ios - UIImagePickerController 相机不提供​​ URL

Question

我有一个应用程序，直到几天前我还在开发它，它只允许某人从他们的照片库上传照片。它运行良好，检查 GPS 坐标的元数据并进行条件比较。

然而，现在我决定让他们通过应用程序拍摄照片，但遇到了麻烦。用于照片选择的代码在使用相机时效果不佳。

这是我上传照片按钮的原始代码:

    @IBAction func pickPhoto(sender: AnyObject) {
    if UIImagePickerController.isSourceTypeAvailable(UIImagePickerControllerSourceType.SavedPhotosAlbum) {
        imageView.hidden = true

        let imagePicker = UIImagePickerController()
        imagePicker.delegate = self
        imagePicker.sourceType = UIImagePickerControllerSourceType.SavedPhotosAlbum
        imagePicker.allowsEditing = false

        presentViewController(imagePicker, animated: true, completion: nil)
    }
}

然后是 UIImagePickerController:

    func imagePickerController(picker: UIImagePickerController, didFinishPickingMediaWithInfo info: [NSObject : AnyObject]) {

    imageView.image = info[UIImagePickerControllerOriginalImage] as? UIImage

    println("imagePickerController")

    let library = ALAssetsLibrary()

    let url: NSURL = info[UIImagePickerControllerReferenceURL] as! NSURL

    library.assetForURL(url, resultBlock: {
        (asset: ALAsset!) in

        if  asset.valueForProperty(ALAssetPropertyLocation) != nil {

            println("!= nil")
            let imageLongitude = (asset.valueForProperty(ALAssetPropertyLocation) as! CLLocation!).coordinate.longitude
            let imageLatitude = (asset.valueForProperty(ALAssetPropertyLocation) as! CLLocation!).coordinate.latitude

            println("imageLatitude = \(imageLatitude)")
            println("imageLongitude = \(imageLongitude)")

            println("Starting the Location check")

... 然后继续进行更多检查。

我复制了 pickPhoto 按钮并将其变成了 takePhoto 按钮，如下所示:

    @IBAction func takePhoto(sender: AnyObject) {
    if UIImagePickerController.isSourceTypeAvailable(UIImagePickerControllerSourceType.Camera) {
        imageView.hidden = true

        let imagePicker = UIImagePickerController()
        imagePicker.delegate = self
        imagePicker.sourceType = UIImagePickerControllerSourceType.Camera
        imagePicker.allowsEditing = false

        presentViewController(imagePicker, animated: true, completion: nil)
    } else {

        notifyUser("No Camera", message: "This device doesn't have a camera!")
    }
}

然后打开相机让我拍照。但是，当我使用照片时，我立即在这一行的 UIImagePickerController 中崩溃:

let url: NSURL = info[UIImagePickerControllerReferenceURL] as! NSURL

我假设这是因为作为相机图像，它在技术上没有保存在任何地方，因此它没有 URL 或路径。

所以我的问题是，如何保存图像(临时或在相机胶卷中)，保持相机的 GPS 元数据完好无损(假设位置服务处于事件状态)，这样我就可以传递它并让它很好地播放？

Answer 1

编辑:

这篇文章已过时

---

swift 2.x

正如您已经注意到的，您需要将文件保存到磁盘才能获取其 url。您必须将其保存到您的相机 SavedPhotosAlbum 或您的文档文件夹中。您还需要保存图像元数据信息 )UIImagePickerControllerMediaMetadata) 如下:

func imagePickerController(picker: UIImagePickerController, didFinishPickingMediaWithInfo info: [String : AnyObject]) {
    
    print(info["UIImagePickerControllerOriginalImage"] ?? "NO IMAGE")
    print(info["UIImagePickerControllerReferenceURL"] ?? "NO URL")
    print(info["UIImagePickerControllerMediaMetadata"] ?? "NO METADATA")

    if let image = info["UIImagePickerControllerOriginalImage"] as? UIImage {
        ALAssetsLibrary().writeImageToSavedPhotosAlbum(image.CGImage!, metadata: info[UIImagePickerControllerMediaMetadata]! as! [NSObject : AnyObject], completionBlock: { (url, error) -> Void in
            
            print("photo saved to asset")
            print(url)   // assets-library://asset/asset.JPG?id=CCC70B9F-748A-43F2-AC61-8755C974EE15&ext=JPG
            
            
            // you can load your UIImage that was just saved to your asset as follow
            let assetLibrary = ALAssetsLibrary()
            assetLibrary.assetForURL(url,
                resultBlock: { (asset) -> Void in
                    if let asset = asset {
                        let assetImage =  UIImage(CGImage:  asset.defaultRepresentation().fullResolutionImage().takeUnretainedValue())
                        print(assetImage)
                    }
                }, failureBlock: { (error) -> Void in
                if let error = error { print(error.description) }
            })

            if let error = error { print(error.description) }
            self.dismissViewControllerAnimated(true, completion: nil)
        })
    }
}