android string null 比较大小,在Android SQLite中选择NULL值

由于selectArgs不正确,我正在执行以下方法但没有成功(至少这是我所相信的.

找到所有：

public Collection findAllByCodigoSetorOrderByStatusWhereDataAgendamentoIsNull(Integer vendedor) {

Collection objects = null;

String selection = Object.FIELDS[20] + "=?" + " OR " + Object.FIELDS[20] + "=?" + " OR " + Object.FIELDS[20] + "=?" + " AND " + Object.FIELDS[6] + "=?";

String[] selectionArgs = new String[] { "''", "'null'", "NULL", String.valueOf(vendedor) };

Collection results = findAllObjects(Object.TABLE_NAME, selection, selectionArgs, Object.FIELDS, null, null, Object.FIELDS[4]);

objects = new ArrayList();

for (ContentValues result : results) {

objects.add(new Object(result));

}

return objects;

}

findAllObjects：

protected Collection findAllObjects(String table, String selection, String[] selectionArgs, String[] columns, String groupBy, String having, String orderBy) {

Cursor cursor = null;

ContentValues contentValue = null;

Collection contentValues = null;

try {

db = openRead(this.helper);

if (db != null) {

cursor = db.query(table, columns, selection, selectionArgs, groupBy, having, orderBy);

contentValues = new ArrayList();

for (int i = 0; i < cursor.getCount(); i++) {

cursor.moveToPosition(i);

contentValue = new ContentValues();

for (int c = 0; c < cursor.getColumnCount(); c++) {

contentValue.put(cursor.getColumnName(c), cursor.getString(c));

}

contentValues.add(contentValue);

cursor.moveToNext();

}

}

return contentValues;

} finally {

close(db);

}

}

如何使用db.query正确选择和比较列 – null,’null’和”？