由网友(青橘栀耳)分享简介：我试图让我的第一个Android应用程序。我听说，如果没有创建数据库的SQLiteOpenHelper.onCreate（）方法的过程创建表。然而，以为我试图调试的onCreate（）方法没有奏效，甚至。请看看下面的code和给我任何建议。任何帮助将AP preciated。======================...

我试图让我的第一个Android应用程序。我听说，如果没有创建数据库的SQLiteOpenHelper.onCreate（）方法的过程创建表。然而，以为我试图调试的onCreate（）方法没有奏效，甚至。请看看下面的code和给我任何建议。任何帮助将AP preciated。

  ====================================== ===========================
公共类NameToPinyinActivity延伸活动{

    DatabaseOpenHelper帮手= NULL;

    @覆盖
    公共无效的onCreate（包savedInstanceState）{
        super.onCreate（savedInstanceState）;
        的setContentView（R.layout.nametopinyin）;

        按钮searchButton =（按钮）findViewById（R.id.search）;
        sea​​rchButton.setOnClickListener（新ButtonClickListener（））;

        辅助=新DatabaseOpenHelper（NameToPinyinActivity.this）;
    }

================================================== ===============
公共类DatabaseOpenHelper扩展SQLiteOpenHelper {

    / ** DB名称* /
    私有静态最后弦乐DB_NAME =拼音;

    / ** CREATE TABLE SQL * /
    私有静态最后弦乐CREATE_TABLE_SQL =CREATE TABLE UNI code_PINYIN
            +（ID INTEGER PRIMARY KEY AUTOINCREMENT，
            +UNI code文本NOT NULL，拼音文字NOT NULL）;

    公共DatabaseOpenHelper（上下文的背景下）{
        超级（上下文，DB_NAME，空，1）;
    }

    @覆盖
    公共无效的onCreate（SQLiteDatabase DB）{
        db.beginTransaction（）;
        尝试 {
            db.execSQL（CREATE_TABLE_SQL）;
            db.setTransactionSuccessful（）;
        }赶上（例外五）{
            e.printStackTrace（）;
        } 最后 {
            db.endTransaction（）;
        }
    }

================================================== ===============
 

解决方案

我也有过与 SQLiteOpenHelper 的麻烦。什么工作对我来说是存储成员变量

  SQLiteDatabase分贝;
 

在SQLiteOpenHelper的子类，并要求

  DB = getWritableDatabase（）;
 

在构造函数中。

在回答这个问题还包括有用的信息：SQLiteOpenHelper不能打电话的onCreate？

我希望这有助于！

I am trying to make my first android app. I heard that the SQLiteOpenHelper.onCreate() method process to create tables if the database is not created. However, the onCreate() method did not work even thought I tried to debug. Please look at the code below and give me any suggestions. Any help will be appreciated.

=================================================================
public class NameToPinyinActivity extends Activity {

    DatabaseOpenHelper helper = null;

    @Override
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.nametopinyin);

        Button searchButton = (Button) findViewById(R.id.search);
        searchButton.setOnClickListener(new ButtonClickListener());

        helper = new DatabaseOpenHelper(NameToPinyinActivity.this);
    }

=================================================================
public class DatabaseOpenHelper extends SQLiteOpenHelper {

    /** DB Name */
    private static final String DB_NAME = "pinyin";

    /** CREATE TABLE SQL */
    private static final String CREATE_TABLE_SQL = "CREATE TABLE UNICODE_PINYIN"
            + "(ID INTEGER PRIMARY KEY AUTOINCREMENT, "
            + "UNICODE TEXT NOT NULL, PINYIN TEXT NOT NULL)";

    public DatabaseOpenHelper(Context context) {
        super(context, DB_NAME, null, 1);
    }

    @Override
    public void onCreate(SQLiteDatabase db) {
        db.beginTransaction();
        try {
            db.execSQL(CREATE_TABLE_SQL);
            db.setTransactionSuccessful();
        } catch (Exception e) {
            e.printStackTrace();
        } finally {
            db.endTransaction();
        }
    }

=================================================================

解决方案

I have also had trouble with the SQLiteOpenHelper. What worked for me was storing a member variable

SQLiteDatabase db;

In the SQLiteOpenHelper subclass and calling

 db = getWritableDatabase();

in the constructor.

The answer to this question also includes helpful information: SQLiteOpenHelper failing to call onCreate?

I hope this helps!