Sqlite Insert Zwei Tabelle - Android

Question

Ich versuche, zwei Tabellen gleichzeitig in Android hinzuzufügen. Ich versuche wie 2 Stunden, es zu tun, aber ich bin jetzt verloren. Ich habe 2 Tische wie gesagt. Sie sind Körper- und Personentabellen. Person Tabelle funktioniert gut, aber Körper nicht. Der Körper hat einen Fremdschlüssel, der auf die PK der Person verweist. Diese

ist, wo im verloren.

public class DatabaseHandler extends SQLiteOpenHelper { 
private static final int DATABASE_VERSION = 1; 

// Database Name 
private static final String DATABASE_NAME = "healthMonitor.db"; 

// Contacts table name 
private static final String TABLE_NAME = "Person"; 
private static final String TABLE_NAME2 = "Body"; 
// Contacts Table Columns names 
private static final String KEY_ID = "id"; 
private static final String BODY_KEY_ID = "bId"; 
private static final String FOREIGN_KEY_ID = "personId"; 
private static final String KEY_NAME = "name"; 
private static final String KEY_ARM = "arm"; 
private static final String KEY_BRAIN = "brain"; 
private static final String KEY_EYE = "eye"; 
private static final String KEY_HEART = "heart"; 
private static final String KEY_LEG = "leg"; 
private static final String KEY_SURNAME = "surname"; 
private static final String KEY_AGE = "age"; 
private static final String KEY_EYECOLOR = "eyeColor"; 

public DatabaseHandler(Context context) { 
    super(context, DATABASE_NAME , null, DATABASE_VERSION); 
} 

public boolean isExternalStorageWritable() { 
    String state = Environment.getExternalStorageState(); 
    if (Environment.MEDIA_MOUNTED.equals(state)) { 
     return true; 
    } 
    return false; 
} 

private String createFirstTable(){ 
    String query = "CREATE TABLE " + TABLE_NAME + " (" + 
      KEY_ID + " INTEGER PRIMARY KEY AUTOINCREMENT, " + 
      KEY_NAME + " TEXT, " + 
      KEY_SURNAME + " TEXT, "+ 
      KEY_AGE + " INTEGER, " + 
      KEY_EYECOLOR + " TEXT);"; 

    return query; 
} 

private String createSecondTable(){ 
    String query = "CREATE TABLE " + TABLE_NAME2 + " (" + 
      BODY_KEY_ID + " INTEGER PRIMARY KEY AUTOINCREMENT, " + 
      KEY_ARM + " INTEGER, " + 
      KEY_BRAIN + " INTEGER, "+ 
      KEY_EYE + " INTEGER, " + 
      KEY_HEART + " INTEGER, "+ 
      KEY_LEG +" INTEGER, " + 
      FOREIGN_KEY_ID + " INTEGER, "+ 
      FOREIGN_KEY_ID + " FOREIGN KEY REFERENCES "+TABLE_NAME+ "("+KEY_ID+");"; 

    return query; 
} 

// Creating Tables 
@Override 
public void onCreate(SQLiteDatabase db) { 
    db.execSQL(createFirstTable()); 
    db.execSQL(createSecondTable()); 
} 

// Upgrading database 
@Override 
public void onUpgrade(SQLiteDatabase db, int oldVersion, int newVersion) { 
    // Drop older table if existed 
    db.execSQL("DROP TABLE IF EXISTS " + TABLE_NAME + "," + TABLE_NAME2); 
    // Create tables again 
    onCreate(db); 
} 

boolean addContact(MainActivity.Person person) { 
    SQLiteDatabase db = this.getWritableDatabase(); 

    ContentValues values = new ContentValues(); 
    values.put(KEY_NAME, person.getName()); // Person Name 
    values.put(KEY_SURNAME, person.getSurname()); // Person Surname 
    values.put(KEY_AGE, person.getAge()); // Person Age 
    values.put(KEY_EYECOLOR, person.getEyeColor()); // Person Surname 

    // Inserting Row 
    db.insert(TABLE_NAME, null, values); 
    //db.close(); // Closing database connection 
    return true; 
} 

boolean addBody() { 
    SQLiteDatabase db = this.getWritableDatabase(); 

    ContentValues values = new ContentValues(); 
    values.put(KEY_ARM, 5); // Person Name 
    values.put(KEY_BRAIN, 4); // Person Surname 
    values.put(KEY_EYE, 3); // Person Age 
    values.put(KEY_HEART, 2); // Person Surname 
    values.put(KEY_LEG, 1); // Person Age 
    values.put(FOREIGN_KEY_ID, 1); // Person Surname 

    // Inserting Row 
    db.insert(TABLE_NAME2, null, values); 
    //db.close(); // Closing database connection 
    return true; 
} 

Bitte überprüfen onCreate() Funktion. Ich bin wirklich verloren. Dank

"FOREIGN KEY (" + FOREIGN_KEY_ID + ") REFERENCES "+ + TABLE_NAME "(" + KEY_ID +")";

Und

FOREIGN_KEY_ID + "INTEGER Informationsquellen "+ TABLE_NAME + "(" + KEY_ID +");"

Dieser ist nicht funktioniert entweder

richtige

FOREIGN_KEY_ID + "INTEGER REFERENCES "+ TABLE_NAME + "(" + KEY_ID +"));";

Answer 1

verpassten Sie die schließende Klammer in Ihrem zweiten Abfrage

Code: FOREIGN_KEY_ID + " FOREIGN KEY REFERENCES "+TABLE_NAME+ "("+KEY_ID+");";

aktualisiert: FOREIGN_KEY_ID + " FOREIGN KEY REFERENCES "+TABLE_NAME+ "("+KEY_ID+"));";