在Typescript中声明并初始化一个Dictionary

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:此后已在最新的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).我在VS中遇到这些错误,正如您所期望的那样:Index signatures are incompatible. Type '{ firstName: string; }' is not assignable to type 'IPerson'. Property 'lastName' is missing in type '{ firstName: string; }'.

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.

您可以通过在声明和初始化中拆分示例来使用类型化字典,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.此实现更长,但它添加了其他功能,如a 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()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typescript中使用字典对象,您可以使用如下界面:

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

并将其用于您的类属性类型.

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

使用和初始化这个类,

getUsers(): Observable<any> {
        var searchParams = new SearchParameters();
        searchParams.SearchFor['userId'] = '1';
        searchParams.SearchFor['userName'] = 'xyz';

        return this.http.post(searchParams, 'users/search')
            .map(res => {
                return res;
            })
            .catch(this.handleError.bind(this));
    }

Answer 4

Typescript 在您的情况下失败，因为它期望所有字段都存在。使用Record 和 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" }
};

解释。

1. 记录类型创建一个字典/哈希图。
2. 部分类型表示某些字段可能丢失。

备用。

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

lastName?: string;

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

Answer 5

Record<Tkey, Tobject> 有点像 C# 字典

\n

let myRecord: Record<string, number> = {}; \n\n//Add\nmyRecord[\xe2\x80\x9dkey1\xe2\x80\x9d] = 1;\n\n//Remove\ndelete myRecord[\xe2\x80\x9dkey1"];\n\n//Loop\nfor (var key in myRecord) {\n    var value = myRecord[key];\n}\n

\n

Answer 6

这是一个更通用的字典实现，灵感来自@dmck

    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

如果您正在寻找一种即使在打字稿中也能创建字典的简单方法，那就是使用 Map 对象。文档链接https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Map。Map对象有添加、检索、删除和移除所有元素的主要方法。

dictionary= new Map<string, string>();
dictionary.set("key", "value");
dictionary.get("key");
dictionary.delete("key");
dictionary.clear(); //Removes all key-value pairs

Answer 8

如果您想忽略某个属性，请通过添加问号将其标记为可选：

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