Question

给出以下代码

interface IPerson {
   firstName: string;
   lastName: string;
}

var persons: { [id: string]: IPerson; } = {
   "p1": { firstName: "F1", lastName: "L1" },
   "p2": { firstName: "F2" }
};

为什么不初始化被拒绝？毕竟，第二个对象没有“lastName”属性。

Answer 1

编辑：此版本已在最新的TS版本中修复。引用@ Simon_Weaver对OP帖子的评论：

注意：此后已经修复（不确定哪个TS版本）。一世如您所料，在VS中获取这些错误：Index signatures are incompatible. Type '{ firstName: string; }' is not assignable to type 'IPerson'. Property 'lastName' is missing in type '{ firstName: string; }'.

<小时/> 显然，在声明时传递初始数据时，这不起作用。我想这是TypeScript中的一个错误，所以你应该在项目站点提出一个错误。

您可以通过在声明和初始化中拆分示例来使用键入的字典，例如：

var persons: { [id: string] : IPerson; } = {};
persons["p1"] = { firstName: "F1", lastName: "L1" };
persons["p2"] = { firstName: "F2" }; // will result in an error

Answer 2

我同意thomaux初始化类型检查错误是TypeScript错误。但是，我仍然希望找到一种方法，在单个语句中使用正确的类型检查来声明和初始化Dictionary。此实现时间较长，但它添加了其他功能，例如containsKey(key: string)和remove(key: string)方法。我怀疑，一旦0.9版本中提供了泛型，这可以简化。

首先我们声明基本的Dictionary类和接口。索引器需要该接口，因为类无法实现它们。

interface IDictionary {
    add(key: string, value: any): void;
    remove(key: string): void;
    containsKey(key: string): bool;
    keys(): string[];
    values(): any[];
}

class Dictionary {

    _keys: string[] = new string[];
    _values: any[] = new any[];

    constructor(init: { key: string; value: any; }[]) {

        for (var x = 0; x < init.length; x++) {
            this[init[x].key] = init[x].value;
            this._keys.push(init[x].key);
            this._values.push(init[x].value);
        }
    }

    add(key: string, value: any) {
        this[key] = value;
        this._keys.push(key);
        this._values.push(value);
    }

    remove(key: string) {
        var index = this._keys.indexOf(key, 0);
        this._keys.splice(index, 1);
        this._values.splice(index, 1);

        delete this[key];
    }

    keys(): string[] {
        return this._keys;
    }

    values(): any[] {
        return this._values;
    }

    containsKey(key: string) {
        if (typeof this[key] === "undefined") {
            return false;
        }

        return true;
    }

    toLookup(): IDictionary {
        return this;
    }
}

现在我们声明Person特定类型和Dictionary / Dictionary接口。在PersonDictionary中注意我们如何覆盖values()和toLookup()以返回正确的类型。

interface IPerson {
    firstName: string;
    lastName: string;
}

interface IPersonDictionary extends IDictionary {
    [index: string]: IPerson;
    values(): IPerson[];
}

class PersonDictionary extends Dictionary {
    constructor(init: { key: string; value: IPerson; }[]) {
        super(init);
    }

    values(): IPerson[]{
        return this._values;
    }

    toLookup(): IPersonDictionary {
        return this;
    }
}

这是一个简单的初始化和使用示例：

var persons = new PersonDictionary([
    { key: "p1", value: { firstName: "F1", lastName: "L2" } },
    { key: "p2", value: { firstName: "F2", lastName: "L2" } },
    { key: "p3", value: { firstName: "F3", lastName: "L3" } }
]).toLookup();


alert(persons["p1"].firstName + " " + persons["p1"].lastName);
// alert: F1 L2

persons.remove("p2");

if (!persons.containsKey("p2")) {
    alert("Key no longer exists");
    // alert: Key no longer exists
}

alert(persons.keys().join(", "));
// alert: p1, p3

Answer 3

要在打字稿中使用字典对象，您可以使用如下界面：

[]

，并将其用于您的类属性类型。

interface Dictionary<T> {
    [Key: string]: T;
}

使用并初始化此类，

export class SearchParameters {
    SearchFor: Dictionary<string> = {};
}

Answer 4

在您的情况下，Typescript失败，因为它希望所有字段都存在。使用 Record and Partial 实用程序类型来解决它。

Record<string, Partial<IPerson>>

interface IPerson {
   firstName: string;
   lastName: string;
}

var persons: Record<string, Partial<IPerson>> = {
   "p1": { firstName: "F1", lastName: "L1" },
   "p2": { firstName: "F2" }
};

说明。

记录类型创建字典/哈希图。
部分类型表示某些字段可能会丢失。

备用。

如果您希望姓氏是可选的，可以附加一个？ Typescript知道它是可选的。

lastName?: string;

https://www.typescriptlang.org/docs/handbook/utility-types.html

Answer 5

如果要忽略属性，请通过添加问号将其标记为可选：

interface IPerson {
    firstName: string;
    lastName?: string;
}

Answer 6

这是来自@dmck的更通用的Dictionary实现

    interface IDictionary<T> {
      add(key: string, value: T): void;
      remove(key: string): void;
      containsKey(key: string): boolean;
      keys(): string[];
      values(): T[];
    }

    class Dictionary<T> implements IDictionary<T> {

      _keys: string[] = [];
      _values: T[] = [];

      constructor(init?: { key: string; value: T; }[]) {
        if (init) {
          for (var x = 0; x < init.length; x++) {
            this[init[x].key] = init[x].value;
            this._keys.push(init[x].key);
            this._values.push(init[x].value);
          }
        }
      }

      add(key: string, value: T) {
        this[key] = value;
        this._keys.push(key);
        this._values.push(value);
      }

      remove(key: string) {
        var index = this._keys.indexOf(key, 0);
        this._keys.splice(index, 1);
        this._values.splice(index, 1);

        delete this[key];
      }

      keys(): string[] {
        return this._keys;
      }

      values(): T[] {
        return this._values;
      }

      containsKey(key: string) {
        if (typeof this[key] === "undefined") {
          return false;
        }

        return true;
      }

      toLookup(): IDictionary<T> {
        return this;
      }
    }

Answer 7

现在，有一个库可以在打字稿中提供强类型的可查询集合。

集合是：

列表
字典

该库称为 ts-generic-collections 。

GitHub上的源代码：

https://github.com/VeritasSoftware/ts-generic-collections

使用此库，您可以创建集合（如List<T>）并按如下所示查询它们。

    let owners = new List<Owner>();

    let owner = new Owner();
    owner.id = 1;
    owner.name = "John Doe";
    owners.add(owner);

    owner = new Owner();
    owner.id = 2;
    owner.name = "Jane Doe";
    owners.add(owner);    

    let pets = new List<Pet>();

    let pet = new Pet();
    pet.ownerId = 2;
    pet.name = "Sam";
    pet.sex = Sex.M;

    pets.add(pet);

    pet = new Pet();
    pet.ownerId = 1;
    pet.name = "Jenny";
    pet.sex = Sex.F;

    pets.add(pet);

    //query to get owners by the sex/gender of their pets
    let ownersByPetSex = owners.join(pets, owner => owner.id, pet => pet.ownerId, (x, y) => new OwnerPet(x,y))
                               .groupBy(x => [x.pet.sex])
                               .select(x =>  new OwnersByPetSex(x.groups[0], x.list.select(x => x.owner)));

    expect(ownersByPetSex.toArray().length === 2).toBeTruthy();

    expect(ownersByPetSex.toArray()[0].sex == Sex.F).toBeTruthy();
    expect(ownersByPetSex.toArray()[0].owners.length === 1).toBeTruthy();
    expect(ownersByPetSex.toArray()[0].owners.toArray()[0].name == "John Doe").toBeTruthy();

    expect(ownersByPetSex.toArray()[1].sex == Sex.M).toBeTruthy();
    expect(ownersByPetSex.toArray()[1].owners.length == 1).toBeTruthy();
    expect(ownersByPetSex.toArray()[1].owners.toArray()[0].name == "Jane Doe").toBeTruthy();

在Typescript中声明并初始化一个Dictionary

7 个答案: