1

我需要保存从服务器选择的图像的路径，然后访问该路径以通过数据库检索在用户单击时显示该图像。

-(void)imagePickerController:(UIImagePickerController *)picker didFinishPickingMediaWithInfo:(NSDictionary *)info
{
      NSString *urlPath =  [info objectForKey:@"UIImagePickerControllerReferenceURL"]absoluteString];
}

并将其保存到数据库中。

它保存为 assets-library://asset/asset.JPG?id=E1136225-97DE-4BF4-A864-67E33D60737A&ext=JPG

然后我想导入到Imageview

 UIImageView *iv = [[UIImageView alloc]init];
 iv.image = [UIimage imageWithData:[NSData dataWithContentsOfURL:[NSURL URLWithString:imagepath]]];

但它不工作

4

2 回答 2

2

试试这个： 

  typedef void (^ALAssetsLibraryAssetForURLResultBlock)(ALAsset *asset);
    typedef void (^ALAssetsLibraryAccessFailureBlock)(NSError *error);    

    ALAssetsLibraryAssetForURLResultBlock resultblock = ^(ALAsset *myasset){

     ALAssetRepresentation *rep = [myasset defaultRepresentation];
     CGImageRef iref = [rep fullResolutionImage];

     if (iref){

            UIImage *myImage = [UIImage imageWithCGImage:iref scale:[rep scale] orientation:(UIImageOrientation)[rep orientation]];
            [fileImage addObject:myImage];

             }      
    };      

    ALAssetsLibraryAccessFailureBlock failureblock  = ^(NSError *myerror){

         //failed to get image.
    };                          

    ALAssetsLibrary* assetslibrary = [[[ALAssetsLibrary alloc] init] autorelease];
    [assetslibrary assetForURL:[filePath objectAtIndex:0] resultBlock:resultblock failureBlock:failureblock];

注意：确保你的[filePath objectAtIndex:0]will 是一个NSUrl对象。否则将其转换为NSUrl

例子：

ALAssetsLibrary* assetslibrary = [[[ALAssetsLibrary alloc] init] autorelease];

NSURL myAssetUrl = [NSURL URLWithString:[filePath objectAtIndex:0]];

assetslibrary assetForURL:myAssetUrl resultBlock:resultblock failureBlock:failureblock];
于 2013-01-17T12:09:43.817 回答
1

使用assetForURL:resultBlock:failureBlock:ALAssetsLibrary 类的方法通过 URL 检索图像。还有更多信息：http: //developer.apple.com/library/ios/#documentation/AssetsLibrary/Reference/ALAssetsLibrary_Class/Reference/Reference.html#//apple_ref/doc/uid/TP40009722

升级版： 

ALAssetsLibrary *lib = [[ALAssetsLibrary alloc] init];
[lib assetForURL: url resultBlock: ^(ALAsset *asset) {
    ALAssetRepresentation *r = [asset defaultRepresentation];
    self.imgView.image = [UIImage imageWithCGImage: r.fullResolutionImage];
}
    failureBlock: nil];

url您的缓存网址在哪里

于 2013-01-17T12:00:50.833 回答