android numberpicker 索引越界旋转

Question

我正在尝试创建自定义时间选择器。但是运行的问题是当我旋转屏幕时。另外，如果我按下主页按钮离开应用程序，当我返回相同的错误时它会崩溃。

public class MyTimePicker extends RelativeLayout {

private NumberPicker hour;
private NumberPicker minute;
private NumberPicker am;
private int interval = 5;

public MyTimePicker(Context context) {
    super(context);
    init();}

public MyTimePicker(Context context, AttributeSet attrs) {
    super(context, attrs);
    init();}

public MyTimePicker(Context context, AttributeSet attrs, int  defStyleAttr) {
    super(context, attrs, defStyleAttr);
    init();}  

    private void init() {
    LayoutInflater inflater = LayoutInflater.from(getContext());
    if (inflater != null) {

        String[] ams = new String[2];
        ams[0]="AM";
        ams[1]="PM";

        root = inflater.inflate(R.layout.my_time_picker, this, true);
        hour = (NumberPicker) root.findViewById(R.id.hour);
        minute = (NumberPicker) root.findViewById(R.id.minute);
        am = (NumberPicker) root.findViewById(R.id.am);  am.setDisplayedValues(ams);
        am.setWrapSelectorWheel(true);
        am.setMinValue(0);
        am.setMaxValue(1);

       String[] hours = new String[24];

       for (int i = 0; i < 24; i++) {
        hours[i]= "" + (i < 12 ? i + 1 : i - 11);
      }

    String[] mins = new String[60 / interval];
    NumberFormat formatter = new DecimalFormat("00");

    for (int i = 0; i < mins.length; i++) {
        mins[i] = formatter.format(i * interval);
    }

    hour.setDisplayedValues(hours);//This is what causes the error
    hour.setWrapSelectorWheel(true);

    hour.setMinValue(0);
    hour.setMaxValue(hours.length - 1);

    minute.setDisplayedValues(mins);//This is what causes the error

    minute.setWrapSelectorWheel(true);

    minute.setMinValue(0);
    minute.setMaxValue(mins.length - 1); 
     }
   }
}

这是我收到的错误

06-24 15:15:50.509    2006-2006/com.E/AndroidRuntime﹕ FATAL EXCEPTION: main
Process: com.mye.appo, PID: 2006
java.lang.IndexOutOfBoundsException: setSpan (2 ... 2) ends beyond length 0
        at android.text.SpannableStringBuilder.checkRange(SpannableStringBuilder.java:1018)
        at android.text.SpannableStringBuilder.setSpan(SpannableStringBuilder.java:611)
        at android.text.SpannableStringBuilder.setSpan(SpannableStringBuilder.java:607)
        at android.text.Selection.setSelection(Selection.java:76)
        at android.widget.EditText.setSelection(EditText.java:92)
        at android.widget.NumberPicker$SetSelectionCommand.run(NumberPicker.java:2189)
        at android.os.Handler.handleCallback(Handler.java:739)
        at android.os.Handler.dispatchMessage(Handler.java:95)
        at android.os.Looper.loop(Looper.java:135)
        at android.app.ActivityThread.main(ActivityThread.java:5221)
        at java.lang.reflect.Method.invoke(Native Method)
        at java.lang.reflect.Method.invoke(Method.java:372)
        at com.android.internal.os.ZygoteInit$MethodAndArgsCaller.run(ZygoteInit.java:899)
        at com.android.internal.os.ZygoteInit.main(ZygoteInit.java:694)

我发现如果设置片段的显示值或 activity 仅当 if(this.isVisible())

时我可以防止此崩溃

根据我的发现，这似乎是一个生命周期问题

Answer 1

由于 bug - 在使用 NumberPicker 对象之前尝试设置：

NumberPicker.setSaveFromParentEnabled(false);       
NumberPicker.setSaveEnabled(false);

我打开了这个和其他数字选择器错误的问题：

reported picker bug at code.google

android numberpicker 索引越界旋转

android numberpicker index outofbounds rotation

lifecycle

android

screen-rotation

numberpicker