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我似乎無法做到正確。我有一個listbox與類別項目。我還有一個數據模型，用於定義ID，名稱，姓氏，類別以及包含數據的ObservableList。我試圖更新現有對象，以防用戶點擊listbox中的同一項目並更改名稱，姓氏。防止ObservableList中的重複條目
這裏是我的代碼：
public class FXMLDocController implements Initializable {
ObservableList<String> listitems = FXCollections.observableArrayList(
"Visual Basic", "ASP.net", "JavaFX");
ObservableList<Persons> personData = FXCollections.observableArrayList();
Persons pp = new Persons();
private Label label;
@FXML
private TextField txtName;
@FXML
private TextField txtLastName;
@FXML
private Button btnSave;
@FXML
private TextArea txtArea;
@FXML
private ListView<String> listview = new ListView<String>();
@FXML
private Button btnTest;
@FXML
private Label lblCategory;
@FXML
private Label lblIndex;
@Override
public void initialize(URL url, ResourceBundle rb) {
listview.setItems(listitems);
}
@FXML
private void handleSave(ActionEvent event) {
String category = lblCategory.getText();
boolean duplicate = false;
//Add data. Check first if personData is empty
if (personData.isEmpty()){
pp = new Persons(Integer.valueOf(lblIndex.getText()),category,txtName.getText(),txtLastName.getText());
personData.add(pp);
}else{
for(int i = 0 ; i<personData.size() ; i ++){
if(Integer.toString(personData.get(i).getID()).equals(lblIndex.getText())){
duplicate = true;
}else{
duplicate = false;
}
}
System.out.println(duplicate);
if (duplicate == false){
pp = new Persons(Integer.valueOf(lblIndex.getText()),category,txtName.getText(),txtLastName.getText());
personData.add(pp);
}else{
System.out.println("Duplicate");
// Do Update later.
}
}
//Show data to Test
System.out.println("-- START OF LIST --");
for (Persons person : personData){
System.out.println(person.getID() + " " + person.getCategory()+ " " + person.getName() + " " + person.getLastname() + " ");
}
System.err.println(" ");
}
@FXML
private void handleListClick(MouseEvent event) {
lblCategory.setText(listview.getSelectionModel().getSelectedItem());
lblIndex.setText(String.valueOf(listview.getSelectionModel().getSelectedIndex()));
}
}
如果我點擊它運作良好的listbox一個項目進行了多次..但如果比如我點擊的Visual Basic，然後ASP.net的回它仍然接受Visual Basic。 :( 需要諮詢和幫助。請
這是基於菲爾的建議代碼
@FXML
private void handleSave(ActionEvent event) {
String category = lblCategory.getText();
boolean duplicate = false;
int x = 0;
//Add data
Persons newPerson = new Persons(Integer.valueOf(lblIndex.getText()), category, txtName.getText(), txtLastName.getText());
if (!personData.contains(newPerson)) {
personData.add(newPerson);
}else{
System.out.println("Duplicate!");
}
//Show data
System.out.println("-- START OF LIST --");
for (Persons person : personData){
System.out.println(person.getID() + " " + person.getCategory()+ " " + person.getName() + " " + person.getLastname() + " ");
}
}
，這裏是我的人類
public class Persons {
private SimpleIntegerProperty id;
private SimpleStringProperty name;
private SimpleStringProperty lastname;
private SimpleStringProperty category;
public Persons(){}
public Persons(int id, String category, String name, String lastname){
this.id = new SimpleIntegerProperty(id);
this.name = new SimpleStringProperty(name);
this.lastname = new SimpleStringProperty(lastname);
this.category = new SimpleStringProperty(category);
}
public Persons(String name, String lastname){
this.name = new SimpleStringProperty(name);
this.lastname = new SimpleStringProperty(lastname);
}
@Override
public boolean equals(Object o){
if (o == this) return true;
if (!(o instanceof Persons)){
return false;
}
Persons persons = (Persons) o;
return persons.id.equals(id) &&
persons.name.equals(name) &&
persons.lastname.equals(lastname) &&
persons.category.equals(category);
}
//SETTERS
public void setID(int id) {
this.id = new SimpleIntegerProperty(id);
}
public void setName(String name) {
this.name = new SimpleStringProperty(name);
}
public void setLastname(String lastname) {
this.lastname = new SimpleStringProperty(lastname);
}
public void setCategory(String category) {
this.category = new SimpleStringProperty(category);
}
//GETTERS
public int getID() {
return id.getValue();
}
public String getName() {
return name.getValue();
}
public String getLastname() {
return lastname.getValue();
}
public String getCategory(){
return category.getValue();
}
// PROPERTIES
public SimpleIntegerProperty idProperty(){
return this.id;
}
public SimpleStringProperty nameProperty(){
return this.name;
}
public SimpleStringProperty lastnameProperty(){
return this.lastname;
}
public SimpleStringProperty categoryProperty(){
return this.category;
}
}
您好，我想我明白了。我用'休息';如果（Integer.toString（personData.get（i）.getID（））。equals（lblIndex.getText（））成爲一個整數，則返回值爲int（int i = 0; i
CuteGirlyGeek