对于具有discount_percent s和quantity s的多个对象的一般情况，.find不是正确的方法，因为它会在找到匹配项后立即停止。考虑.reduce-如果正在迭代的元素通过了测试 * 并且 * 它的discount_percent大于累加器中的当前元素（如果累加器中有任何内容可以作为开始），则将其返回。

let tiers = [
   {id: 1, discount_percent:0, quantity:6, title:"Tier 1"},
   {id: 2, discount_percent:5, quantity:8, title:"Tier 2"},
   {id: 3, discount_percent:10, quantity:10, title:"Tier 3"},
   {id: 4, discount_percent:12, quantity:12, title:"Tier 4"},
]
function calculateDiscount(){
   const ordersQuanity = 10;
   const bestTier = tiers.reduce((a, tier) => (
     tier.quantity <= ordersQuanity && (!a || tier.discount_percent > a.discount_percent)
       ? tier
       : a
   ), null) || tiers[0]; // alternate with the first element of the array
   // if you want to default to that tier even if the quantity isn't sufficient
   console.log(bestTier);
}
calculateDiscount();

如果你碰巧能够假设每增加一个discount_percent就带来一个更大的数量，并且数组是排序的，那么如果你先反转数组，你就可以使用.find（这样，具有最大discount_percent的项会先被迭代）。

let tiers = [
   {id: 1, discount_percent:0, quantity:6, title:"Tier 1"},
   {id: 2, discount_percent:5, quantity:8, title:"Tier 2"},
   {id: 3, discount_percent:10, quantity:10, title:"Tier 3"},
   {id: 4, discount_percent:12, quantity:12, title:"Tier 4"},
];
const tiersReversed = [...tiers].reverse();
function calculateDiscount(){
   const ordersQuanity = 10;
   const tier = tiersReversed
     .find((_tier) => _tier.quantity <= ordersQuanity)
      || tiers[0]; // alternate with the first element of the array
   // if you want to default to that tier even if the quantity isn't sufficient
   console.log(tier);
}
calculateDiscount();