0
I have two variables:
int trabalhadorFormal;
int dispensadoIndireta;
And I’m wearing them JOptionPane.showConfirmDialog to confirm as a kind of Boolean, yes or no, for the user.
trabalhadorFormal = JOptionPane.showConfirmDialog(null,
"É trabalhador formal?",
"Por favor selecione",
JOptionPane.YES_NO_OPTION);
dispensadoIndireta = JOptionPane.showConfirmDialog(null,
"Foi dispensado indiretamente?",
"Por favor selecione",
JOptionPane.YES_NO_OPTION);
And with this I have the following code:
if (trabalhadorFormal == JOptionPane.YES_OPTION) {
if (dispensadoIndireta == JOptionPane.NO_OPTION) {
do {
opcao = Integer.parseInt(JOptionPane.showInputDialog("[1] - É empregado doméstico" +
"\n[2] - Não é empregado doméstico"));
switch (opcao) {
case 1:
JOptionPane.showMessageDialog(null, "É empregado doméstico!");
break;
case 2:
JOptionPane.showMessageDialog(null, "Não é empregado doméstico!");
break;
default:
JOptionPane.showMessageDialog(null, "Opção inválida!");
break;
}
} while (opcao != 1 && opcao != 2);
} else {
JOptionPane.showMessageDialog(null, "Teste!");
}
}
The only problem is that when I select the options that will lead me to the else after the second if, nothing happens and the program closes me by returning successfully, as simply ignoring the else.
What is going wrong here? In building the if-else?
What the Bugger says about the value of
dispensadoIndireta?– Oralista de Sistemas
I replicated your code here and it worked right, according to your code if you choose not it enters your do while, and while if you indicate a valid condition it breaks and exits the loop. In the second if you give a yes it goes to the Else and prints test.
– Edjane