ios - 如何检查我的应用是否可以访问手机图库

Question

我有一个应用程序，我在其中用相机拍照并将该图像存储到本机图库中。但是，如果应用程序没有权限，我希望用户知道这一点。那么我该如何检查呢？

顺便说一句：我将图像存储到图库中：

UIImageWriteToSavedPhotosAlbum(image, nil, nil, nil);

Answer 1

您需要检查状态ALAssetLibrary 以确保您已AssetsLibrary/AssetsLibrary.h包含在文件中

  ALAuthorizationStatus status = [ALAssetsLibrary authorizationStatus];

// 检查ALAuthorizationStatusAuthorizedor的状态，ALAuthorizationStatusDenied例如

    if (status != ALAuthorizationStatusAuthorized) {
        //show alert for asking the user to give permission

    }

Answer 2

斯威夫特 3

import photos

PHPhotoLibrary.requestAuthorization { status in
     switch status {
     case .authorized:
          self.processSnapShotPhotos()
     case .restricted:
          print("handle restricted")
     case .denied:
          print("handle denied")     
     default:
       // place for .notDetermined - in this callback status is already determined so should never get here
            break
     }
}

Answer 3

如果您正在使用照片框架，因为 ALAsset 库已从 ios 9 中弃用，您可以使用 PHAuthorizationStatus 检查图库访问权限。您还需要导入照片框架。

  #import <Photos/Photos.h>

- (BOOL)hasGalleryPermission
{
    BOOL hasGalleryPermission = NO;
    PHAuthorizationStatus authorizationStatus = [PHPhotoLibrary authorizationStatus];

    if (authorizationStatus == PHAuthorizationStatusAuthorized) {
        hasGalleryPermission = YES;
    }
    return hasGalleryPermission;
}

Answer 4

注意：仅限 iOS 6

这是你想要的

[ALAssetsLibrary authorizationStatus] == ALAuthorizationStatusAuthorized;

authorizationStatus 的其他值是

ALAuthorizationStatusRestricted,        // This application is not authorized to access photo data.
                                            // The user cannot change this application’s status, possibly due to active restrictions
                                            //  such as parental controls being in place.
    ALAuthorizationStatusDenied,            // User has explicitly denied this application access to photos data.
    ALAuthorizationStatusAuthorized         // User has authorized this application to access photos data.