Question

我有一个相同类的Observable of Objects数组= Observable <[MyClass]>。 MyClass包含一个Observable属性sortAttribute = Observable。我想根据可观察的sort属性对数组进行排序。

class MyClass {

    let title:String
    let sortAttribute:Observable<Int>

    init(withTitle title:String, andSortValue sortValue: Int) {
        self.title = title
        self.sortAttribute = Observable.just(sortValue)
    }

}

let arrayToSort:Observable<[MyClass]> = Observable.just([
    MyClass(withTitle: "A", andSortValue: 4),
    MyClass(withTitle: "B", andSortValue: 2),
    MyClass(withTitle: "C", andSortValue: 42),
    MyClass(withTitle: "D", andSortValue: 1337),
    MyClass(withTitle: "E", andSortValue: 24)
])

arrayToSort
    .subscribe(onNext: { ar in
        for element in ar {
            print(element.title)
        }
    })

实际结果：

A //4
B //2
C //42 
D //1337 
E //24

预期结果：

B //2
A //4
E //24
C //42
D //1337

Answer 1

arrayToSort
    .flatMap { classes in
        Observable.combineLatest(
            classes.map { elem in
                elem.sortAttribute.map { (elem, $0) }
            }
        )
    }
    .map {
        $0.sorted(by: { a, b in a.1 < b.1 }).map { $0.0.title }
    }
    .subscribe(onNext: {
        print($0)
    })

Answer 2

尝试一下：

arrayToSort
.subscribe(onNext: { ar in
        let sortedAr = ar.sorted(by: { $0.title < $1.title })
        for element in sortedAr {
            print(element.title)
        }
    })

更新：设置您的排序属性NOT Observable ... Just Int

class MyClass {

    let title:String
    let sortAttribute: Int

    init(withTitle title:String, andSortValue sortValue: Int) {
        self.title = title
        self.sortAttribute = sortValue
    }

}

然后：

arrayToSort
    .subscribe(onNext: { ar in
            let sortedAr = ar.sorted(by: { $0.sortAttribute < $1.sortAttribute })
            for element in sortedAr {
                print(element.title)
            }
        })

如果将其保留为Observable，则必须订阅每个sortAttribute提取其Int，然后以某种方式对其进行排序，这对于简单的排序来说浪费了处理能力...

如何通过Observable属性对项目数组进行排序？

2 个答案: