blocks|key|328478|text|使用接口获取强类型：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|328479|//+Creating+
var+foo:any+=+{};
foo.x+=+3;
foo.y='123';

var+jsonString+=+JSON.stringify(foo);
alert(jsonString);


//+Reading
interface+Bar{
++++x:number;
++++y?:string;+
}

var+baz:Bar+=+JSON.parse(jsonString);
alert(baz.y);|code-block|syntax|javascript|328480|如果需要，可以使用类型断言"<>“。|328481|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Use Interfaces to get strong types:

<pre><code>// Creating 
var foo:any = {};
foo.x = 3;
foo.y='123';

var jsonString = JSON.stringify(foo);
alert(jsonString);


// Reading
interface Bar{
 x:number;
 y?:string; 
}

var baz:Bar = JSON.parse(jsonString);
alert(baz.y);
</code></pre>

And use type assertion "&lt;>" if you need to.

blocks|key|111093|text|我认为处理这个问题的更好的方法是使用Object.assign+(但是需要ECMAScript+2015)。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|111094|给定一个类|111095|class+Pet+{
++++name:+string;
++++age:+number;
++++constructor(name?:+string,+age?:+number)+{
++++++++this.name+=+name;
++++++++this.age+=+age;
++++}
++++getDescription():+string+{
++++++++return+"My+pet+"+%2B+this.name+%2B+"+is+"+%2B+this.age+%2B+"+years+old.";
++++}
++++static+fromJSON(d:+Object):+Pet+{
++++++++return+Object.assign(new+Pet(),+d);
++++}
}|code-block|syntax|javascript|111096|像这样序列化和反序列化...|111097|var+p0+=+new+Pet("Fido",+5);
var+s+=+JSON.stringify(p0);
var+p1+=+Pet.fromJSON(JSON.parse(s));
console.log(p1.getDescription());|111098|要将此示例带到下一个级别，请考虑嵌套对象...|111099|class+Type+{
++++kind:+string;
++++breed:+string;
++++constructor(kind?:+string,+breed?:+string)+{
++++++++this.kind+=+kind;
++++++++this.breed+=+breed;
++++}
++++static+fromJSON(d:+Object)+{
++++++++return+Object.assign(new+Type(),+d);
++++}
}
class+Pet+{
++++name:+string;
++++age:+number;
++++type:+Type;
++++constructor(name?:+string,+age?:+number)+{
++++++++this.name+=+name;
++++++++this.age+=+age;
++++}
++++getDescription():+string+{
++++++++return+"My+pet+"+%2B+this.name+%2B+"+is+"+%2B+this.age+%2B+"+years+old.";
++++}
++++getFullDescription():+string+{
++++++++return+"My+"+%2B+this.type.kind+%2B+",+a+"+%2B+this.type.breed+%2B+",+is+"+%2B+this.age+%2B+"+years+old.";
++++}
++++static+fromJSON(d:+Object):+Pet+{
++++++++var+o+=+Object.assign(new+Pet(),+d);
++++++++o.type+=+Type.fromJSON(o['type']);
++++++++return+o;
++++}
}|111100|111101|var+q0+=+new+Pet("Fido",+5);
q0.type+=+new+Type("dog",+"Pomeranian");
var+t+=+JSON.stringify(q0);
var+q1+=+Pet.fromJSON(JSON.parse(t));
console.log(q1.getFullDescription());|111102|因此，与使用接口不同，这种方法保留了方法。|111103|entityMap^0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|X|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|Y|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|Z|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|10|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|11|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|12|8|@]|9|@]|A|$]]|$1|O|3|P|5|F|7|13|8|@]|9|@]|A|$G|H]]|$1|Q|3|J|5|6|7|14|8|@]|9|@]|A|$]]|$1|R|3|S|5|F|7|15|8|@]|9|@]|A|$G|H]]|$1|T|3|U|5|6|7|16|8|@]|9|@]|A|$]]|$1|V|3|-4|5|6|7|17|8|@]|9|@]|A|$]]]|W|$]]

I think a better way to handle this is to use Object.assign (which however requires ECMAScript 2015).

Given a class

<pre><code>class Pet {
 name: string;
 age: number;
 constructor(name?: string, age?: number) {
 this.name = name;
 this.age = age;
 }
 getDescription(): string {
 return "My pet " + this.name + " is " + this.age + " years old.";
 }
 static fromJSON(d: Object): Pet {
 return Object.assign(new Pet(), d);
 }
}
</code></pre>

Serialize and deserialize like this...

<pre><code>var p0 = new Pet("Fido", 5);
var s = JSON.stringify(p0);
var p1 = Pet.fromJSON(JSON.parse(s));
console.log(p1.getDescription());
</code></pre>

To take this example to the next level, consider nested objects...

<pre><code>class Type {
 kind: string;
 breed: string;
 constructor(kind?: string, breed?: string) {
 this.kind = kind;
 this.breed = breed;
 }
 static fromJSON(d: Object) {
 return Object.assign(new Type(), d);
 }
}
class Pet {
 name: string;
 age: number;
 type: Type;
 constructor(name?: string, age?: number) {
 this.name = name;
 this.age = age;
 }
 getDescription(): string {
 return "My pet " + this.name + " is " + this.age + " years old.";
 }
 getFullDescription(): string {
 return "My " + this.type.kind + ", a " + this.type.breed + ", is " + this.age + " years old.";
 }
 static fromJSON(d: Object): Pet {
 var o = Object.assign(new Pet(), d);
 o.type = Type.fromJSON(o['type']);
 return o;
 }
}
</code></pre>

Serialize and deserialize like this...

<pre><code>var q0 = new Pet("Fido", 5);
q0.type = new Type("dog", "Pomeranian");
var t = JSON.stringify(q0);
var q1 = Pet.fromJSON(JSON.parse(t));
console.log(q1.getFullDescription());
</code></pre>

So unlike using an interface, this approach preserves methods.

blocks|key|328543|text|AQuirky+answer适用于我。您可能在使用Object.assign方法时遇到了一些问题。我不得不修改我的tsconfig.json以包含：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|328544|"compilerOptions":+{
++++...
++++"lib":+["es2015"],
++++...
}|code-block|syntax|javascript|328545|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/52237888/4629842^0|0|E|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

The <a href="https://stackoverflow.com/a/52237888/4629842">AQuirky answer</a> works for me.
You may have some troubles with the Object.assign method. I had to modify my tsconfig.json to include:

<pre><code>"compilerOptions": {
 ...
 "lib": ["es2015"],
 ...
}
</code></pre>

blocks|key|328599|text|首先，您需要为您的源实体创建一个接口，该接口作为JSON从API接收：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|328600|interface+UserEntity+{
++name:+string,
++age:+number,
++country_code:+string+
};|code-block|syntax|javascript|328601|第二，用构造器实现你的模型，你可以定制(驼峰化)一些字段名：|328602|++class+User++{+
++++constructor({+name,+age,+country_code:+countryCode+}:+UserEntity)+{
++++++Object.assign(this,+{+name,+age,+countryCode+});
++++}
++}|328603|最后使用JS+object+jsonUser创建用户模型的实例“|328604|++++const+jsonUser+=+{name:+'Ted',+age:+2,+country_code:+'US'};

++++const+userInstance+=+new+User(jsonUser);
++++
++++console.log({+userInstance+})|328605|​|328606|📷|atomic|offset|length|328607|328608|Here+is+the+link+to+playground|328609|entityMap|0|IMAGE|mutability|IMMUTABLE|imageUrl|https://ask.qcloudimg.com/http-save/yehe-900000/9d34f48091e7a50128ea51fae35b0872.png|imageAlt|1|LINK|MUTABLE|url|https://www.typescriptlang.org/play?#code/JYOwLgpgTgZghgYwgAgKoGdoFFzDAT2QG8BYAKGWRDgFsIAuZdMKUAcwBpzK42GqArjQBG0LhWQIA9gPBR8AfWkATfs1Yg2ycgF8A3OXIIANnHTo0mKJSLaJlaSHUCEYKVAAUt6nQ7JeEH7SsiyKKvzBcvgAwlKqyDqMGNi4BACUxNyU2dkA8sIAVhCuAHRm6MBsIB5gABbA6H7etIH%2520fEEyUbHxOmkG9gm6hmSOzMgF6FIgydYAvMQ%2520-ADkACoQysgAMnAADsDKAuhLfgGMAEwdIfJKccuoAMpL%2520sOjYMiH0ACSTmBwIEjIeYgCAAd0s0A8EymMz6wyMU0mxggJWMUjYXg%2520UG%2520zD%2520SF65CAA^0|0|0|0|0|0|0|0|0|1|0|0|0|0|U|1|0^^$0|@$1|2|3|4|5|6|7|1C|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|1D|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|1E|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|1F|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|1G|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|1H|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|1I|8|@]|9|@]|A|$]]|$1|Q|3|R|5|S|7|1J|8|@]|9|@$T|1K|U|1L|1|1M]]|A|$]]|$1|V|3|P|5|6|7|1N|8|@]|9|@]|A|$]]|$1|W|3|X|5|6|7|1O|8|@]|9|@$T|1P|U|1Q|1|1R]]|A|$]]|$1|Y|3|-4|5|6|7|1S|8|@]|9|@]|A|$]]]|Z|$10|$5|11|12|13|A|$14|15|16|-4]]|17|$5|18|12|19|A|$1A|1B]]]]

First, you need to create an interface of your source entity which you receive from the API as JSON:
<pre><code>interface UserEntity {
 name: string,
 age: number,
 country_code: string 
};
</code></pre>
Second implement your model with constructor where you can customize (camelize) some field names:
<pre><code> class User { 
 constructor({ name, age, country_code: countryCode }: UserEntity) {
 Object.assign(this, { name, age, countryCode });
 }
 }
</code></pre>
Last create instance of your User model using JS object jsonUser&quot;
<pre><code> const jsonUser = {name: 'Ted', age: 2, country_code: 'US'};

 const userInstance = new User(jsonUser);
 
 console.log({ userInstance })
</code></pre>
<a href="https://i.stack.imgur.com/NdYVY.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NdYVY.png" alt="Result log" /></a>
<a href="https://www.typescriptlang.org/play?#code/JYOwLgpgTgZghgYwgAgKoGdoFFzDAT2QG8BYAKGWRDgFsIAuZdMKUAcwBpzK42GqArjQBG0LhWQIA9gPBR8AfWkATfs1Yg2ycgF8A3OXIIANnHTo0mKJSLaJlaSHUCEYKVAAUt6nQ7JeEH7SsiyKKvzBcvgAwlKqyDqMGNi4BACUxNyU2dkA8sIAVhCuAHRm6MBsIB5gABbA6H7etIH%20fEEyUbHxOmkG9gm6hmSOzMgF6FIgydYAvMQ%20-ADkACoQysgAMnAADsDKAuhLfgGMAEwdIfJKccuoAMpL%20sOjYMiH0ACSTmBwIEjIeYgCAAd0s0A8EymMz6wyMU0mxggJWMUjYXg%20UG%20zD%20SF65CAA" rel="nofollow noreferrer">Here is the link to playground</a>

Are there any means for JSON serialization/deserialization of Typescript objects so that they don't loose type information? Simple <code>JSON.parse(JSON.stringify)</code> has too many caveats.

Or I should use adhoc solutions?

Typescript objects serialization?

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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有没有办法对Typescript对象进行JSON序列化/反序列化，这样它们就不会丢失类型信息？简单的JSON.parse(JSON.stringify)有太多的警告。或者我应该使用即席解决方案？

问Typescript对象序列化？
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问Typescript对象序列化？
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