RxSwift和UIPickerView

Question

有没有办法将UIPickerView与Observable绑定？

例如对于UITableView，我会这样做：

myObservableArray.bindTo(tableView.rx.items(cellIdentifier: "Identifier", cellType: MyCustomTableViewCell.self)) { (row, title, cell) in
        cell.textLabel?.text = title
    }
    .disposed(by: disposeBag)

UIPickerView是否有类似内容？

Answer 1

事实上，RxCocoa库中有：

示例：

let items = Observable.just([
        "First Item",
        "Second Item",
        "Third Item"
    ])

items
    .bind(to: pickerView.rx.itemTitles) { (row, element) in
        return element.title
    }
    .disposed(by: disposeBag)

还有：

items
   .bind(to: pickerView.rx.items) { (row, element, view) in
       guard let myView = view as? MyView else {
           let view = MyView()
           view.configure(with: element)
           return view
       }
       myView.configure(with: element)
       return myView
   }
   .disposed(by: disposeBag)

Answer 2

提供的选择器数据源如下所示：

let pickerDataSource: [[String]] = [ ["asdadadad", "sffgddfg"],
                                     ["sfsdasgag", "sdfasdfasfsf", "sdsfgagagaggs"] ]

你可以用这种方式实现你需要的“绑定”：

pickerView.rx.itemSelected.subscribe(onNext: { [weak self] row, component in
    guard let s = self else { return }
    s.label.text = s.pickerDataSource[component][row]
}).addDisposableTo(disposeBag)