我从对REST服务器的AJAX调用中接收到一个JSON对象。这个对象的属性名与我的TypeScript类相匹配(这是这个问题的后续)。
初始化它的最佳方法是什么?我不认为这将工作,因为类(& JSON对象)的成员是对象的列表和成员是类,而这些类的成员是列表和/或类。
但我更喜欢一种方法,查找成员名和分配他们,创建列表和实例化类的需要,所以我不必为每个类中的每个成员写显式代码(有很多!)
我从对REST服务器的AJAX调用中接收到一个JSON对象。这个对象的属性名与我的TypeScript类相匹配(这是这个问题的后续)。
初始化它的最佳方法是什么?我不认为这将工作,因为类(& JSON对象)的成员是对象的列表和成员是类,而这些类的成员是列表和/或类。
但我更喜欢一种方法,查找成员名和分配他们,创建列表和实例化类的需要,所以我不必为每个类中的每个成员写显式代码(有很多!)
当前回答
这是我的方法(非常简单):
const jsonObj: { [key: string]: any } = JSON.parse(jsonStr);
for (const key in jsonObj) {
if (!jsonObj.hasOwnProperty(key)) {
continue;
}
console.log(key); // Key
console.log(jsonObj[key]); // Value
// Your logic...
}
其他回答
选项#5:使用Typescript构造函数和jQuery.extend
这似乎是最可维护的方法:添加一个以json结构作为参数的构造函数,并扩展json对象。这样就可以将json结构解析为整个应用程序模型。
不需要创建接口,或者在构造函数中列出属性。
export class Company
{
Employees : Employee[];
constructor( jsonData: any )
{
jQuery.extend( this, jsonData);
// apply the same principle to linked objects:
if ( jsonData.Employees )
this.Employees = jQuery.map( jsonData.Employees , (emp) => {
return new Employee ( emp ); });
}
calculateSalaries() : void { .... }
}
export class Employee
{
name: string;
salary: number;
city: string;
constructor( jsonData: any )
{
jQuery.extend( this, jsonData);
// case where your object's property does not match the json's:
this.city = jsonData.town;
}
}
在你的ajax回调中,你收到一个公司来计算工资:
onReceiveCompany( jsonCompany : any )
{
let newCompany = new Company( jsonCompany );
// call the methods on your newCompany object ...
newCompany.calculateSalaries()
}
我的方法略有不同。我没有将属性复制到新的实例中,我只是改变了现有pojo的原型(在旧的浏览器上可能不太好用)。每个类负责提供一个setprototype方法来设置任何子对象的原型,这些子对象反过来提供它们自己的setprototype方法。
(我也使用_Type属性来获取未知对象的类名,但在这里可以忽略)
class ParentClass
{
public ID?: Guid;
public Child?: ChildClass;
public ListOfChildren?: ChildClass[];
/**
* Set the prototypes of all objects in the graph.
* Used for recursive prototype assignment on a graph via ObjectUtils.SetPrototypeOf.
* @param pojo Plain object received from API/JSON to be given the class prototype.
*/
private static SetPrototypes(pojo: ParentClass): void
{
ObjectUtils.SetPrototypeOf(pojo.Child, ChildClass);
ObjectUtils.SetPrototypeOfAll(pojo.ListOfChildren, ChildClass);
}
}
class ChildClass
{
public ID?: Guid;
public GrandChild?: GrandChildClass;
/**
* Set the prototypes of all objects in the graph.
* Used for recursive prototype assignment on a graph via ObjectUtils.SetPrototypeOf.
* @param pojo Plain object received from API/JSON to be given the class prototype.
*/
private static SetPrototypes(pojo: ChildClass): void
{
ObjectUtils.SetPrototypeOf(pojo.GrandChild, GrandChildClass);
}
}
下面是ObjectUtils.ts:
/**
* ClassType lets us specify arguments as class variables.
* (where ClassType == window[ClassName])
*/
type ClassType = { new(...args: any[]): any; };
/**
* The name of a class as opposed to the class itself.
* (where ClassType == window[ClassName])
*/
type ClassName = string & {};
abstract class ObjectUtils
{
/**
* Set the prototype of an object to the specified class.
*
* Does nothing if source or type are null.
* Throws an exception if type is not a known class type.
*
* If type has the SetPrototypes method then that is called on the source
* to perform recursive prototype assignment on an object graph.
*
* SetPrototypes is declared private on types because it should only be called
* by this method. It does not (and must not) set the prototype of the object
* itself - only the protoypes of child properties, otherwise it would cause a
* loop. Thus a public method would be misleading and not useful on its own.
*
* https://stackoverflow.com/questions/9959727/proto-vs-prototype-in-javascript
*/
public static SetPrototypeOf(source: any, type: ClassType | ClassName): any
{
let classType = (typeof type === "string") ? window[type] : type;
if (!source || !classType)
{
return source;
}
// Guard/contract utility
ExGuard.IsValid(classType.prototype, "type", <any>type);
if ((<any>Object).setPrototypeOf)
{
(<any>Object).setPrototypeOf(source, classType.prototype);
}
else if (source.__proto__)
{
source.__proto__ = classType.prototype.__proto__;
}
if (typeof classType["SetPrototypes"] === "function")
{
classType["SetPrototypes"](source);
}
return source;
}
/**
* Set the prototype of a list of objects to the specified class.
*
* Throws an exception if type is not a known class type.
*/
public static SetPrototypeOfAll(source: any[], type: ClassType): void
{
if (!source)
{
return;
}
for (var i = 0; i < source.length; i++)
{
this.SetPrototypeOf(source[i], type);
}
}
}
用法:
let pojo = SomePlainOldJavascriptObjectReceivedViaAjax;
let parentObject = ObjectUtils.SetPrototypeOf(pojo, ParentClass);
// parentObject is now a proper ParentClass instance
你可以像下面这样做
export interface Instance {
id?:string;
name?:string;
type:string;
}
and
var instance: Instance = <Instance>({
id: null,
name: '',
type: ''
});
我已经创建了一个生成TypeScript接口和运行时“类型映射”的工具,用于对JSON的结果执行运行时类型检查。解析:ts.quicktype.io
例如,给定这个JSON:
{
"name": "David",
"pets": [
{
"name": "Smoochie",
"species": "rhino"
}
]
}
quicktype生成以下TypeScript接口和类型映射:
export interface Person {
name: string;
pets: Pet[];
}
export interface Pet {
name: string;
species: string;
}
const typeMap: any = {
Person: {
name: "string",
pets: array(object("Pet")),
},
Pet: {
name: "string",
species: "string",
},
};
然后检查JSON的结果。根据类型映射解析:
export function fromJson(json: string): Person {
return cast(JSON.parse(json), object("Person"));
}
我省略了一些代码,但您可以尝试快速输入细节。
上面描述的第4个选项是一种简单而漂亮的方法,它必须与第二个选项相结合,在这种情况下,您必须处理一个类层次结构,例如成员列表,它是成员超类的任何一个子类,例如Director extends member或Student extends member。在这种情况下,你必须以json格式给出子类类型