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将Json转换为对象_Json_Angular_Typescript - Fatal编程技术网

将Json转换为对象

将Json转换为对象,json,angular,typescript,Json,Angular,Typescript,在我的打字脚本代码中,我有两个类ClassA和ClassB: export class ClassA { name: string; classB: ClassB; getName(): string { return this.name; } } export class ClassB { name: string; getName(): string { return this.name; } }

在我的打字脚本代码中,我有两个类
ClassA
ClassB

export class ClassA {
    name: string;
    classB: ClassB;

    getName(): string {
        return this.name;
    }
}

export class ClassB {
    name: string;

    getName(): string {
        return this.name;
    }
}
我尝试将Json解析为
ClassA
实例,如下所示:

let classA: ClassA = Object.assign(new ClassA(), JSON.parse('{"name":"classA", "classB": {"name":"classB"}}'));
export interface ClassBData {
    name: string;
}

export interface ClassAData {
    name: string;
    classB: ClassBData;
}

export class ClassA {
    name: string;
    classB: ClassB;

    constructor(json: ClassAData) {
        this.name = json.name;
        this.classB = Object.assign(new ClassB(), json.classB);
    }

    getName(): string {
        return this.name;
    }
}
但是只有
ClassA
是实例化的,而在classB属性中不是

我在记录对象时得到以下结果:

console.log(classA); // ClassA {name: "classA", classB: Object}
console.log(classA.getName()); // classA
console.log(classA.classB); // Object {name: "classB"}
console.log(classA.classB.getName()); // EXCEPTION: classA.classB.getName is not a function

可以深入解析Json吗?

您需要执行以下操作:

let classA: ClassA = Object.assign(new ClassA(), JSON.parse('{"name":"classA", "classB": {"name":"classB"}}'));
export interface ClassBData {
    name: string;
}

export interface ClassAData {
    name: string;
    classB: ClassBData;
}

export class ClassA {
    name: string;
    classB: ClassB;

    constructor(json: ClassAData) {
        this.name = json.name;
        this.classB = Object.assign(new ClassB(), json.classB);
    }

    getName(): string {
        return this.name;
    }
}
或者您可以用相同的方法实例化
ClassB

export class ClassB {
    name: string;

    constructor(json: ClassBData) {
        this.name = json.name;
    }

    getName(): string {
        return this.name;
    }
}

到目前为止,我发现最好的方法是:

export class ClassA {
    name: string;
    classB: ClassB;

    constructor(json: any) {
        this.name = json.name;
        if (json.classB !== undefined && json.classB !== null) {
            this.classB = new ClassB(json.classB);
        }
    }
}

export class ClassB {
    name: string;

    constructor(json: any) {
        this.name = json.name;
    }
}
可能重复的