JS：具有多個條件的回圈

我有一組物件（汽車），我想從中只顯示“競爭”的汽車。

const cars = [
  {
    name: "red",
    competes: true,
    category: 1
  },
  {
   name: "blue",
   competes: false,
   category: 1
  },
  {
    name: "green",
    competes: true,
     category: 2
  },
  {
    name: "yellow",
    competes: true,
    category: 3
  }
]

但是從那些競爭的汽車中，我只想展示那些屬于第一類的汽車，這對 for 回圈來說很好。

這些物件現在是靜態的，但由于我想在第二個時刻更改它們，我需要一個代碼來檢查汽車是否競爭，然后檢查“類別”中是否有汽車。

• 如果“類別 1”中有汽車，請列出它們
• 如果“類別” 1 中沒有汽車，則列出“類別” 2（但不是 3）中的汽車
• 等等

我的嘗試是在回圈中使用回圈，因為它多次顯示所有內容，所以不起作用

for (let i = 0; i < cars.length; i  ) {
  if (cars[i].competes === false) continue;
  
  for (let f = 0; f < cars.length; f  ) {
  if (cars[f].category > 1) break;
  console.log(`Cat1: ${cars[f].name}`);
}
}

如果“汽車”中的“類別”屬性例如全部為 2 個或更多，如何解決此問題？

const cars = [
  {
    name: "red",
    competes: true,
    category: 2
  },
  {
   name: "blue",
   competes: false,
   category: 2
  },
  {
    name: "green",
    competes: true,
     category: 3
  },
  {
    name: "yellow",
    competes: true,
    category: 4
  }
]

uj5u.com熱心網友回復：

首先，您根據類別對陣列進行排序，然后篩選出包含將要競爭的汽車的陣列

const cars = [
    {
        name: "red",
        competes: true,
        category: 1
    },
    {
        name: "blue",
        competes: false,
        category: 1
    },
    {
        name: "green",
        competes: true,
        category: 2
    },
    {
        name: "yellow",
        competes: true,
        category: 3
    }
]
const sorted = cars.sort((a, b) => a.category - b.category);
const compete = sorted.filter(x => x.competes && x.category == sorted[0].category);
console.log(compete[0]);

uj5u.com熱心網友回復：

這是一個帶有排序和過濾器的簡單解決方案

const cars = [
  {
    name: "red",
    competes: true,
    category: 1
  },
  {
   name: "blue",
   competes: false,
   category: 1
  },
  {
    name: "green",
    competes: true,
     category: 2
  },
  {
    name: "yellow",
    competes: true,
    category: 3
  }
]

let result = cars.sort((a,b) => a.category - b.category)
                 .filter(car => car.competes && (cars.length && car.category == cars[0].category))

console.log(result)

uj5u.com熱心網友回復：

我們可以通過使用Math.min()Array.filter()獲得最低的競爭類別來開始獲得“競爭類別”

然后我們可以通過使用Array.filter()僅回傳具有competes真實值且也屬于競爭類別的汽車來獲取所有競爭汽車。

    
const cars = [ { name: "red", competes: true, category: 1 }, { name: "blue", competes: false, category: 1 }, { name: "green", competes: true, category: 2 }, { name: "yellow", competes: true, category: 3 } ]

// Only this category is competing.. 
const competingCategory = Math.min(...cars.filter(car => car.competes).map(({category}) => category));
const competingCars = cars.filter(car => car.competes && car.category === competingCategory);

console.log('Competing cars:', competingCars)

.as-console-wrapper { max-height: 100% !important; top: 0; }

此示例僅存在類別 2 和 3：

    
const cars = [ { name: "blue", competes: true, category: 2 }, { name: "green", competes: true, category: 2 }, { name: "yellow", competes: true, category: 3 } ]

// Only this category is competing.. 
const competingCategory = Math.min(...cars.filter(car => car.competes).map(({category}) => category));
const competingCars = cars.filter(car => car.competes && car.category === competingCategory);

console.log('Competing cars (example II):', competingCars)

.as-console-wrapper { max-height: 100% !important; top: 0; }

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