Question

假设

const someList = Immutable.fromJS([
  {'id': 2, foo: 'puppet'}, 
  {'id': 4, foo: 'kitten'}
]);

//Then I filter like so

const entry = someList.filter(elem => {
    return elem.get('id') === 4
});

console.log(entry);

<script src="https://cdnjs.cloudflare.com/ajax/libs/immutable/3.8.2/immutable.js"></script>

这给了我

{'id': 4, foo: 'kitten'}

如何编写过滤器函数，entry等于kitten，而无需附加行。

我尝试了

const entry = someList.filter(elem => {
    return elem.get('id') === 4
}).get('foo');

但是没有运气。

Answer 1

您可以使用Array#find而不是Array#filter找到该项目。

使用find，您将获得第一个找到的谓词在其上返回true的元素。

const entry = someList.find(elem => elem.get('id') === 4).get('foo');

如果不确定要查找元素，则还需要一个默认对象。

const entry = (someList.find(elem => elem.get('id') === 4) || {}).get('foo');

Answer 2

使用解构，find不filter：

const someList = [{'id':2,foo:'puppet'},{'id':4,foo:'kitten'}];
const { foo: entry } = someList.find(({ id }) => id == 4);
console.log(entry);

ImmutableJS版本：

const someList = Immutable.fromJS([
    {'id': 2, foo: 'puppet'}, 
    {'id': 4, foo: 'kitten'}
]);

const entry = someList.find(elem => elem.get('id') === 4).get("foo");

console.log(entry);

<script src="https://cdnjs.cloudflare.com/ajax/libs/immutable/3.8.2/immutable.js"></script>

Answer 3

您可以在过滤器后映射数组：

    
const someList = Immutable.fromJS([
  {'id': 2, foo: 'puppet'}, 
  {'id': 4, foo: 'kitten'}
]);

const entry = someList.filter(elem => {
  return elem.get('id') === 4
}).map(el => el.get('foo'));

console.log(entry);

<script src="https://cdnjs.cloudflare.com/ajax/libs/immutable/3.8.2/immutable.js"></script>

在一行中写一个嵌套的过滤器

3 个答案: