Question

我尝试做的只是过滤商店列表，将其存储在名为grapefruitStores的新变量对象中，然后用JavaScript将其打印到控制台。我允许使用filter，map和reduce。这就是我到目前为止所拥有的：

const stores = [{
   name: "Kents",
   foods: [
       {name: 'bagels', type: 'grain'},
       {name: 'bread', type: 'grain'},
       {name: 'cereal', type: 'grain'},
       {name: 'milk', type: 'dairy'},
   ]
   },{
   name: "Maceys",
   foods: [
       {name: 'bagels', type: 'grain'},
       {name: 'bread', type: 'grain'},
       {name: 'cereal', type: 'grain'},
       {name: 'grapefruit', type: 'fruit'},
       {name: 'milk', type: 'dairy'},
   ]
}];

//Filtering code

function sellsGrapefruit(stores) {
   return stores.foods.name === 'grapefruit'; //Don't think this is correct.
}

var grapefruitStores = store.filter(sellsGrapefruit);

console.log(grapefruitStores);

因此，只有Maceys的对象信息应该打印到控制台，因为Maceys出售葡萄柚而Kents不出售。一个空的数组/对象继续打印到屏幕，我不知道为什么。我在这做错了什么？任何帮助将不胜感激。

Answer 1

您可以使用Array.some()：

Caused by: org.jboss.weld.exceptions.IllegalProductException: WELD-000053: Producers cannot declare passivating scope and return a non-serializable class: Producer for Producer Method [FacesContext] with qualifiers [@Any @Default] declared as [[BackedAnnotatedMethod] @Produces @ViewScoped public br.com.dropper.web.factory.FacesContextFactory.getFacesContext()] declared on Managed Bean [class br.com.dropper.web.factory.FacesContextFactory] with qualifiers [@Any @Default]
    at br.com.dropper.web.factory.FacesContextFactory.getFacesContext(FacesContextFactory.java:16)

Answer 2

因此使用filter方法非常简单。这样做包括数组中的元素，具体取决于传递给它的函数的布尔返回值。你几乎和你发布的内容一样，但食物是一个对象数组，每个对象都有一个name属性，你当前正在调用商店的（不存在的）name属性。

因此，对于这个例子，通过工作迭代，一个可行的解决方案是：

var grapefruitStores = stores.filter(function(store) {
    for (var i = 0; i < store.foods.length; i++) {
        if (store.foods[i].name === 'grapefruit') {
            return true;
        }
    }
});

Answer 3

您的过滤器逻辑存在一些缺陷。您的算法正在检查foods数组的stores属性。它应该检查foods数组中每个元素的stores属性。试试这个：

//Filtering code
function sellsGrapefruit(store) {
  for (var i in store.foods) {
    var food = store.foods[i];
    if (food.name === 'grapefruit')
      return true;
  }
  return false;
}

这是一个有效的例子：

jsFiddle：https://jsfiddle.net/mspinks/8vzfc8uL/3/

Answer 4

Foods数组嵌套在stores数组中的对象中，因此您必须使用filter两次才能获得所需的结果：

const stores = [{
   name: "Kents",
   foods: [
       {name: 'bagels', type: 'grain'},
       {name: 'bread', type: 'grain'},
       {name: 'cereal', type: 'grain'},
       {name: 'milk', type: 'dairy'}
   ]
   },{
   name: "Maceys",
   foods: [
       {name: 'bagels', type: 'grain'},
       {name: 'bread', type: 'grain'},
       {name: 'cereal', type: 'grain'},
       {name: 'grapefruit', type: 'fruit'},
       {name: 'milk', type: 'dairy'}
   ]
}];

function sellsGrapefruit(store) {
    var x = store.foods.filter(function(i) {
        return i.name === 'grapefruit';
    });
    return x.length > 0;
}

var grapefruitStores = stores.filter(sellsGrapefruit); // typo error: store -> stores

console.log(grapefruitStores[0]);

Answer 5

由于您只能使用map，reduce，filter。我采用了这种方法。

var grapeStores = stores.filter(function (store) {
  return store.foods.filter(function (food) {
    return food.name === 'grapefruit';
  }).length;
});

console.log('grapeStores', grapeStores);

记录

grapeStores [ { name: 'Maceys',
    foods: [ [Object], [Object], [Object], [Object], [Object] ] } ]

JavaScript：过滤数组元素/对象

5 个答案: