在UIDocumentInteractionController或UIActivityViewController中自动选择Instagram

Question

UIDocumentInteractionController和UIActivityViewController都提供了用于将图像共享到其他网络的菜单。点击Instagram后，您会看到一个很好的模态，允许您在不离开应用程序的情况下发布到Instgram。我的问题是，如何在不显示菜单的情况下自动显示该模态？

此应用程序名为Sounds，因此我知道它可行，但我无法在线查找有关其完成方式的任何文档。以下是我显示菜单的当前代码：

    NSString *documentDirectory=[NSHomeDirectory() stringByAppendingPathComponent:@"Documents"];
    NSString *saveImagePath=[documentDirectory stringByAppendingPathComponent:@"Image.ig"];
    NSData *imageData=UIImagePNGRepresentation(cardImage);
    [imageData writeToFile:saveImagePath atomically:YES];
    NSURL *imageURL=[NSURL fileURLWithPath:saveImagePath];

    UIDocumentInteractionController *docController = [UIDocumentInteractionController interactionControllerWithURL:imageURL];
    docController.UTI = @"com.instagram.exclusivegram";
    docController = [self setupControllerWithURL:imageURL usingDelegate:self];
    [docController presentOpenInMenuFromRect:CGRectMake(1, 1, 1, 1) inView:self.view animated:YES];

这是菜单：

这是我想要自动显示的内容：

Answer 1

只要活动项目中只有UIImage，您就可以使用UIActivityViewController来显示Instagram。 如果您添加更多项目，如文本，网址等，它将不会显示。 这是Instagram的一个不好的限制。

像这样：

NSMutableArray *items = [NSMutableArray array];
[items addObject:[UIImage ...]];

// show view
UIActivityViewController *activityVC = [[UIActivityViewController alloc] initWithActivityItems:items
                                                                             applicationActivities:nil];
[vc presentViewController:activityVC animated:YES completion:nil];

Answer 2

我用来在Instagram上分享图片的代码

import UIKit
import Photos

class SocialShare: NSObject {
    static let shared = SocialShare()

    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
            UIApplication.sharedApplication().openURL(NSURL(string: u)!)
        }
    }
}

没有任何地方你只想打电话

SocialShare.shared.postImageToInstagram(UIImage(named: "1"))

Answer 3

这是一个实现Zuhair答案的Swift 4类。

import Foundation
import Photos

class InstagramSharer: NSObject {
    static let shared = InstagramSharer()

    func post(image: UIImage) {
        UIImageWriteToSavedPhotosAlbum(image, self, #selector(image(image:didFinishSavingWithError:contextInfo:)), nil)
    }

    @objc
    private func image(image: UIImage, didFinishSavingWithError error: NSError?, contextInfo: UnsafeRawPointer) {
        guard error == nil else {
            return
        }

        let fetchOptions = PHFetchOptions()
        fetchOptions.sortDescriptors = [NSSortDescriptor(key: "creationDate", ascending: false)]
        let fetchResult = PHAsset.fetchAssets(with: .image, options: fetchOptions)

        if let lastAsset = fetchResult.firstObject {
            let localIdentifier = lastAsset.localIdentifier
            let url = "instagram://library?LocalIdentifier=" + localIdentifier
            UIApplication.shared.openURL(URL(string: url)!)
        }
    }
}

用法

InstagramSharer.shared.post(image: #YOUR_IMAGE#)

Answer 4

要删除其他应用程序，请更改以下代码行（只需翻转它们）：

docController.UTI = @"com.instagram.exclusivegram";
docController = [self setupControllerWithURL:imageURL usingDelegate:self];

这样在覆盖docController对象后设置.UTI变量。

编辑：如果您尝试直接打开Instagram应用程序并填写完信息并准备共享，请查看iPhone挂钩here。