JS：具有多个条件的循环

Question

我有一组对象（汽车），我想从中只显示“竞争”的汽车。

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
  }
]

但是从那些竞争的汽车中，我只想展示那些属于第一类的汽车，用 for 循环就可以了。

对象现在是静态的，但由于我想在第二个时刻更改它们，我需要一个代码来检查汽车是否竞争，然后检查“类别”中是否有汽车。

• 如果有属于“1 类”的汽车，请列出它们
• 如果“类别”1 中没有汽车，请列出“类别”2（但不是 3）中的汽车
• 等等

我的尝试是在一个循环中使用一个循环，由于它多次显示所有内容，所以有点不起作用

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“汽车”中的“类别”属性例如全部为 2 个或更多，如何解决此问题？

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
  }
]

Answer 1

首先你根据类别对数组进行排序，然后过滤出一个包含要竞争的汽车的数组

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]);

Answer 2

这是一个简单的排序和过滤解决方案

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)

Answer 3

我们可以通过使用 Math.min() 和 Array.filter() 来获得最低竞争类别来获得“竞争类别”

然后我们可以通过使用Array.filter() 只返回具有真实值且也属于竞争类别的competes 值的汽车来获取所有竞争汽车。

    
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)

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此示例仅存在类别 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)

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