java Android Studio SQLite数据库创建不起作用?
对于Android Studio和/或Java编程,我不是一个有经验的用户。我正在从事一个项目,该项目要求应用程序使用SQLite DB
根据一些指南,我创建了一个Java类,以便在本地管理我将使用的非常特定的DB,代码如下:
public class gestionedb {
static final String KEY_RIGAID = "_id";
static final String KEY_EMAIL = "email";
static final String KEY_PASSWORD = "password";
static final String KEY_NICKNAME = "nickname";
static final String TAG = "gestionedb";
static final String DATABASE_NOME = "superchat5db";
static final String DATABASE_TABELLA = "users";
static final int DATABASE_VERSIONE = 1;
static final String DATABASE_CREAZIONE = "create table clienti (_id integer primary key autoincrement, " + "email text not null, password text not null, nickname text not null);";
final Context context;
DatabaseHelper DBHelper;
SQLiteDatabase db;
public gestionedb(Context ctx){
this.context = ctx;
DBHelper = new DatabaseHelper(context);
}
private static class DatabaseHelper extends SQLiteOpenHelper {
DatabaseHelper(Context context) {
super(context, DATABASE_NOME, null, DATABASE_VERSIONE);
}
@Override
public void onCreate(SQLiteDatabase db) {
try {
db.execSQL(DATABASE_CREAZIONE);
}
catch (SQLException e) {
e.printStackTrace();
}
}
@Override
public void onUpgrade(SQLiteDatabase db, int oldVersion, int newVersion) {
Log.d("Test", "in onUpgrade. Old is: " + oldVersion + " New is: " + newVersion);
}
}
public gestionedb open() throws SQLException {
db = DBHelper.getWritableDatabase();
return this;
}
public void close() {
DBHelper.close();
}
public Cursor obtainallusers() {
return db.query(DATABASE_TABELLA, new String[] {KEY_RIGAID, KEY_EMAIL, KEY_NICKNAME}, null, null, null, null, null);
}
public Cursor obtainuser(long rigaId) throws SQLException {
Cursor mCursore = db.query(true, DATABASE_TABELLA, new String[] {KEY_RIGAID, KEY_EMAIL, KEY_NICKNAME}, KEY_RIGAID + "=" + rigaId, null, null, null, null, null);
if (mCursore != null) {
mCursore.moveToFirst();
}
return mCursore;
}
public long adduser(String email, String password, String nickname) {
ContentValues initialValues = new ContentValues();
initialValues.put(KEY_EMAIL, email);
initialValues.put(KEY_PASSWORD, password);
initialValues.put(KEY_NICKNAME, nickname);
return db.insert(DATABASE_TABELLA, null, initialValues);
}
public boolean deleteuser(long rigaId) {
return db.delete(DATABASE_TABELLA, KEY_RIGAID + "=" + rigaId, null) > 0;
}
public boolean updatenickname(long rigaId, String nickname) {
ContentValues args = new ContentValues();
args.put(KEY_NICKNAME, nickname);
return db.update(DATABASE_TABELLA, args, KEY_RIGAID + "=" + rigaId, null) > 0;
}
public boolean updatepsw(long rigaId, String password){
ContentValues args = new ContentValues();
args.put(KEY_PASSWORD, password);
return db.update(DATABASE_TABELLA, args, KEY_RIGAID + "=" + rigaId, null) > 0;
}
}
当然,为了使这个类适应我想要使用的DB,我必须稍微调整变量。我使用的指南是意大利语的,因此变量的名称是意大利语的,对此表示抱歉
我决定通过从MainActivity创建一个按钮来测试所有这些,该按钮启动一个名为“ProvaDB”(TestDB)的全新活动,该活动执行以下操作:
gestionedb db = new gestionedb(this);
db.open();
long id = db.adduser("prova@gmail.com", "prova123", "prova1");
id = db.adduser("prova2@gmail.com", "prova123", "prova2");
db.close();
db.open();
Cursor c = db.obtainallusers();
if (c.moveToFirst()){
do{
Toast.makeText(this,
"id: " + c.getString(0) + "\n" +
"Email: " + c.getString(1) + "\n" +
"Nickname: " + c.getString(2),
Toast.LENGTH_LONG).show();
} while (c.moveToNext());
}
db.close();
}
它应该创建数据库并添加几个用户,然后用一个长的Toast文本列出他们。当我尝试运行应用程序时,单击按钮转到“ProvaDB”后,应用程序崩溃。查看日志,我注意到数据库丢失了,这很奇怪。我不明白这有什么不对,因为正如我前面提到的,我不是一个有经验的用户。有什么想法吗
日志的一部分:
2018-12-10 19:11:05.175 17701-17701/com.example.lorenzo.superchat5 E/AndroidRuntime: FATAL EXCEPTION: main
Process: com.example.lorenzo.superchat5, PID: 17701
java.lang.RuntimeException: Unable to start activity ComponentInfo{com.example.lorenzo.superchat5/com.example.lorenzo.superchat5.ProvaDB}: 安卓.database.sqlite.SQLiteException: no such table: users (code 1): , while compiling: SELECT _id, email, nickname FROM users
#################################################################
Error Code : 1 (SQLITE_ERROR)
Caused By : SQL(query) error or missing database.
(no such table: users (code 1): , while compiling: SELECT _id, email, nickname FROM users)
多谢各位
# 1 楼答案
也许当你更改各种名称时,你忘记了:
将其更改为:
并检查代码的其他部分,以确定您可能已硬编码了此类名称,并将其更改为新名称