Question

我有以下课程：

export class Son 
{
    public Field8_1 : number;
    public Field8_2 : number;
}

export class Father
{
    constructor (){
        this.Field8 = new Son;
    }
    public Field1 : number;
    public Field2 : number;
    public Field3 : number;
    public Field4 : number;
    public Field5 : number;
    public Field6 : number;
    public Field7 : number;

    Field8 : Son;
}

我必须使用Angular的HttpClient发送它。所以我跑了：

var body = JSON.stringify(this.Father);
    this.http.post (TGT_IP,body).subscribe (...)

我的尸体是：

{"Field8":{"Field8_1":81,"Field8_2":82},"Field1":1,"Field2":2,"Field3":3,"Field4":4,"Field5":5,"Field6":6,"Field7":7}

如何根据自己的需要“定制”字符串化以创建字符串：

"Field1":1,"Field2":2,"Field3":3,"Field4":4,"Field5":5,"Field6":6,"Field7","Field8_1":81, "Field8_2":82

我要发送“原始”字符串。

Answer 1

您可以通过两种方式实现这一目标。方法一是使用散布运算符和儿子映射父亲的所有键，然后删除不需要的字段：

// First option
let mappedFather1 = {
  ...this.father,
  Field8_1: this.father.Field8.Field8_1,
  Field8_2: this.father.Field8.Field8_2
};
delete mappedFather1.Field8;

另一种选择是从本质上“挑选”所需的对象键：

// Second option
let mappedFather2 = {
  Field1: this.father.Field1,
  Field2: this.father.Field2,
  Field3: this.father.Field3,
  Field4: this.father.Field4,
  Field5: this.father.Field5,
  Field6: this.father.Field6,
  Field7: this.father.Field7,
  Field8_1: this.father.Field8.Field8_1,
  Field8_2: this.father.Field8.Field8_2
}

如果不需要JSON的大括号，可以这样删除它们：

let noBracesFather = JSON.stringify(mappedFather1).substr(1).slice(0, -1);

这是一个堆栈闪电，将两个选项的输出记录到控制台，并删除括号：https://stackblitz.com/edit/angular-5wmz4c

显然，这假定父子对象的结构是已知的并且它们是静态的。

将嵌套类转换为“原始” json

1 个答案: