Question

我是Android的新手，刚开始使用我的第一个应用程序。任何帮助是极大的赞赏。我创建了一个dbhelper类，但它失败并出现NullPointer异常。当我尝试调试时，上下文始终显示为null，因此mDatabase也为null。

以下是我的代码。

public class PharmaExpDBHelper {

    static final String DB_NAME = "pharmaApp.db";
    static final int VERSION = 1;
    private SQLiteDatabase db;
    private Context context;

    public PharmaExpDBHelper(Context context) {
        this.context = context;
        OpenHelper openHelper = new OpenHelper(this.context);
        this.db = openHelper.getWritableDatabase(); 
    }


    private static class OpenHelper extends SQLiteOpenHelper {

        OpenHelper(Context context) {
            super(context, DB_NAME, null, VERSION);
        }

        @Override
        public void onCreate(SQLiteDatabase database) {
            database.execSQL("CREATE TABLE pharmacy_list(id integer primary key autoincrement, name text);");
        }

        @Override
        public void onUpgrade(SQLiteDatabase db, int oldVersion, int newVersion) {
            Log.w(PharmaExpDBHelper.class.getName(),
                    "Upgrading database from version " + oldVersion + " to "
                            + newVersion + ", which will destroy all old data");
            db.execSQL("DROP TABLE IF EXISTS" + " pharmacy_list");
            onCreate(db);
        }


    }

}

Answer 1

发布的代码我没有看到任何明显错误。由于您说Context为null，因此应确保实例化此类的位置正在传递有效的Context实例。

如果有人通过null IllegalArgumentException作为完整性检查，则抛出Context会有所帮助。

Answer 2

代替此代码

public PharmaExpDBHelper(Context context) {
    this.context = context;
    OpenHelper openHelper = new OpenHelper(this.context);
    this.db = openHelper.getWritableDatabase(); 
}

使用一个返回SQliteDatabase db的方法

 public SQLiteDatabase getDBHelper(Context context) {
    this.context = context;
    OpenHelper openHelper = new OpenHelper(this.context);
    this.db = openHelper.getWritableDatabase(); 
    return this.db
}

现在在您的活动中访问它

 SQLiteDatabase db=PharmaExpDBHelper class Obejct.getDBHelper(context);

Answer 3

公共类PharmaExpDBHelper {

static final String DB_NAME = "pharmaApp.db"; 
static final int VERSION = 1; 
private SQLiteDatabase db; 

private PharmaExpDBHelper(Context context) {  
        super(context, DB_NAME, null, VERSION);  
    }  
 @Override    
    public void onCreate(SQLiteDatabase database) {    
        database.execSQL("CREATE TABLE pharmacy_list(id integer primary key autoincrement, name text);");    
    }    

    @Override    
    public void onUpgrade(SQLiteDatabase db, int oldVersion, int newVersion) {    
        Log.w(PharmaExpDBHelper.class.getName(),    
                "Upgrading database from version " + oldVersion + " to "    
                        + newVersion + ", which will destroy all old data");    
        db.execSQL("DROP TABLE IF EXISTS" + " pharmacy_list");    
        onCreate(db);    
    }    

    public PharmaExpDBHelper open(){
              db=PharamExpDBHelper.getWriteableDatabase();
          return db;
     }
}

现在通过

访问您的Activity中的此数据库

PharmaExpDBHelper dbhHelper=new PharmaExpDBHelper(this);
db.open();

试试这段代码

android - 无法访问db - context始终为null

3 个答案: