable comparison would be made with ==:

//inicialização de variáveis
montagem = "";
palavras = 0;

for(cont = 0; cont < texto.length(); cont++){
    a = texto.charAt(cont);
    
    if(!a.equals(" ")){
        montagem += a;
    } else {
        if(montagem == palavraBusca){
            palavras++;
        }

        //montagem deve ser reiniciada independente de ter sido encontrada
        montagem = "";
    }
}

//validação da última palavra do texto
if(montagem.equals(palavraBusca)){
    palavras++;
}

lblResultado.setText(Integer.toString(palavras));

detail: I am not with java compiler, the answer admits that your code is correct.

Answer 2

The logic for treatment is: If the counter reads a blank space it has two alternatives, it is either the word searched or not. So after checking the two will return the mounting variable to zero. If it is not a blank space the word has not yet finished. Follows excerpt from the code.

    for(cont = 0; cont < texto.length(); cont++){  
    a = texto.charAt(cont);
    if(Character.toString(a).equals(" ")){
        // verifica
        if(montagem == palavraBusca){
            palavras++;
        } 
        montagem = "";
    } else {
        montagem+= a;
    }
    if(montagem.equals(palavraBusca)) palavras++;
    }
    lblResultado.setText(Integer.toString(palavras));