我有一个带有键的对象
```js
const People = {
    groupA: ['Amy', 'Beth'],
    groupB: ['Ben', 'Matt'],
    groupC: ['Jen', 'Eli'],
};

const Person = Object.entries(People).find(([, values]) => values.includes('Amy'))[0];
console.log(Person); //输出 'groupA'

const Person1 = Object.entries(People).find(([, values]) => values.includes('Brad'))[0];
console.log(Person1); //输出应该是 undefined

<details>
<summary>英文:</summary>
I have an object with keys

Now I am declaring a varieable where I will pass the values dynamical. for example

But when In pass something thats not in the object, its throwing an undefined error. I was not able to understand where to put the if condition.

I tried putting if conditions but was not able to handle both scenerios.
</details>
# 答案1
**得分**: 0
尝试使用[Nullish 合并运算符][1]，这样你可以返回默认值，如果它是未定义的，你可以将其放在 '=&gt;' 后的返回语句中。
或者在整个Object.entries语句之后的变量中，这取决于实际返回的是未定义的什么。
  [1]: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Nullish_coalescing
<details>
<summary>英文:</summary>
Try using the [Nullish Coalescing Operator][1], this way you can returns. Default value if it is undefined, you would put this in the return statement after the ‘=&gt;’
Or after the entire Object.entries statement in the variable depending on what is actually returning undefined.
  [1]: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Nullish_coalescing
</details>
# 答案2
**得分**: 0
这是由于函数 `Array.prototype.find` 未找到条目，此函数在没有元素满足处理程序条件时返回 `undefined`。
另一种方法是使用 OR（`||`）运算符，如下所示，该方法将在函数 `find` 返回 `undefined` 时返回一个空数组。
在以下代码片段中，您可以看到其他备选方案。
```javascript
const Person = (Object.entries(People).find(([,values]) => values.includes('Amy')) || [])[0]

const People = {    groupA : ['Amy','Beth'],    groupB : ['Ben','Matt'],    groupC : ['Jen','Eli']};
//OR运算符
let Person = (Object.entries(People).find(([,values]) => values.includes('Ele')) || [])[0];
console.log(Person);
//可选链运算符 (?.)
Person = Object.entries(People).find(([,values]) => values.includes('Ele'))?.[0];
console.log(Person);
//空值合并运算符
Person = (Object.entries(People).find(([,values]) => values.includes('Ele')) ?? [])[0];
console.log(Person);