如何从对象值中获取键？

Question

我正在尝试构建一个和弦字典，它以和弦的名称为键，并以一个数组作为值，其中包含组成和弦的 MIDI 数字（音符）。但是我遇到了一个问题，一旦我将midi数字数组作为函数getkey（）的输入，我就无法获得字典的键。我该怎么办？在此先感谢:)

 var dictionary = {
  "Cmaj7": [60,64,67,71]
};

const getKey = (obj,val) => Object.keys(obj).find(key => obj[key] === val);

console.log(getKey(dictionary,[60,64,67,71]));

Answer 1

var dictionary = {
  "Cmaj7": [60,64,67,71]
};

function getKey(obj, arr) {

  // Find the first object that satisfies the condition...
  const found = Object.entries(obj).find(([key, notes]) => {

    // ...that all the notes in the object are included in the
    // array that was passed in. As long as the notes array and the
    // query array are the same length, the notes can be in any order
    return arr.length === notes.length && arr.every(note => notes.includes(note));
  });

  // If `found` is an Object.entries array return the key
  // otherwise return the error
  return found && found[0] || 'Key not found';
}

// Same order
const key = getKey(dictionary, [60, 64, 67, 71]);
console.log(key);

// Different order
const key2 = getKey(dictionary, [71, 64, 67, 60]);
console.log(key2);

// Missing note
const key3 = getKey(dictionary, [71, 64, 60]);
console.log(key3);

Answer 2

您不能使用=== 运算符来比较两个数组。使用以下函数：

function compareArrays(a, b) {
  if(a.length != b.length) return false;
  for(let i = 0; i < a.length; i++) {
    if(a[i] !== b[i]) return false;
  }
  return true;
}

只需将obj[key] === val 替换为compareArrays(obj[key], val)，它应该可以工作，假设所有值都将是简单的数组。如果没有，您想在函数中检查它。

Answer 3

假设您正在搜索数组，您可以使用函数every 来检查数组val 中的每个值是否为included。

这是假设我们正在处理数组

var dictionary = {"Cmaj7": [60,64,67,71]},
    getKey = (obj,val) => Object.keys(obj).find(key => obj[key].length === val.length && obj[key].every((kn) => val.includes(kn)));

console.log(getKey(dictionary,[60,64,71,67]));