我想用SQLite打开我的数据库。但是,当尝试从TextInputEditText获取数据并将其放入数据库时,它会在db = this.getWritableDatabase();中抛出NPE。
在大多数情况下,这似乎是由context为null造成的。但是它不是null,我用调试器检查了它。(myDB = new DatabaseHelper(this);)
(我已经尝试过关于这个问题的所有其他问题,我的情况肯定是不同的,因为它们没有帮助。)
MainActivity.java
public class MainActivity extends AppCompatActivity {
DatabaseHelper myDB;
TextInputEditText inputEmail;
TextInputEditText inputPass;
Button btn_login;
Button btn_signup;
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
try {
myDB = new DatabaseHelper(this);
} catch (Exception e) {
System.out.print("Exception occurred");
}
btn_signup = findViewById(R.id.btn_signup);
btn_login = findViewById(R.id.btn_login);
addData();
}
public void addData() {
btn_signup.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View v) {
inputEmail = findViewById(R.id.editTextEmail);
inputPass = findViewById(R.id.editTextPass);
//try {
myDB.insertData(inputEmail.getText().toString(),
inputPass.getText().toString());
Toast.makeText(MainActivity.this, "Data Inserted",
Toast.LENGTH_LONG).show();
// } catch (Exception e) {
// e.printStackTrace();
// }
}
});
}DatabaseHelper.java
public class DatabaseHelper extends SQLiteOpenHelper {
public static final String DATABASE_NAME = "Login_android.db";
public static final String TABLE_NAME = "login_table.db";
public static final String COL1 = "ID";
public static final String COL2 = "EMAIL";
public static final String COL3 = "PASSWORD";
public DatabaseHelper(Context context) {
super(context, DATABASE_NAME, null, 1);
//SQLiteDatabase db = this.getWritableDatabase();
}
@Override
public void onCreate(SQLiteDatabase db) {
db.execSQL("CREATE TABLE " + TABLE_NAME +" (ID INTEGER PRIMARY KEY
AUTOINCREMENT, EMAIL TEXT, PASSWORD TEXT)");
}
@Override
public void onUpgrade(SQLiteDatabase db, int oldVersion, int newVersion) {
db.execSQL("DROP TABLE IF EXISTS " + TABLE_NAME);
onCreate(db);
}
public boolean insertData(String email, String password) {
SQLiteDatabase db = null;
try {
db = this.getWritableDatabase();
} catch (Exception e) {
e.printStackTrace();
}
//db.isOpen();
ContentValues contentValues = new ContentValues();
contentValues.put(COL2, email);
contentValues.put(COL3, password);
long result = db.insert(TABLE_NAME, null, contentValues);
if (result == -1) {
return false;
} else {
return true;
}
}
}异常
12-06 23:39:12.799 26304 -26304/com.example.ryan.timearyname E/AndroidRuntime:致命异常:主进程: com.example.ryan.temporaryname,PID: 26304 java.lang.NullPointerException:尝试调用虚拟方法的long java.lang.NullPointerException java.lang.String,com.example.ryan.temporaryname.DatabaseHelper.insertData(DatabaseHelper.java:45) at com.example.ryan.temporaryname.MainActivity$1.onClick(MainActivity.java:64) at android.view.View.performClick(View.java:6294) at android.view.View$PerformClick.run(View.java:24770) at android.os.Handler.handleCallback(Handler.java:790)上的空对象引用在android.os.Handler.dispatchMessage(Handler.java:99) at android.os.Looper.loop(Looper.java:164) at android.app.ActivityThread.main(ActivityThread.java:6494) at java.lang.reflect.Method.invoke(原生方法)在com.android.internal.os.RuntimeInit$MethodAndArgsCaller.run(RuntimeInit.java:438) at com.android.internal.os.ZygoteInit.main(ZygoteInit.java:807)
发布于 2018-12-07 10:57:15
我认为问题在于TABLE_NAME -- "login_table.db“指的是名为"login_table”的数据库中一个名为"db“的表,该表不存在,并且返回一个null数据库对象。
请试着改名字。
发布于 2018-12-07 04:46:39
我认为您应该(如果这不是一个测试sqlite应用程序)通过创建App()类并在there.Try中初始化数据库来创建一个单例,以解决您的问题
https://stackoverflow.com/questions/53663291
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