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我正在使用 Objectbox (1.3.0) 在 Flutter 上构建我的数据库。

我尝试创建一个由自定义类型（枚举）组成的实体，如下所示：

type_enum.dart

/// Type Enumeration.
enum TypeEnum { one, two, three }

方法.dart

@Entity()
class Method {
  /// ObjectBox 64-bit integer ID property, mandatory.
  int id = 0;

  /// Custom Type. 
  TypeEnum type;

  /// Constructor.
  Method(this.type);

  /// Define a field with a supported type, that is backed by the state field.
  int get dbType {
    _ensureStableEnumValues();
    return type.index;
  }

  /// Setter of Custom type. Throws a RangeError if not found.
  set dbType(int value) {
    _ensureStableEnumValues();
    type = TypeEnum.values[value];
  }

  void _ensureStableEnumValues() {
    assert(TypeEnum.one.index == 0);
    assert(TypeEnum.two.index == 1);
    assert(TypeEnum.three.index == 2);
  }
}

前面的代码导致此错误（运行此命令后dart run build_runner build：


[WARNING] objectbox_generator:resolver on lib/entity/method.dart:
  skipping property 'type' in entity 'Method', as it has an unsupported type: 'TypeEnum'
[WARNING] objectbox_generator:generator on lib/$lib$:
Creating model: lib/objectbox-model.json
[SEVERE] objectbox_generator:generator on lib/$lib$:

Cannot use the default constructor of 'Method': don't know how to initialize param method - no such property.

我想通过构造函数参数给定类型来构造方法。怎么了？

如果我删除构造函数，我应该在类型字段前面添加后期标识符。我不想这样做。可能吧，我不明白。我没有找到任何例子。

我的解决方案：

方法.dart

@Entity()
class Method {
  /// ObjectBox 64-bit integer ID property, mandatory.
  int id = 0;

  /// Custom Type. 
  late TypeEnum type;

  /// Constructor.
  Method(int dbType){
     this.dbType = dbType;
  }

  /// Define a field with a supported type, that is backed by the state field.
  int get dbType {
    _ensureStableEnumValues();
    return type.index;
  }

  /// Setter of Custom type. Throws a RangeError if not found.
  set dbType(int value) {
    _ensureStableEnumValues();
    type = TypeEnum.values[value];
  }
}
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1 回答 1

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要使 ObjectBox 能够构造从数据库读取的对象，实体必须具有默认构造函数或具有与属性匹配的参数名称的构造函数。

例如，要在这种情况下获取默认构造函数，请将参数设为可选并提供默认值：

Method({this.type = TypeEnum.one});

或者添加一个默认构造函数并将需要类型的构造函数设为命名构造函数：

Method() : type = TypeEnum.one;
Method.type(this.type);
于 2022-01-25T07:10:58.703 回答