IndexOf() find word in a phrase

The method String#indexOfwill try to find the full text of what you indicate within the text string. Since your string has the value "sql"and you're looking for "lenguaje sql", it won't find it since there's no trace of "lenguaje ".

What you're trying to do is kind of forced and maybe you shouldn't do it. On the other hand, maybe whatever you want to do is look to see if any of the "words" within your text string might be embedded in the string. To do this, you could use the following algorithm:

String cadenaDondeBuscar = "sql";
String loQueQuieroBuscar = "lenguaje sql";
String[] palabras = loQueQuieroBuscar.split("\\s+");
for (String palabra : palabras) {
    if (cadenaDondeBuscar.contains(palabra)) {
        System.out.println("Encontrado");
        //aquí tu lógica en caso que se haya encontrado...
    }
}

Per Mariano's comment, if we wanted to evaluate the string by removing not only whitespace but also any other character that is not a vowel or consonant, we use \\W+to separate the string:

String cadenaDondeBuscar = "sql";
String loQueQuieroBuscar = "lenguaje sql";
String[] palabras = loQueQuieroBuscar.split("\\W+");
for (String palabra : palabras) {
    if (cadenaDondeBuscar.contains(palabra)) {
        System.out.println("Encontrado");
        //aquí tu lógica en caso que se haya encontrado...
    }
}

The method works the indexOfother way around.

You must apply the method to the string in which you want to search, passing the text to search for as a parameter.

String a = "sql";
String texto = "lenguaje sql";
int intIndex = texto.indexOf(a);

You can also use contains:

String palabra = "sql";
String texto = "lenguaje sql";
boolean resultado = texto.contains(palabra);

if(resultado){
    System.out.println("palabra encontrada");
}else{
    System.out.println("palabra no encontrada");
}

All the other answers are the fastest way to see if one string is part of another. However, when searching for words , you may want to meet the following conditions:

• Match ignoring case .
  Finding "palabra"within "ejemplo de ¿PaLaBrA?".

• Match the whole word .
  And that it is NOT considered "la"as part of "dos palabras",
  but that it does coincide with "(la primera)".

For that, we'll use a regular expression , with wholeword bounds ( \b) around the search word. That is, it is not preceded or followed by letters, numbers or an underscore.

Example

import java.util.regex.*;

String aguja = "sql";             //palabra buscada
String pajar = "lenguaje SQL";    //texto

//escapar y agregar limites de palabra completa - case-insensitive
Pattern regex = Pattern.compile("\\b" + Pattern.quote(aguja) + "\\b", Pattern.CASE_INSENSITIVE);
Matcher match = regex.matcher(pajar);

//la palabra está en el texto??
if (match.find()) {  //si se quiere encontrar todas las ocurrencias: cambiar el if por while
    System.out.println("Encontrado: '" + match.group() 
                     + "' dentro de '" + pajar 
                     + "' en la posición " + match.start());
} else {
    System.out.println("'" + aguja + "' NO está dentro de '" + pajar);
}

Result

Encontrado: 'SQL' dentro de 'lenguaje SQL' en la posición 9

demo en ideone.com

This way you will be able to count how many times the second string appears within the first:

    String cadena1 = in.nextLine();
    cadena1 = cadena1.toLowerCase();
    String cadena2 = in.nextLine();
    cadena2 = cadena2.toLowerCase();
    System.out.println(" ");

    int contador = 0;

    while (cadena1.indexOf(cadena2) > -1) {
        cadena1 = cadena1.substring(cadena1.indexOf(cadena2) + cadena2.length(), cadena1.length());
        contador++;
    }

    System.out.println("La segunda cadena se incluye " + contador + " veces en la primera cadena.");

What you should do is something like the following.

public static void main(String[] args) {
            // TODO code application logic here
            String a="lenguaje sql";
            int intIndex = a.indexOf("sql");
          if(intIndex == - 1){
             System.out.println("palabra no encontrada");
          }else{
             System.out.println("palabra encontrada a partir del carácter en la posicion: "
             + intIndex);
          }

        }

If we wanted to evaluate the string by removing not only whitespace but also any other characters that are not vowels or consonants, we use \W+ to separate the string:

public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    // TODO code application logic here
    String cadenaDondeBuscar = "sql";
    String loQueQuieroBuscar = sc.nextLine();
    String[] palabras = loQueQuieroBuscar.split("\\W+"); 

    for (String palabra : palabras) {
        if (cadenaDondeBuscar.contentEquals(palabra)) {
            System.out.println("Encontrado");
            //aquí tu lógica en caso que se haya encontrado...
        }
    }
}

With the contentEquals function it is evaluated that the word is completely present, while in a contains it validates that at least one character of the requests to be searched is in the String. Greetings.