JSON.parse(str)以某種方式檢索了原始物件的__proto__非常具體的資訊，該資訊不在JSON字串(=str)中。如何？

在 JavaScript 中，JSON.parse(str);將 JSON 字串轉換為 JavaScript 物件。我發現奇特的是，盡管JSON字串不包含任何資訊__proto__的Object，JSON.parse(str)有時設法恢復的嵌套屬性__proto__屬性。

示例（直接在 Brave 瀏覽器的開發者工具控制臺中運行）：

{

    let user1 = {
      name: "John",
      surname: "Smith",
      isAdmin: false,
      "birthday": new Date(2000, 2, 3),
      "friends": [0,1,2,3]
    };

    user1.__proto__.age = 30;

    const str1 = JSON.stringify(user1, null, 2);

    user1.__proto__ = null;
    user1 = null;
    const user2 = JSON.parse(str1);

    console.log(str1);
    console.log(user2);

    const str2 = `{
      "name": "John",
      "surname": "Smith",
      "isAdmin": false,
      "birthday": "2000-03-02T15:00:00.000Z",
      "friends": [
        0,
        1,
        2,
        3
      ]
    }`;

    const user3 = JSON.parse(str2);

    console.log(user3);

}

在這里， user1 是用這樣的屬性創建的。然后在它的proto 上我分配了 age : 30 屬性。從 user1，我用 制作了它的 JSON 字串JSON.stringify，并將它保存在str1. 現在我洗掉了user1它的__proto__. 只是為了確保，我列印str1出來并沒有關于__proto__和 age:30 的資訊，正如預期的那樣。但是，當我user2從str1with創建時JSON.parse，__proto__的 age:30 恢復了。這怎么可能？現在我什至使用變數名 str2 制作完全相同的 JSON 字串，并使用 str2 創建另一個物件。結果是一樣的。這太瘋狂了。

console.log(str1) 的輸出；

{
  "name": "John",
  "surname": "Smith",
  "isAdmin": false,
  "birthday": "2000-03-02T15:00:00.000Z",
  "friends": [
    0,
    1,
    2,
    3
  ]
}

console.log(user2) 的輸出；

{name: 'John', surname: 'Smith', isAdmin: false, birthday: '2000-03-02T15:00:00.000Z', friends: Array(4)}
birthday: "2000-03-02T15:00:00.000Z"
friends: (4) [0, 1, 2, 3]
isAdmin: false
name: "John"
surname: "Smith"
[[Prototype]]: Object
age: 30
constructor: ? Object()
hasOwnProperty: ? hasOwnProperty()
isPrototypeOf: ? isPrototypeOf()
propertyIsEnumerable: ? propertyIsEnumerable()
toLocaleString: ? toLocaleString()
toString: ? toString()
valueOf: ? valueOf()
__defineGetter__: ? __defineGetter__()
__defineSetter__: ? __defineSetter__()
__lookupGetter__: ? __lookupGetter__()
__lookupSetter__: ? __lookupSetter__()
__proto__: (...)
get __proto__: ? __proto__()
set __proto__: ? __proto__()

console.log(user3) 的輸出；

{name: 'John', surname: 'Smith', isAdmin: false, birthday: '2000-03-02T15:00:00.000Z', friends: Array(4)}
birthday: "2000-03-02T15:00:00.000Z"
friends: (4) [0, 1, 2, 3]
isAdmin: false
name: "John"
surname: "Smith"
[[Prototype]]: Object
age: 30
constructor: ? Object()
hasOwnProperty: ? hasOwnProperty()
isPrototypeOf: ? isPrototypeOf()
propertyIsEnumerable: ? propertyIsEnumerable()
toLocaleString: ? toLocaleString()
toString: ? toString()
valueOf: ? valueOf()
__defineGetter__: ? __defineGetter__()
__defineSetter__: ? __defineSetter__()
__lookupGetter__: ? __lookupGetter__()
__lookupSetter__: ? __lookupSetter__()
__proto__: (...)
get __proto__: ? __proto__()
set __proto__: ? __proto__()

By the way I turned the browser off and on and reran the code block to make sure it wasn't a result of some pre assigned variables. The result was the same and it (age: 30) wasn't coming from pre assigned variables or cached value whatsoever.

Also I opened a fresh tab and executed the following smaller code block.

{
    const str2 = `{
      "name": "John",
      "surname": "Smith",
      "isAdmin": false,
      "birthday": "2000-03-02T15:00:00.000Z",
      "friends": [
        0,
        1,
        2,
        3
      ]
    }`;

    const user3 = JSON.parse(str2);

    console.log(user3);
}

In this case, age: 30 property in the proto is not there, which is the original expected result to have happened with first example, at least for me.

Output with smaller code block:

{name: 'John', surname: 'Smith', isAdmin: false, birthday: '2000-03-02T15:00:00.000Z', friends: Array(4)}
birthday: "2000-03-02T15:00:00.000Z"
friends: (4) [0, 1, 2, 3]
isAdmin: false
name: "John"
surname: "Smith"
[[Prototype]]: Object
constructor: ? Object()
hasOwnProperty: ? hasOwnProperty()
isPrototypeOf: ? isPrototypeOf()
propertyIsEnumerable: ? propertyIsEnumerable()
toLocaleString: ? toLocaleString()
toString: ? toString()
valueOf: ? valueOf()
__defineGetter__: ? __defineGetter__()
__defineSetter__: ? __defineSetter__()
__lookupGetter__: ? __lookupGetter__()
__lookupSetter__: ? __lookupSetter__()
__proto__: (...)
get __proto__: ? __proto__()
set __proto__: ? __proto__()

uj5u.com熱心網友回復：

所以看起來我誤用了原型。洗掉年齡的正確方法：30 是

任何一個

delete user1.__proto__.age;

或者

delete Object.prototype.age;

而這里user1=null無關緊要。

由于某種原因user1.__proto__ = null一事無成，如果有人暗示我為什么會很好

編輯：user1.__proto__ = null確實洗掉了 user1 的原型，但它不影響 Object.prototype。我認為它只會將 user1 從原型鏈中分離出來。

標籤：javascript json parsing prototype

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