I compare two Strings and I do not get into the IF

Question

I compare two Strings and I do not get into the IF

Navigation

#1 by (0 votes)
#2 by (0 votes)

1

System.out.println("Dime el tablero al que te quieres conectar :");

tablero = sc.nextLine(); // AQUI LO LEO COMO UN STRING EN EL CLIENTE

tablero = tablero.trim();

byte buffer_tablero[];

buffer_tablero = tablero.getBytes();

DatagramPacket mensaje_tablero = new DatagramPacket(buffer_tablero, buffer_tablero.length, InetAddress.getLocalHost(), 50000);


puerto.send(mensaje_tablero); // Aqui envio el mensaje con el numero de tablero al que me quiero conectar


//------------------------------------------------------------------------------

puerto.receive(tablero);
buffer = tablero.getData();
datos = new String(buffer, buffer.length); // AQUI PASO A STRING LO QUE RECIBO DEL PAQUETE

System.out.println("He recibido una peticion para el tablero " + datos);

//-------------------------------------------------------------------------------
if (datos.equals("1") || datos.equals("2") || datos.equals("3")) {}

In this if it does not enter me sending from the client any of these 3 values. Debugueando I look at the value of the variable and it has the same value, when it reaches the if it does not enter.

java string netbeans

asked by Ruben Gil Gomez 13.12.2016 в 20:00

source

2 answers

0

In java it is best to use the equals function when it is a string within the if.

But Wait

To be completely sure of what you are going to compare if it is a String, as a good practice of programming it is better that you compare in the following way.

if (("1".equals(datos)) || ("2".equals(datos)) || ("3".equals(datos))) {}

Now why I say that to be completely sure that it is a String , because if you compare the data sent to you with the string to compare. The data sent to you may come as null , if you see. In your case if you compare like this:

if ((datos.equals("1")) || (datos.equals("2")) || (datos.equals("3"))) {}

The data variable may come as null so once the condition is not met.

Now if you do as I say, you're completely sure that the ones you're going to compare is a string with anything else that is supposed to be a string.

Greetings

answered by 14.12.2016 в 18:47

Problem with the Bootstrap modal [closed] execute query TOP 1, JDBC

score 0 · Accepted Answer

If the variable datos is of type String even if it is an object ( new String() ), it is difficult not to comply with the sentence, possibly your variable datos contain space (s):

if (datos.trim().equals("1") || datos.trim().equals("2") || datos.trim().equals("3")) {
...
...