@H_301_1@是否可以绕过行动共享表将照片分享给Instagram？
请注意,我知道UIDocumentInteractionController和钩子,事实上它工作正常.使用他们的示例代码,您可以获得Copy to Instagram选项(如果您使用独家UTI或可以处理JPG / PNG的大型应用程序列表,以及Instagram,则可以是唯一的).
这很好,但我想知道是否有办法执行“复制到Instagram”操作,而无需在iOS 9中显示UIDocumentInteractionController菜单.
为了记录,这是完美运行的代码的简化版本.假设你有一个有效的NSURL ……
guard let data: NSData = NSData(contentsOfURL: url),image = UIImage(data: data) else {
return
}
let imageData = UIImageJPEGRepresentation(image,100)
let captionString = "caption"
let writePath = (NSTemporaryDirectory() as NSString).stringByAppendingPathComponent("instagram.ig")
guard let _ = imageData?.writeToFile(writePath,atomically: true) else {
return
}
let fileURL = NSURL(fileURLWithPath: writePath)
self.documentController = UIDocumentInteractionController(URL: fileURL)
self.documentController.delegate = self
self.documentController.UTI = "com.instagram.photo"
self.documentController.annotation = NSDictionary(object: captionString,forKey: "InstagramCaption")
self.documentController.presentOpenInMenuFromRect(viewController.view.frame,inView: viewController.view,animated: true)
问题是,这将提供一个“操作表”,我想避免这样做,如果可能的话,我想使用instagram.ige(或任何名称使其独占)并跳过此ActionSheet.
那可能吗？
更新：我还没有找到解决方案,但似乎Instagram最终添加/添加扩展：“Instagram最近为其iOS应用程序添加了共享扩展功能.现在,您可以直接将第三方应用程序的照片分享到Instagram”
来源：http://www.macworld.com/article/3080038/data-center-cloud/new-instagram-feature-for-ios-makes-it-easier-to-share-photos-from-other-apps.html
解决方法
import Photos
...
func postImageToInstagram(image: UIImage) {
UIImageWriteToSavedPhotosAlbum(image,self,#selector(SocialShare.image(_:didFinishSavingWithError:contextInfo:)),nil)
}
func image(image: UIImage,didFinishSavingWithError error: NSError?,contextInfo:UnsafePointer<Void>) {
if error != nil {
print(error)
}
let fetchOptions = PHFetchOptions()
fetchOptions.sortDescriptors = [NSSortDescriptor(key: "creationDate",ascending: false)]
let fetchResult = PHAsset.fetchAssetsWithMediaType(.Image,options: fetchOptions)
if let lastAsset = fetchResult.firstObject as? PHAsset {
let localIdentifier = lastAsset.localIdentifier
let u = "instagram://library?LocalIdentifier=" + localIdentifier
let url = NSURL(string: u)!
if UIApplication.sharedApplication().canOpenURL(url) {
UIApplication.sharedApplication().openURL(NSURL(string: u)!)
} else {
let alertController = UIAlertController(title: "Error",message: "Instagram is not installed",preferredStyle: .Alert)
alertController.addAction(UIAlertAction(title: "OK",style: .Default,handler: nil))
self.presentViewController(alertController,animated: true,completion: nil)
}
}
}
