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java - 值未插入到 SQLite 数据库

转载 作者:搜寻专家 更新时间:2023-11-01 09:30:00 24 4
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我使用两个编辑文本(Emailpassword)和 2 个按钮(提交)创建了一个用于测试的应用程序和 viewdata)。我调试了这个应用程序,它工作正常。

我想通过再添加一个 Edittext (Name) 来更新此应用程序,并在 onUpgrade() 中升级 SQLite 数据库 方法。代码如下所示。

public class DatabaseHelper extends SQLiteOpenHelper {
private static final String DATABASE_NAME="upgrade";
private static final int DATABASE_VERSION=2;

//details table details
public static final String TABLE_NAME_Details="details";
public static final String USERNAME="USERNAME";
public static final String PASSWORD="PASSWORD";
public static final String Name="Name"; //newly added column in version 2


public static final String TABLE_NAME_Details_temp="temp";
public static final String USERNAME_temp="USERNAME";
public static final String PASSWORD_temp="PASSWORD";
public static final String Name_temp="Name";

//create table statements for version 2
public static final String Create_Table_Details = "CREATE TABLE "
+ TABLE_NAME_Details + " (" + USERNAME + " TEXT PRIMARY KEY,"
+ PASSWORD + " TEXT ,"
+ Name + " TEXT"
+")";

/* creation for version 1
private static final String Create_Table_Details = "CREATE TABLE " +
TABLE_NAME_Details + "("
+ USERNAME + " TEXT PRIMARY KEY,"
+ PASSWORD + " TEXT"
+ ")";*/
//Alter table statements FOR onUpgrade()
//private static final String ALTER_Details = "ALTER TABLE " +
TABLE_NAME_Details + " ADD COLUMN" + Name + " TEXT";
public DatabaseHelper(Context context)
{
super(context, DATABASE_NAME, null, DATABASE_VERSION);
}
@Override
public void onCreate(SQLiteDatabase db) {
try{
db.execSQL(Create_Table_Details);
Log.d("database","installed successfully");
}catch (Exception e)
{
e.printStackTrace();
Log.e("/test","Exception due to"+e.toString());
}
}
@Override
public void onUpgrade(SQLiteDatabase db, int oldVersion, int newVersion)
{

String TEMP_CREATE_CONTACTS_TABLE = "CREATE TABLE " +
TABLE_NAME_Details_temp + "("
+ USERNAME_temp + " TEXT," + PASSWORD_temp + " TEXT )";
db.execSQL(TEMP_CREATE_CONTACTS_TABLE);

db.execSQL("INSERT INTO " + TABLE_NAME_Details_temp + " SELECT " +
USERNAME + ", "
+ PASSWORD + " FROM " + TABLE_NAME_Details);

db.execSQL("DROP TABLE "+ TABLE_NAME_Details);

String CREATE_CONTACTS_TABLE = "CREATE TABLE " + TABLE_NAME_Details + "("
+ USERNAME + " TEXT," + PASSWORD + " TEXT," + Name + " TEXT )";
db.execSQL(CREATE_CONTACTS_TABLE);
db.execSQL("INSERT INTO " + TABLE_NAME_Details + " SELECT " +
USERNAME_temp + ", "
+ PASSWORD_temp + ", " + Name_temp + ", " + null + " FROM " +
TABLE_NAME_Details_temp);
db.execSQL("DROP TABLE " + TABLE_NAME_Details);
}

public boolean insertData(String username, String password,String name)
{
try{
SQLiteDatabase db = this.getWritableDatabase();
ContentValues contentValues=new ContentValues();
contentValues.put(USERNAME,username);
contentValues.put(PASSWORD,password);
contentValues.put(Name,name);
long result=db.insertOrThrow(TABLE_NAME_Details, null, contentValues);
if(result==-1)
return false;
else
return true;
}catch(Exception e)
{
e.printStackTrace();
Log.e("/test","Exception due to"+e.toString());
return false;
}
}
//update value
public boolean updatePassword(String LoggedUsername)
{
SQLiteDatabase db = null;
try {
db = this.getWritableDatabase();
ContentValues values = new ContentValues();
values.put(PASSWORD, "test");
if (db.isOpen()) {
db.update(TABLE_NAME_Details, values, USERNAME + "='" +
LoggedUsername+"'",null);
return true;
}
else
return false;
} catch (Exception e) {
Log.d("eEmp/DBUpdateUser ", e.toString());
return false;
}
}
//getting data from database
public Cursor getAllData()
{
SQLiteDatabase db = this.getWritableDatabase();
Cursor res = db.rawQuery("select * from "+TABLE_NAME_Details,null);
return res;
}
}

我安装了第一个版本并且工作正常,然后卸载了。

卸载后,安装第二个版本并存储数据。它也工作正常。

当我在 Debug模式下尝试用第二个版本更新第一个版本时,应用程序更新成功,并且在第一个版本中输入的数据可见。当我尝试将数据存储在第二个版本中时,数据没有插入到 sqlite 数据库中。为什么?

调试时,插入方法被调用,但这里 if(result==-1) 我得到的是 -1 而不是 1。因此,数据控制来自该方法,而数据不是插入数据库。

例如:在第一个版本中:我输入了用户名:stackoverflow 和密码 stackoverflow。我将应用程序更新到第二版并成功更新。现在我输入用户名:hello 密码:hello 姓名:hello 然后点击提交按钮。

之后,如果我点击viewdata 按钮:它显示用户名:stackoverflow 密码stackoverflow name:null。但不显示在第二个版本中插入的数据。

如有任何帮助,我们将不胜感激。请帮助我找到正确的解决方案。

谢谢。

最佳答案

试试这个..

      // Upgrading database
@Override
public void onUpgrade(SQLiteDatabase db, int oldVersion, int newVersion) {

String TEMP_CREATE_CONTACTS_TABLE = "CREATE TABLE " + TEMP_TABLE_CONTACTS + "("
+ KEY_ID + " INTEGER PRIMARY KEY," + KEY_NAME + " TEXT," + KEY_ADDRESS + " TEXT)";
db.execSQL(TEMP_CREATE_CONTACTS_TABLE);

// Create an temporaty table that can store data of older version

db.execSQL("INSERT INTO " + TEMP_TABLE_CONTACTS + " SELECT " + KEY_ID + ", "
+ KEY_NAME + ", " + KEY_ADDRESS + " FROM " + TABLE_CONTACTS);

// Insert data into temporary table from existing older version database (that doesn't contains ADDRESS2 //column)

db.execSQL("DROP TABLE "+ TABLE_CONTACTS);
// Remove older version database table

String CREATE_CONTACTS_TABLE = "CREATE TABLE " + TABLE_CONTACTS + "("
+ KEY_ID + " INTEGER PRIMARY KEY," + KEY_NAME + " TEXT," + KEY_ADDRESS + " TEXT," + KEY_ADDRESS2 + " TEXT )";
db.execSQL(CREATE_CONTACTS_TABLE);

// Create new table with ADDRESS2 column
db.execSQL("INSERT INTO " + TABLE_CONTACTS + " SELECT " + KEY_ID + ", "
+ KEY_NAME + ", " + KEY_ADDRESS + ", " + null + " FROM " + TEMP_TABLE_CONTACTS);
// Insert data ffrom temporary table that doesn't have ADDRESS2 column so left it that column name as null.
db.execSQL("DROP TABLE " + TEMP_TABLE_CONTACTS);
}

关于java - 值未插入到 SQLite 数据库,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/47898999/

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