Question

错误：android.database.sqlite.SQLiteException：table用户没有名为name（code 1）的列:,编译时：INSERT INTO用户（名称，余额，密码，年龄）VALUES（？，？，？，？）

<input type="text" id="pesos" name="pesos" value="">

}

public class NewUserTable {

    int _id;
    String _name;
    String _password;
    int _balance;
    int _age;
    public NewUserTable(){

    }
    public NewUserTable( String name, String password, int balance, int age){
        this._name = name;
        this._password = password;
        this._balance = balance;
        this._age = age;
    }
    public NewUserTable( int id, String name, String password, int balance, int age){
        this._id = id;
        this._name = name;
        this._password = password;
        this._balance = balance;
        this._age = age;
    }
    public  int getID(){
        return this._id;
    }
    public void setID(int id){
        this._id = id;
    }
    public String getName(){
        return this._name;
    }
    public void setName(String name){
        this._name = name;
    }
    public String getPassword(){
        return this._password;
    }
    public void setPassword(String password){
        this._password = password;
    }
    public int getBalance(){
        return _balance;
    }
    public void setBalance(int balance){
        this._balance = balance;
    }
    public int getAge(){
        return _age;
    }
    public void setAge(int age){
        this._age = age;
    }
}
public class DatabaseHandler extends SQLiteOpenHelper {
//----------- TABLE COLUMNS -----------//
public static final String KEY_ID = "id"; // eash user has unique id
public static final String KEY_NAME = "name";
public static final String KEY_PASSWORD = "password";
public static final String KEY_BALANCE = "balance";
public static final String KEY_AGE ="age";
//----------- TABLE COLUMNS -----------//
private static final String DATABASE_NAME = "newUser";
private static final String TABLE_NAME = "User";
public DatabaseHandler(Context context){
    super(context, DATABASE_NAME, null, DATABASE_VERSION);
}
private static final int DATABASE_VERSION = 1;
@Override
public void onCreate(SQLiteDatabase db) {
String CREATE_NEWUSER_TABLE = "CREATE TABLE " + TABLE_NAME + "("
        + KEY_ID + "INTEGER PRIMARY KEY," + KEY_NAME +
        "TEXT," + KEY_PASSWORD + "TEXT," + KEY_BALANCE + "INTEGER,"
        + KEY_AGE + "INTEGER" + ")";
    db.execSQL(CREATE_NEWUSER_TABLE);
}

@Override
public void onUpgrade(SQLiteDatabase db, int oldVersion, int newVersion) {
db.execSQL("DROP TABLE IF EXISTS" + TABLE_NAME);
    onCreate(db);
}

public int getUsersCount() {
    String countQuery = "SELECT  * FROM " + TABLE_NAME;
    SQLiteDatabase db = this.getReadableDatabase();
    Cursor cursor = db.rawQuery(countQuery, null);
    cursor.close();
    // return count
    return cursor.getCount();
}
// добавить новую запись
public void addUser(NewUserTable newUserTable){
    SQLiteDatabase db = this.getWritableDatabase();
    ContentValues values = new ContentValues();
    values.put(KEY_NAME,newUserTable.getName());
    values.put(KEY_PASSWORD,newUserTable.getPassword());
    values.put(KEY_BALANCE,newUserTable.getBalance());
    values.put(KEY_AGE,newUserTable.getAge());
    db.insert(TABLE_NAME, null, values);
    db.close();
}
// считать записи
public NewUserTable getContact(int id) {
    SQLiteDatabase db = this.getReadableDatabase();

    Cursor cursor = db.query(TABLE_NAME, new String[] { KEY_ID,
                    KEY_NAME, KEY_PASSWORD, KEY_BALANCE, KEY_AGE }, KEY_ID + "=?",
            new String[] { String.valueOf(id) }, null, null, null, null);
    if (cursor != null)
        cursor.moveToFirst();

    NewUserTable newUserTable = new NewUserTable(Integer.parseInt(cursor.getString(0)),cursor.getString(1),cursor.getString(2), Integer.parseInt(cursor.getString(3)),
            Integer.parseInt(cursor.getString(4)));
    // return contact
    return newUserTable;
}

} 问题出在哪儿？我怎样才能看到数据库结果？

Answer 1

更改

String CREATE_NEWUSER_TABLE = "CREATE TABLE " + TABLE_NAME + "("
    + KEY_ID + "INTEGER PRIMARY KEY," + KEY_NAME +
    "TEXT," + KEY_PASSWORD + "TEXT," + KEY_BALANCE + "INTEGER,"
    + KEY_AGE + "INTEGER" + ")";

带

String CREATE_NEWUSER_TABLE = "CREATE TABLE " + TABLE_NAME + " ("
    + KEY_ID + " INTEGER PRIMARY KEY, " + KEY_NAME +
    " TEXT, " + KEY_PASSWORD + " TEXT, " + KEY_BALANCE + " INTEGER, "
    + KEY_AGE + " INTEGER" + ")";

您在创建表脚本中遗漏了一些空格。

Answer 2

String CREATE_NEWUSER_TABLE = "CREATE TABLE " + TABLE_NAME + "("
    + KEY_ID + "INTEGER PRIMARY KEY," + KEY_NAME +
    "TEXT," + KEY_PASSWORD + "TEXT," + KEY_BALANCE + "INTEGER,"
    + KEY_AGE + "INTEGER" + ")";

您需要在列名和列类型之间使用空格。

添加空格后，请卸载您的应用，以便再次调用您的SQLite帮助程序onCreate()。

在android studio中创建表

2 个答案: