JavaFX(数组)中的java创建和登录程序
因此,我想创建一个程序,为用户创建一个新帐户,用户应该能够使用相同的帐户“登录”。现在,我已经设法完成了第一部分(创建一个带有数组的帐户),但现在我不知道如何使用该帐户登录
PS:当我说登录时,它只是一个弹出的警报框
以下是代码:
import javafx.application.Application;
import javafx.event.ActionEvent;
import javafx.event.EventHandler;
import javafx.scene.Scene;
import javafx.scene.control.Button;
import javafx.scene.control.TextField;
import javafx.scene.control.PasswordField ;
import javafx.scene.control.Label;
import javafx.scene.layout.VBox;
import javafx.stage.Stage;
public class Login extends Application {
Button btnAanmaken,btnInloggen;
TextField txtUsername;
PasswordField Password;
Label lblUsername,lblPassword;
int index;
String[] user = new String[10];
String[] pas = new String[10];
public static void main(String[] args) {
launch(args);
}
@Override
public void start(Stage primaryStage) {
primaryStage.setTitle("Aanmelden");
btnAanmaken = new Button();
btnAanmaken.setText("Aanmaken");
btnInloggen = new Button();
btnInloggen.setText("Inloggen");
txtUsername=new TextField();
txtUsername.setMaxWidth(200);
Password=new PasswordField();
Password.setMaxWidth(200);
lblUsername=new Label();
lblUsername.setText("Gebruikersnaam");
lblPassword=new Label();
lblPassword.setText("Paswoord");
btnAanmaken.setOnAction(e ->{
if (index < 10){
index++;
user[index]=txtUsername.getText();
pas[index]=Password.getText();
AlertBox.display("Accountdetails","Account werd succesvol aangemaakt: \n Gebruikersnaam: " + user[index] + " \n Paswoord: " + pas[index] );
txtUsername.clear();
Password.clear();
}else{
AlertBox.display("Error", "U heeft het maximaal aantal accounts bereikt.");
}
});
btnInloggen.setOnAction(e ->{
for (int c = 0; c < 10; c++)
{
if ((user[c] == txtUsername.getText()) && (pas[c] == Password.getText())){
AlertBox.display("OKEE", "goed.");
}
else{
AlertBox.display("Error", "NIET OK.");
}
}
});
VBox layout = new VBox();
layout.getChildren().addAll(lblUsername,txtUsername,lblPassword,Password,btnAanmaken,btnInloggen);
Scene scene = new Scene(layout, 300, 250);
primaryStage.setScene(scene);
primaryStage.show();
}
}
# 1 楼答案
这种情况永远不会得到验证:
不能通过
==
比较String
,但必须使用equals
: