我有这样的SQLite数据库表:
_id || Column1 || Column2 ||
1 || test1 || a,b,c,d,e,f ||
2 || test2 || g,h,i,j,k,l ||
3 || test3 || m,n,o,p,q,r ||
如何从此表中选择Column2的一个随机数?
我想要这样的结果:
_id || Column1 || Column2 ||
1 || test1 || d ||
2 || test2 || k ||
3 || test3 || r ||
假设d,k,r是每个记录的Column2的随机值。
由于
更新: 我创建了CustomAdapter并添加了一些这样的View:
@Override
public View newView(Context context, Cursor cursor, ViewGroup parent) {
Cursor c = getCursor();
final LayoutInflater inflater = LayoutInflater.from(context);
View v = inflater.inflate(layout, parent, false);
int column2Name= c.getColumnIndex(DBAdapter.COLUMN2);
String col2Name= c.getString(column2Name);
String[] temp;
Random random = new Random();
temp = col2Name.split(",");
String col2 = temp[random.nextInt(temp.length)];
TextView col2_name = (TextView) v.findViewById(R.id.column2_label);
if (col2_name != null) {
col2_name .setText(col2);
}
return v;
}
@Override
public void bindView(View v, Context context, Cursor c) {
int column2Name= c.getColumnIndex(DBAdapter.COLUMN2);
String col2Name= c.getString(column2Name);
String[] temp;
Random random = new Random();
temp = col2Name.split(",");
String col2 = temp[random.nextInt(temp.length)];
TextView col2_name = (TextView) v.findViewById(R.id.column2_label);
if (col2_name != null) {
col2_name .setText(col2);
}
}
现在,只需在您的活动中使用该customAdapter,以便column2可以显示为1个随机值。
答案 0 :(得分:2)
一旦获得数据库中的数据,这在Java中就会容易得多。
Random RANDOM = new Random(System.currentTimeMillis());
int index = cursor.getColumnIndex("Column2");
String value = cursor.getString(index);
String[] values = value.split(",");
String randomValue = values[RANDOM.nextInt(values.length)];