1
我有一个Person类,并在设置其属性后,找出将该类转换为json对象的最佳方法。从Typscript类创建JSON对象
class Person {
firstName: string;
lastName: string;
}
let person = new Person();
person.firstName = "FirstName";
person.lastName = "LastName";
,如果我做person.getJson()也应该给JSON对象下面
{
"firstName": "FirstName",
"lastName": "LastName"
}
视为给定的柜面lastName的也不会设置JSON对象应该只有的firstName
{
"firstName": "FirstName"
}