從對象數組中獲取最大值和最小值並返回與相同對象關聯的另一個值 object. JavaScript

Question

我在完成我正在參加的教育課程的任務時遇到問題，highestValShoe 和 lowestValShoe 函數返回的結果相同。

我已盡力而為，但我不明白我哪里出錯了。 我真的很感激一些指示。 謝謝！

//First I will establish an empty array to push() the shoes in later.
shoeArray = [];

//Now I will create a class for the shoes.
class Shoes {
  constructor(name, productCode, quantity, valuePerItem) {
    this.name = name;
    this.productCode = productCode;
    this.quantity = quantity;
    this.valuePerItem = valuePerItem;
  }

  //This code will enable us to update the quantity.
  updateQuantity(newQuantity) {
    this.quantity = newQuantity;
  }
}

//Now I will create 5 instances for the class.
converse = new Shoes("Converse", 010405, 3, 55.5);
adidas = new Shoes("Adidas", 030602, 5, 85.0);
nike = new Shoes("Nike", 052656, 2, 165.0);
vans = new Shoes("Vans", 745023, 6, 95.5);
fila = new Shoes("Fila", 034567, 3, 45.0);

//This will push all instances into the shoeArray.
shoeArray.push(converse, adidas, nike, vans, fila);

//This function will enable us to search for any shoe within the array.
function searchShoes(shoeName, shoeArray) {
  for (i = 0; i < shoeArray.length; i++) {
    if (shoeArray[i].name === shoeName) {
      return shoeArray[i];
    }
  }
}

//This function will enable us to search for the lowest value shoe. 
function lowestValShoe (shoeArray) {
  for (i=0; i<shoeArray.length; i++){
    lowestVal = Math.min(shoeArray[i].valuePerItem)
    shoe = shoeArray[i].name
}
  return shoe
}

//This function will enable us to search for the highest value shoe.
function highestValShoe (shoeArray){
  for (i=0; i<shoeArray.length; i++){
    highestVal = Math.max(shoeArray[i].valuePerItem)
    shoe1 = shoeArray[i].name
  }
  return shoe1
}

我試圖返回最大值“Shoe”和最小值“Shoe”，當我測試 lowestValShoe function 時我認為它有效，但是當我將其轉換為最大值時，它返回相同的值。

我更改了其中一只鞋的數值，將另一只鞋設為最低值，然后意識到我的最低值 ValShoe function 也沒有按預期工作。

這兩個函數都返回 'Fila'

Answer 1

您需要存儲 object 而不是僅用於查找最小值或最大值鞋的名稱。

取第一項並從第二項迭代到結束。 然后檢查值是更大還是更小並拿走該項目。 稍后僅返回名稱。

function highestValShoe(shoeArray) {
    let shoe = shoeArray[0];
    for (let i = 1; i < shoeArray.length; i++) {
        if (shoeArray[i].valuePerItem > shoe.valuePerItem) shoe = shoeArray[i];
    }
    return shoe.name;
}

順便說一句，請聲明所有變量。 如果沒有，你會得到全局的，這可能會導致有趣的結果。 並使用分號。 總是。

 //Now I will create a class for the shoes. class Shoes { constructor(name, productCode, quantity, valuePerItem) { this.name = name; this.productCode = productCode; this.quantity = quantity; this.valuePerItem = valuePerItem; } //This code will enable us to update the quantity. updateQuantity(newQuantity) { this.quantity = newQuantity; } } //This function will enable us to search for any shoe within the array. function searchShoes(shoeName, shoeArray) { for (let i = 0; i < shoeArray.length; i++) { if (shoeArray[i].name === shoeName) return shoeArray[i]; } } //This function will enable us to search for the lowest value shoe. function lowestValShoe(shoeArray) { let shoe = shoeArray[0]; for (let i = 1; i < shoeArray.length; i++) { if (shoeArray[i].valuePerItem < shoe.valuePerItem) shoe = shoeArray[i]; } return shoe.name; } function highestValShoe(shoeArray) { let shoe = shoeArray[0]; for (let i = 1; i < shoeArray.length; i++) { if (shoeArray[i].valuePerItem > shoe.valuePerItem) shoe = shoeArray[i]; } return shoe.name; } //Now I will create 5 instances for the class. const shoeArray = [], converse = new Shoes("Converse", "010405", 3, 55.5), adidas = new Shoes("Adidas", "030602", 5, 85.0), nike = new Shoes("Nike", "052656", 2, 165.0), vans = new Shoes("Vans", "745023", 6, 95.5), fila = new Shoes("Fila", "034567", 3, 45.0); //This will push all instances into the shoeArray. shoeArray.push(converse, adidas, nike, vans, fila); console.log(lowestValShoe(shoeArray)); console.log(highestValShoe(shoeArray));

Answer 2

因為你需要重寫你的 min 和 max 函數，所以他們不會只使用 Math.min 和 Math.max 你可以使用Array.reduce使它們成為一個班輪：

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/reduce

function lowestValShoe (shoeArray) {
  return shoeArray.reduce((prev, curr) => (prev.valuePerItem < curr.valuePerItem ? prev : curr));
}

function highestValShoe (shoeArray){
   return shoeArray.reduce((prev, curr) => (prev.valuePerItem > curr.valuePerItem ? prev : curr));   
}

在這里你可以看到它分別返回Fila和Nike的最小值和最大值：

 //First I will establish an empty array to push() the shoes in later. shoeArray = []; //Now I will create a class for the shoes. class Shoes { constructor(name, productCode, quantity, valuePerItem) { this.name = name; this.productCode = productCode; this.quantity = quantity; this.valuePerItem = valuePerItem; } //This code will enable us to update the quantity. updateQuantity(newQuantity) { this.quantity = newQuantity; } } //Now I will create 5 instances for the class. converse = new Shoes("Converse", 010405, 3, 55.5); adidas = new Shoes("Adidas", 030602, 5, 85.0); nike = new Shoes("Nike", 052656, 2, 165.0); vans = new Shoes("Vans", 745023, 6, 95.5); fila = new Shoes("Fila", 034567, 3, 45.0); //This will push all instances into the shoeArray. shoeArray.push(converse, adidas, nike, vans, fila); console.log(lowestValShoe(shoeArray)); console.log(highestValShoe(shoeArray)); function lowestValShoe(shoeArray) { return shoeArray.reduce((prev, curr) => (prev.valuePerItem < curr.valuePerItem? prev: curr)); } function highestValShoe(shoeArray){ return shoeArray.reduce((prev, curr) => (prev.valuePerItem > curr.valuePerItem? prev: curr)); }