Question

我已经在对象列表中加载了 json 数据。每当我尝试将数据插入表中时，我的控制台都会说“未处理的异常：DatabaseException（没有这样的表：相册（代码 1 SQLITE_ERROR）：，编译时：插入或替换到相册......”。我在插入数据之前知道我必须创建数据库和表。我已经在我的代码中完成了。但是由于某种未知的原因，该表没有被构建。这是我的数据库助手类 -

class DatabaseHelper {
  static final _dbName = "myDatabase.db";
  static final _dbVersion = 1;

  static final _tableAlbum = "Album";
  static final columnAlbum = '_fetched_album';



  DatabaseHelper._privateConstructor();

  static final DatabaseHelper instance = DatabaseHelper._privateConstructor();

  static Database? _database;

  Future<Database?> get database async {
    if (_database == null) {
    print("instance.db null");
    _database ??= await _initiateDatabase();
    return _database;
    }else{
    print("instance.db not null");
    return _database; }


     } 

    Future<Database> _initiateDatabase() async {
    String directory = await getDatabasesPath();
    String path = join(directory, _dbName);

    return await openDatabase(
    path,
    version: _dbVersion,
    onCreate:(Database db,int version) async {

    await db.execute('''
    CREATE TABLE $_tableAlbum (
    $columnAlbum TEXT
    )''');


    },
    );


    }




    Future<int?> insertAlbum(ModelAlbum list) async {
    Database? db = await instance.database;
    return await db?.insert(
    _tableAlbum,list.toMap(),conflictAlgorithm: ConflictAlgorithm.replace);
    }
    Future<int?> insertPhoto(ModelPhoto list) async {
    Database? db = await instance.database;
    return await db?.insert(
    _tablePhoto,list.toMap(),conflictAlgorithm: ConflictAlgorithm.replace);
    }


    Future<List<Map<String, dynamic>>> queryAlbum() async {
    Database? db = await instance.database;
    return await db!.query(_tableAlbum);
    }
    Future<List<Map<String, dynamic>>> queryPhoto() async {
    Database? db = await instance.database;
    return await db!.query(_tableAlbum);
    }

    Future<List<ModelAlbum>> retrieveAlbum() async {
    final Database? db = await database;
    final List<Map> maps = await db!.query(_tableAlbum);
    return List.generate(maps.length, (i) {
    return ModelAlbum(
    id: maps[i]['id'],
    title: maps[i]['title'],
    userId: maps[i]['userId'],
    );
    });
    }
    Future<List<ModelPhoto>> retrievePhoto() async {
    final Database? db = await database;
    final List<Map> maps = await db!.query(_tablePhoto);
    return List.generate(maps.length, (i) {
    return ModelPhoto(
    albumId: maps[i]['albumId'],
    thumbnailUrl: maps[i]['thumbnailUrl'],
    url: maps[i]['url'],
    id: maps[i]['id'],
    title: maps[i]['title'],
    );
    });
    } 
    }

每当 Database.instance.insert(object);被称为数据库和表应该自动创建。但它没有。

Answer 1

您指的是实例方法中的 static 字段。这可能是一个问题尝试从字段中删除静态关键字，

final _dbName = "myDatabase.db";
final _dbVersion = 1;

final _tableAlbum = "Album";
final columnAlbum = '_fetched_album';

Sqflite 数据库异常在颤振中没有这样的表

1 个答案: