如何在 Javascript 的数组中返回仅包含某个 object 的对象数组？

Question

我有一个称为演员的 object 数组：

const actors =  [{
   "name": "John Doe",
   "rating": 1234,
   "alternative_name": null,
   "objectID": "551486300"
},
{
   "name": "Jane Doe",
   "rating": 4321,
   "alternative_name": "Monica Anna Maria Bellucci",
   "objectID": "551486310"
}];

我特别想知道演员的名字和收视率。 我尝试通过制作一个名为 actorNameRating 的 function 来获得它，但我的代码不起作用。

const nameAndRating = function() {

    const actorName = nameAndRating.filter(name);
    return actorName;

    const actorRating = nameAndRating.filter(rating);
    return actorRating;
 };

Answer 1

返回后的任何事情都不会执行，您可以将两个值作为一个字符串返回，然后将其分开，例如：

const nameAndRating = function() {
    const actorName = nameAndRating.filter(name);
    const actorRating = nameAndRating.filter(rating);
    return actorName+","+actorRating;
 };

或您认为更适合您使用的任何其他方式

Answer 2

您可以通过map和 function 并仅返回名称和评级：

 const actors = [{ "name": "John Doe", "rating": 1234, "alternative_name": null, "objectID": "551486300" }, { "name": "Jane Doe", "rating": 4321, "alternative_name": "Monica Anna Maria Bellucci", "objectID": "551486310" }]; const nameAndRating = function() { return actors.map(actor => ({ name: actor.name, rating: actor.rating })) }; console.log(nameAndRating())

Answer 3

您可以为此简单地使用Array.map ：

 const actors = [{ "name": "John Doe", "rating": 1234, "alternative_name": null, "objectID": "551486300" }, { "name": "Jane Doe", "rating": 4321, "alternative_name": "Monica Anna Maria Bellucci", "objectID": "551486310" }]; let result = actors.map(({name, rating}) => ({name, rating})); console.log(result);

或者，如果您希望它们在两个单独的 arrays 中，则可以使用通用 function 来返回给定道具的值数组：

 const actors = [{ "name": "John Doe", "rating": 1234, "alternative_name": null, "objectID": "551486300" }, { "name": "Jane Doe", "rating": 4321, "alternative_name": "Monica Anna Maria Bellucci", "objectID": "551486310" }]; let getPropValues = (arr, prop) => arr.map(o => o[prop]); console.log("Names = ",getPropValues(actors, "name")); console.log("Ratings = ",getPropValues(actors, "rating"));