我确实有一个可以满足您需求的可行解决方案。并且可能是一种通过在不需要时避免使用可选项来改进代码的方法。如果您的 Web 服务将始终为您的对象返回名称和分数,那么不要使用可选项,这将使您的生活更轻松。当我没有任何选项时,您可以在我的解决方案中看到代码更易于阅读并且这样的操作很简单。
// -------------- Solution 1 avoiding optionals ----------------
struct MyObject {
var name: String
var score: Score
}
struct Score {
var number: Double
}
let data: [MyObject] = [MyObject(name: "obj1", score: Score(number: 12)),
MyObject(name: "obj2", score: Score(number: 2)),
MyObject(name: "obj3", score: Score(number: 24)),
MyObject(name: "obj4", score: Score(number: 4)),
MyObject(name: "obj5", score: Score(number: 1))]
let sortedObj = data.sorted(by: { $0.score.number > $1.score.number })
print(sortedObj)
// Print: obj3, obj1, obj4, obj2 and obj5
// -------------- Solution 2 using optionals ----------------
struct MyObject2 {
var name: String?
var score: Score2?
}
struct Score2 {
var number: Double?
}
let data2: [MyObject2] = [MyObject2(name: "obj1", score: Score2(number: 12)),
MyObject2(name: "obj2", score: Score2(number: 2)),
MyObject2(name: "obj3", score: Score2(number: 24)),
MyObject2(name: "obj4", score: Score2(number: 4)),
MyObject2(name: "obj5", score: Score2(number: 1))]
let sortedObj2 = data2.sorted(by: {
guard let num1 = $0.score?.number, let num2 = $1.score?.number else {
return false
}
return num1 > num2
})
print(sortedObj2)
// Print: obj3, obj1, obj4, obj2 and obj5
希望对你有帮助