Typescript JSON 字符串到类
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Typescript JSON string to class
提问by Ben
Let be this JSON string:
让这个 JSON 字符串：
[
{
"id": 1,
"text": "Jon Doe"
},
{
"id": 1,
"text": "Pablo Escobar"
}
]
Let be this class:
让这个班级：
export class MyObject{
id: number;
text: string;
}
How can I cast this JSON string to list of MyObject?
如何将此 JSON 字符串转换为列表MyObject？
If I do:
如果我做：
console.log(<MyObject[]>JSON.parse(json_string));
It returns a list of Objectinstead of a list of MyObject
它返回一个列表Object而不是一个列表MyObject
回答by Titian Cernicova-Dragomir
You don't necessarily need a class here. You can just use an interface
你不一定需要在这里上课。你可以只使用一个接口
export interface MyObject{
id: number;
text: string;
}
Then you can just write:
然后你可以写：
var myObjArray : MyObject[] = [
{
"id": 1,
"text": "Jon Doe"
},
{
"id": 1,
"text": "Pablo Escobar"
}
];
If you data comes from the server, you will probably have it in a variable of type any, and you can just assign it to an array of that type and it will work as expected.
如果您的数据来自服务器，您可能会将它放在 any 类型的变量中，您只需将其分配给该类型的数组，它就会按预期工作。
var data: any = getFromServer();
var myObjectArray:MyObject[] = data;
In typescript you don't need a class implementing an interface. Any object literal that satisfies the interface contract will do.
在打字稿中，您不需要实现接口的类。任何满足接口契约的对象文字都可以。
If your data is still in string for you can just use JSON.parse(jsonString)to parse the string to JavaScript objects.
如果您的数据仍然在字符串中，您可以使用JSON.parse(jsonString)将字符串解析为 JavaScript 对象。
See playground here
看这里的游乐场
回答by toskv
You will need to create a constructor for your class, and call it for each item in the list you receive.
您需要为您的类创建一个构造函数，并为您收到的列表中的每个项目调用它。
export class MyObject{
constructor(public id: number, public text: string) { }
}
let data = [
{
"id": 1,
"text": "Jon Doe"
},
{
"id": 1,
"text": "Pablo Escobar"
}
];
let objects = data.map(o => new MyObject(o.id, o.text));
You can check it out in the playground here.
回答by Anthony Brenelière
There is a problem when MyObject has 50 or more properties...
当 MyObject 有 50 个或更多属性时出现问题...
Add a constructor in your MyObject class so that it extends your json object.
在 MyObject 类中添加一个构造函数，以便它扩展您的 json 对象。
export class MyObject {
constructor( json: any )
{
$.extend(this, json);
}
id : number;
text : string;
methodOnMyObject() {...}
}
In your ajax callback, create the MyObject object from your json Object:
在您的 ajax 回调中，从您的 json 对象创建 MyObject 对象：
let newObject = new MyObject( json );
newObject.methodOnMyObject();
I detailed the solution in that post.
我在那篇文章中详细介绍了解决方案。
回答by OlegI
One more way to achieve this:
实现这一目标的另一种方法：
var data: any = getFromServer();
var myObjectArray = data as MyObject;
Or:
或者：
var data: any = getFromServer();
var myObjectArray = <MyObject>dataMyObject;
