Question

我对以下情况有疑问，因此，请检查以下对象数组：

 let arrayQualityRated = [{
     name: "Jande",
     col: 4
 }, {
     name: "Good",
     col: 4
 }, {
     name: "Good",
     col: 4
 }, {
     name: "Bad",
     col: 4
 }, {
     name: "Elmo",
     col: 2
 }, {
     name: "Bad",
     col: 2
 }, {
     name: "Tiago",
     col: 3
 }, {
     name: "Bad",
     col: 3
 }];

我想根据名称和 col 键进行类似的过滤。

但是我需要动态地进行操作，因此，据我了解，请根据以下数组查找键“ name”：

let persons = ["Jande", "Elmo", "Tiago"]

我希望我对自己的疑问已经清楚了。我指望您的耐心和帮助！：）

我希望这样的输出：

[   
    {
        name: "Jande",
        col: 4
    },
    {
        name: "Good",
        col: 4
    },
    {
        name: "Bad",
        col: 4
    },
    {
        name: "Good",
        col: 4
    }

],

[   
    {
        name: "Elmo",
        col: 2
    },
    {
        name: "Bad",
        col: 2
    }

],

[   
    {
        name: "Tiago",
        col: 3
    },
    {
        name: "Bad",
        col: 3
    }

]

简而言之，我想要一个基于字符串“ person name”（基于数组“ persons”）和“ col”的对象的单独数组。

Answer 1

使用array.filter

var nameArr = ...;
var col = ...;
var found = arrayQualityRated.filter(function() {
  return name.indexOf(this.name) != -1 && this.col === col;
});

Answer 2

我认为这不是一个简单的过滤器，因为预期结果会出现在三个不同的数组中。我会使用 map（），然后使用 filter（），因为 col 值是动态的（在识别出该人之前无法告诉按名称）。

看看结果数组-它具有您要的结构。

let arrayQualityRated = [{
  name: "Jande",
  col: 4
}, {
  name: "Good",
  col: 4
}, {
  name: "Good",
  col: 4
}, {
  name: "Bad",
  col: 4
}, {
  name: "Elmo",
  col: 2
}, {
  name: "Bad",
  col: 2
}, {
  name: "Tiago",
  col: 3
}, {
  name: "Bad",
  col: 3
}];

let persons = ["Jande", "Elmo", "Tiago"]

const classifyArrayItems = (persons, arrayQualityRated) => {
  // mapping the array, so it has all the persons
  return persons.map(person => {
    // first find the col number corresponding to the
    // person in the array
    const col = arrayQualityRated.find(e => e.name === person).col
    // return all the objects that have the same
    // col value
    return arrayQualityRated.filter(e => e.col === col)
  })
}

console.log(classifyArrayItems(persons, arrayQualityRated))

如何通过名称和列关联对象

2 个答案: