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doox911
Asked:2022-02-05 17:13:40 +0000 UTC2022-02-05 17:13:40 +0000 UTC

如何正确创建字符串的 emun 并检查其中是否出现？

  • 772

随着API我得到状态的名称。从他们那里，在工厂的帮助下，我在前面形成了状态实例。附带API一根绳子。为了使代码中没有错误并且可以输入，我创建了enum状态名称：

export enum STATUSES_NAMES {
  CREATED = 'Создано',
  DELETED = 'Удалено',

  // Очень много имён статусов
}

工厂实例化状态：

// Импорты

export default class FactoryStatuses {
  public static createByName(name: STATUSES_NAMES): IStatus {
    const parameters = {
      name,
    };

    return FactoryStatuses._create(parameters);
  }

  public static createFromDB(parameters: IConstructorParameters): IStatus {
    const _parameters = {
      name: parameters.name,
    };

    return FactoryStatuses._create(_parameters);
  }

  private static _create(parameters: IConstructorParameters): IStatus {
    if (!FactoryStatuses._existsName(parameters.name)) {
      throw new Error('Status name dose not exist');
    }

    return new Status(parameters);
  }

  private static _existsName(name: string): boolean {
    return Object.values(STATUSES_NAMES).includes(name);
  }
}

接口：

export interface IStatus {
  name: STATUSES_NAMES;
}

export interface IConstructorParameters {
  name: string;
}

我在方法中existsName收到类型错误：

TS2345：“字符串”类型的参数不可分配给“状态名称”类型的参数。

如何解决这个问题或正确创建它enum？

typescript

2 个回答

  1. Best Answer

    一

    // @ts-ignore

    2

    name as any||name as STATUSES_NAMES

    3

    沙盒

    export enum STATUSES_NAMES {
    CREATED = 'Создано',
    DELETED = 'Удалено',
    // Очень много имён статусов
}

function _existsName(name: string): boolean {
    // return Object.values(STATUSES_NAMES).includes(name);
    return is_STATUSES_NAMES(name)
}

function is_STATUSES_NAMES(name: any): name is STATUSES_NAMES {
    return Object.values(STATUSES_NAMES).includes(name);
}


test: {
    function test(name: string) {
        if (is_STATUSES_NAMES(name)) {
            const t1 = name // const t1: STATUSES_NAMES
            Object.values(STATUSES_NAMES).includes(name); // is OK
        } else {
            const t2 = name // const t2: string
        }
    }
}

    四

    您还可以提出主题的变体：

        enum names {
        CREATED = 'Создано',
        DELETED = 'Удалено',
        // Очень много имён статусов
    }
    class Name extends String {
        constructor(name: names) { super(name) }
    }

    function _existsName(name: Name | string): boolean {
        return name instanceof Name
    }
    const _name = new Name(names.CREATED)
    console.log(_name)
    console.log(_existsName(_name))

    5 但它会是正确的

        enum names {
        CREATED,
        DELETED,
        // Очень много имён статусов
    }

    function ch(name: names | string): boolean {
        return name in names
    }
    console.log(ch(names.CREATED))
    console.log(ch('CREATED'))

    比较：

    namespace A {
    enum names {
        CREATED = 'Создано',
        DELETED = 'Удалено',
    }
}
namespace B {
    enum names {
        CREATED,
        DELETED,
    }
}


    var A;
(function (A) {
    let names;
    (function (names) {
        names["CREATED"] = "\u0421\u043E\u0437\u0434\u0430\u043D\u043E";
        names["DELETED"] = "\u0423\u0434\u0430\u043B\u0435\u043D\u043E";
    })(names || (names = {}));
})(A || (A = {}));
var B;
(function (B) {
    let names;
    (function (names) {
        names[names["CREATED"] = 0] = "CREATED";
        names[names["DELETED"] = 1] = "DELETED";
    })(names || (names = {}));
})(B || (B = {}));
    • 2

  2. 您可以通过添加关联来消除错误enum：

    private static _existsName(name: string): boolean {
  return Object.values(STATUSES_NAMES).includes(name as STATUSES_NAMES);
}

    类似的问题：https ://stackoverflow.com/ 。

    PS 一个 看起来更像拐杖的解决方案，但我没有找到任何其他更好的选择。

    • 0

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