Question

我正在使用PHP脚本填充我的SQLiteDatabase以获取登录凭据和帐户信息。即使登录工作正常，SQLite在我的性别名称列中也有错误我不知道我对此缺少什么。有人可以为我指出它吗？ Cu'z me和我的朋友找不到它，并且都是Android Development的新手。

这是错误

>     E/SQLiteDatabase﹕ Error inserting gender=tuesdaycross04@gmail.com bday=24242 email=Male address=asdasd mobile=1995-03-03
> full_name=Arnold
>         android.database.sqlite.SQLiteException: table accounts has no column named gender (code 1): , while compiling: INSERT INTO
> accounts(gender,bday,email,address,mobile,full_name) VALUES
> (?,?,?,?,?,?)

SQLiteHelper

    private static final int DATABASE_VERSION = 1;
        // Database Name
        private static final String DATABASE_NAME = "brm_dbs";
        // Login table name
        private static final String TABLE_LOGIN = "accounts";
        // Login Table Columns names
        private static final String KEY_ID = "id";
        private static final String KEY_NAME = "full_name";
        private static final String KEY_ADDRESS = "address";
        private static final String KEY_MOBILE = "mobile";
        private static final String KEY_EMAIL = "email";
        private static final String KEY_UID = "uid";
        private static final String KEY_CREATED_AT = "created_at";
        private static final String KEY_BDAY = "bday";
        private static final String KEY_GENDER = "gender";

        public SQLiteHandler(Context context) {
            super(context, DATABASE_NAME, null, DATABASE_VERSION);
        }
        // Creating Tables
        @Override
        public void onCreate(SQLiteDatabase db) {
            String CREATE_LOGIN_TABLE = "CREATE TABLE " + TABLE_LOGIN + "("
                    + KEY_ID + " INTEGER PRIMARY KEY," + KEY_NAME + " TEXT,"
                    + KEY_ADDRESS + " TEXT," + KEY_MOBILE + " INTEGER,"
                    + KEY_EMAIL + " TEXT UNIQUE," + KEY_UID + " TEXT,"
                    + KEY_CREATED_AT + " TEXT," + KEY_BDAY + " TEXT,"
                    + KEY_GENDER + " TEXT" + ");";
            db.execSQL(CREATE_LOGIN_TABLE);

            Log.d(TAG, "Database tables created");
        }

public void getloginUser(String full_name, String address,
                             String bday, String gender,
                             String mobile, String email) {
        SQLiteDatabase db = this.getWritableDatabase();

        ContentValues values = new ContentValues();
        //values.put(KEY_UNIQUE_ID, unique_id);
        values.put(KEY_NAME, full_name); // Name
        values.put(KEY_ADDRESS, address); // address
        values.put(KEY_BDAY, bday); // BDAY
        values.put(KEY_GENDER, gender); // GENDER
        values.put(KEY_MOBILE, mobile); // mobile
        values.put(KEY_EMAIL, email); // Email


        // Inserting Row
        long id = db.insert(TABLE_LOGIN, null, values);
        db.close(); // Closing database connection
        Log.d(TAG, "New user inserted into sqlite: " + id);
    }

通过登录凭据插入SQLite

try {
                    JSONObject jObj = new JSONObject(response);
                    boolean error = jObj.getBoolean("error");
                    // Check for error node in json
                    if (!error) {
                        // user successfully logged in
                        // Create login session
                        session.setLogin(true);
                        JSONObject user = jObj.getJSONObject("user");
                        String full_name = user.getString("full_name");
                        String address = user.getString("address");
                        String mobile = user.getString("mobile");
                        String email = user.getString("email");
                        String bday = user.getString("bday");
                        String gender = user.getString("gender");
                        // Inserting row in users table
                        db.getloginUser(full_name, address, mobile, email, bday, gender);
                        // Launch main activity

                        // Launch main activity
                        Intent intent = new Intent(Loign.this,
                                MainActivity.class);
                        startActivity(intent);
                        finish();

                    } else {
                        // Error in login. Get the error message
                        String errorMsg = jObj.getString("error_msg");
                        Toast.makeText(getApplicationContext(),
                                errorMsg, Toast.LENGTH_LONG).show();
                    }
                } catch (JSONException e) {
                    // JSON error
                    e.printStackTrace();
                }

Answer 1

应插入值，表格的创建顺序为......

Answer 2

打开DDMS并删除应用程序数据，否则擦除您的模拟器并运行您的应用程序，它将正常工作。该表已经存在，没有性别列，当你正在运行的应用程序时，因为DB退出然后它不会调用你的SQLiteOpenHelper的onCreate方法，那么在DB之后你做了什么改变它可能不会影响表的结构。

<强>解决方案：

从模拟器中卸载应用程序并尝试从Android工作室运行该应用程序，
将diff数据库版本设为2或以前的数据库版本+ 1。然后onUpgrade（）方法将调用编写代码来更改现有表的结构，或添加所需的任何表或根据您的要求删除表。

表名没有名称Android SQLite PHP的列

2 个答案: