我認爲可能有辦法做到這一點,而無需構建整個範圍的實際列表。
一種方法是擴展AbstractSpinnerModel
以創建LongNumberModel
,如下所示。另見相關的example。
import java.awt.BorderLayout;
import java.awt.EventQueue;
import java.awt.HeadlessException;
import java.text.ParseException;
import javax.swing.AbstractSpinnerModel;
import javax.swing.JFormattedTextField;
import javax.swing.JFormattedTextField.AbstractFormatter;
import javax.swing.JFrame;
import javax.swing.JSpinner;
import javax.swing.text.DefaultFormatter;
import javax.swing.text.DefaultFormatterFactory;
/**
* @see https://stackoverflow.com/a/13121724/230513
* @see https://stackoverflow.com/a/9758714/230513
*/
public class HexSpinnerTest {
public static void main(String[] args) {
EventQueue.invokeLater(new Runnable() {
@Override
public void run() {
display();
}
});
}
private static void display() throws HeadlessException {
JFrame f = new JFrame();
f.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
JSpinner sp = new JSpinner(new LongNumberModel(0x8000L, 0x8000L, 0xFFFFL, 1L));
JSpinner.DefaultEditor editor = (JSpinner.DefaultEditor) sp.getEditor();
JFormattedTextField tf = editor.getTextField();
tf.setFormatterFactory(new MyFormatterFactory());
f.getContentPane().add(sp, BorderLayout.NORTH);
f.pack();
f.setLocationRelativeTo(null);
f.setVisible(true);
}
private static class LongNumberModel extends AbstractSpinnerModel {
private Long value, stepSize;
private Comparable<Long> minimum, maximum;
public LongNumberModel(Long value, Long minimum, Long maximum, Long stepSize) {
this.value = value;
this.minimum = minimum;
this.maximum = maximum;
this.stepSize = stepSize;
}
@Override
public Object getValue() {
return value;
}
@Override
public void setValue(Object value) {
this.value = (Long) value;
fireStateChanged();
}
@Override
public Object getNextValue() {
long v = value.longValue() + stepSize.longValue();
return bounded(v);
}
@Override
public Object getPreviousValue() {
long v = value.longValue() - stepSize.longValue();
return bounded(v);
}
private Object bounded(long v) {
if ((maximum != null) && (maximum.compareTo(v) < 0)) {
return null;
}
if ((minimum != null) && (minimum.compareTo(v) > 0)) {
return null;
}
return Long.valueOf(v);
}
}
private static class MyFormatterFactory extends DefaultFormatterFactory {
@Override
public AbstractFormatter getDefaultFormatter() {
return new HexFormatter();
}
}
private static class HexFormatter extends DefaultFormatter {
@Override
public Object stringToValue(String text) throws ParseException {
try {
return Long.valueOf(text, 16);
} catch (NumberFormatException nfe) {
throw new ParseException(text, 0);
}
}
@Override
public String valueToString(Object value) throws ParseException {
return Long.toHexString(
((Long) value).intValue()).toUpperCase();
}
}
}
[一些提示爲十六進制的JSpinner的可能重複?我的方法是正確的](http://stackoverflow.com/questions/9758469/some-hints-for-hex-jspinner-i-s-the-approach-correct) – trashgod
我嘗試了上述方法,在試驗解決方案沒有。 1.上面的鏈接只能解決十六進制格式問題,不能顯示負值。我面臨着一個不同的問題,其中包括以負HexBinary表示啓動JSpinner,並向上滾動直到達到上限。 – schinoy
我設法最終通過使用JSpinnerListModel來完成任務,並將其傳遞給由實用程序類構造的範圍內的六進制字符串值列表。當然,我不喜歡這種解決方案,並認爲可能有辦法做到這一點,而無需構建整個範圍的實際列表。 – schinoy