Android SQLite 在选定的微调器上显示特定行

如何解决Android SQLite 在选定的微调器上显示特定行

我从 3 个难度中获得了创纪录的高分：简单、中等、困难。所有数据进入 1 个表。我有 1 个活动来显示所有的高分。

在那个活动中，我创建了一个微调器，让用户选择他们想要显示的难度数据。所有数据都将显示在列表视图（ScoreView）中。

这里我想做的是在微调器上选择用户时显示的具体数据。

有什么办法吗？

这是我的数据库：

 public Data_Collect[] getScore() {
    Data_Collect[] data = null;
    String[] colums = {"Difficult","BRedo","BCorrect"};
    String orderBy = "Difficult ASC";
    Cursor cursor = getReadableDatabase().query(
            "score",colums,null,orderBy);
    if (cursor.getCount()>0 && cursor.moveToNext()) {

        data = new Data_Collect[cursor.getCount()];
        int counter = 0;

        do {

            data[counter] = new Data_Collect(
                    cursor.getString(0),cursor.getInt(1),cursor.getInt(2));
            counter++;

        }
        while (cursor.moveToNext());
    }

    cursor.close();
    return data;
}

这是我的活动：

ArrayAdapter<String> adapterDifficult = new ArrayAdapter<>(
            Score.this,android.R.layout.simple_list_item_1,Difficulty
    );
    SpinnerDifficulty.setAdapter(adapterDifficult);
    SpinnerDifficulty.setOnItemSelectedListener(this);

    new Thread(new Runnable() {
        @Override
        public void run() {
            RecordDB db = new RecordDB(Score.this);
            Data_Collect[] data = db.getScore();
            db.close();

            if (data != null) {
                adapter = new RecordAdapter(Score.this,R.layout.recordlist,R.id.text_Difficulty,data);
                runOnUiThread(new Runnable() {
                    @Override
                    public void run() {
                        ScoreView.setAdapter(adapter);
                    }
                });
            }
        }
    }).start();

}


@Override
public void onItemSelected(AdapterView<?> parent,View view,int position,long id) {
}

@Override
public void onNothingSelected(AdapterView<?> parent) {

}

解决方法

定义困难 = '简单'

Data_Collect[] data = null;
String[] colums = {"Difficult","BRedo","BCorrect"};
String orderBy = "Difficult ASC";
Cursor cursor = getReadableDatabase().query(
        "score",colums,"Difficult ='Easy'",null,orderBy);