Question

我希望我的应用用户使用UIACtivityViewController共享链接。链接不是静态的，我必须从api中获取它。我想在有人按下UIActivityViewController中的图标时立即获取此链接，但我现在如何操作却无法正常工作。请求的完成处理程序始终在方法fetchShareURL（）的末尾执行。有人能告诉我如何解决这个问题吗？

    import UIKit

    class ShareItemSource: NSObject, UIActivityItemSource {


        var objects : [AnyObject]?
        var shareURL : String?

        init(objects : [AnyObject]?){
            self.objects = objects
        }

        func fetchShareURL() -> Bool{
            //fetch share url from api

            let api = Api()
            let semaphore = dispatch_semaphore_create(0);
            let backgroundQueue = dispatch_get_global_queue(DISPATCH_QUEUE_PRIORITY_BACKGROUND, 0)
            dispatch_async(backgroundQueue, {

                api.export(self.groupObjList, receiptObjList: self.receiptObjList){ status, message, url in
                    dispatch_async(dispatch_get_main_queue(), {
                        self.shareURL = url

                        dispatch_semaphore_signal(semaphore);
                    })
                }

            });

            let delayInSeconds = 30.0;
            let delayTime = dispatch_time(DISPATCH_TIME_NOW, Int64(delayInSeconds * Double(NSEC_PER_SEC)))
            dispatch_semaphore_wait(semaphore, delayTime)

            if self.shareURL != nil{
                return true
            } else {
                return false
            }

        }


        @objc func activityViewControllerPlaceholderItem(activityViewController: UIActivityViewController) -> AnyObject {
            return ""
        }

        @objc func activityViewController(activityViewController: UIActivityViewController, itemForActivityType activityType: String) -> AnyObject? {
            if fetchShareURL() == true{
                return "Download files: "+self.shareURL!

            } else {
                return nil
            }

        }

        func activityViewController(activityViewController: UIActivityViewController, subjectForActivityType activityType: String?) -> String {
            return "title of share"
        }

        func activityViewController(activityViewController: UIActivityViewController, thumbnailImageForActivityType activityType: String?, suggestedSize size: CGSize) -> UIImage? {
            //do stuff
    }
}

Answer 1

如果要在UIActivityView中提供的数据需要时间，请使用UIActivityItemProvider。它是一个NSOperation子类，以完全一致的方式解决整个问题。实际上，这种情况正是它的目的。（请观看有关高级NSOperation的WWDC 2015视频。）

Answer 2

暂时将其更改为此内容。后来我会使用NSOperation。

    api.export(self.objects){ status, message, url in
        if(status == true){
            if let url = url{
                alert.dismissViewControllerAnimated(true){
                    let sharingItems = ["Share url", url]
                    let activityViewController = UIActivityViewController(activityItems: [ShareItemSource(url: url)], applicationActivities: nil)
                    activityViewController.excludedActivityTypes = [UIActivityTypeAddToReadingList, UIActivityTypeCopyToPasteboard]
                    self.presentViewController(activityViewController, animated: true, completion: nil)
                    activityViewController.completionWithItemsHandler = { (activity, success, items, error) in
                        //Response
                    }

                }
            }
        } else {
            //handle fail
        }
    }

从UIActivityViewController中的API获取自定义消息

2 个答案: