安卓SQLiteOpenHelper：的onCreate（）方法没有被调用。为什么？

本文介绍了安卓SQLiteOpenHelper：的onCreate（）方法没有被调用。为什么？的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

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!