在使用子类UIActivityItemProvider时没有可用的共享操作

Question

当我只是传入一个NSString时，一切都很完美，但是当我试图传递我的自定义ItemProvider时，我没有任何共享操作可用。

MyCustomProvider

class MyCustomProvider: UIActivityItemProvider {

    var placeholder: String!
    var alternate: String!


    init(placeholder: String, alternate: String) {
        self.placeholder = placeholder
        self.alternate = alternate

        super.init()
    }

    override func activityViewController(activityViewController: UIActivityViewController, itemForActivityType activityType: String) -> AnyObject? {
        if activityType == UIActivityTypeMessage {
            return NSString(string: alternate)
        } else {
            return NSString(string: placeholder)
        }
    }
}

我执行的：

var stringToShare = MyCustomProvider(placeholder: "Placeholder", alternate: "Alternate")

var sharingItems = [AnyObject]()
sharingItems.append(stringToShare)

let activityViewController = UIActivityViewController(activityItems: sharingItems, applicationActivities: nil)

activityViewController.excludedActivityTypes = [UIActivityTypePrint, UIActivityTypeAssignToContact, UIActivityTypeSaveToCameraRoll, UIActivityTypeAddToReadingList, UIActivityTypePostToFlickr, UIActivityTypePostToVimeo, UIActivityTypeAirDrop]

self.presentViewController(activityViewController, animated: true, completion: nil)

当上面提到的stringToShare是NSString(string: "Placeholder")时，一切都很完美。我在执行过程中搞砸了什么？

Answer 1

我用以下方法解决了这个问题：

class MyCustomProvider: NSObject, UIActivityItemSource {

    var placeholder: String!
    var alternate: String!


    init(placeholder: String, alternate: String) {
        self.placeholder = placeholder
        self.alternate = alternate

        super.init()
    }

    func activityViewControllerPlaceholderItem(activityViewController: UIActivityViewController) -> AnyObject {
        return NSString(string: placeholder)
    }

    func activityViewController(activityViewController: UIActivityViewController, itemForActivityType activityType: String) -> AnyObject? {
        if activityType == UIActivityTypeMessage {
            return NSString(string: alternate)
        } else {
            return NSString(string: placeholder)
        }
    }
}

我不知道为什么，但很管用。

Answer 2

我相信您必须将placeholderItem设置为要在item()中返回的类型。

可以代表您计划创建的实际对象的对象。对象的内容可能是空的，但对象的类必须与您计划稍后提供的对象的类匹配。