我写了一个方法,可以让你在数据库中添加一个“单词”。一旦我调用 addWord() 方法,它会在尝试 database.insert() 时引发 Null 指针异常。
public class DatabaseControl {
// columns for table
public static final String KEY_ROWID = "_id";
public static final String KEY_SYNID = "synsetID";
public static final String KEY_WORD = "synsetID";
public static final String KEY_DESCRIPTION = "description";
public static final String KEY_POPULARITY = "popularity";
// table name
public static final String DATABASE_TABLE = "wordnet"; // Name of the Database
private Context context;
private SQLiteDatabase database;
private DatabaseHelper dbHelper;
public DatabaseControl(Context context){
this.context = context;
}
public DatabaseControl open() throws SQLiteException {
dbHelper = new DatabaseHelper(context);
database = dbHelper.getWritableDatabase();
return this;
}
public void close(){
dbHelper.close();
}
public long addWord (String synId, String word, String description, int popularity) {
ContentValues setUpVals = createContentValues(synId, word, description, popularity);
// SEEMS TO THROW NULL POINTER EXCEPTION
return database.insert(DATABASE_TABLE, null , setUpVals);
}
我在这里调用这个方法:
private void populateDatabase(){
try {
long rowId = 0;
rowId = dbControl.addWord("testSyn","","",1); // Add a row to the database for test
}
catch (SQLiteException e){
e.printStackTrace();
}
不太明白为什么会抛出这个错误,而且我不完全理解 insert() 中的第二个参数,所以据我所知可能是这样:/