ALAsset valueForProperty:ALAssetPropertyLocation未声明？

Question

这看起来似乎太容易了,也许这对我来说太困惑了......或许我太累了,无法理性思考.

我正在尝试使用imagePickerController获取用户从相册中选择的UIImage的位置.

- (void)imagePickerController:(UIImagePickerController *)picker didFinishPickingMediaWithInfo:(NSDictionary *)info
{
    UIImage *pic = [info objectForKey:@"UIImagePickerControllerOriginalImage"];
    NSData *data = UIImageJPEGRepresentation(pic, 0.01f);
    UIImage *image1 = [UIImage imageWithData:data];
    bookmarkImage.image = image1;

    NSLog(@"new image size = %i", [data length]);

    NSURL *url = [info objectForKey:@"UIImagePickerControllerReferenceURL"];
    4NSLog(@"%@",url);

    ALAssetsLibraryAssetForURLResultBlock resultBlock = ^(ALAsset *myAsset){
        [myAsset valueForProperty:ALAssetPropertyLocation];
    };

    [self dismissModalViewControllerAnimated:YES];
}

现在,我尝试为ALAsset 调用valueForProperty是我遇到麻烦的地方.这是我得到的错误:

使用未声明的标识符'ALAssetPropertyLocation'.

请帮忙!这真的令人沮丧......我需要照片的位置......

Answer 1

这很傻,但我一直错过了一个导入:

#import <AssetsLibrary/ALAsset.h>

感谢Markus指出来的!