Question

首先抱歉蹩脚的格式化。我不知道如何在这里格式化代码。我在从照片库中检索图像时遇到问题。我粘贴了我的代码请帮帮我。我很多天都遇到这个问题。

这里，我使用referenceURL将图像路径存储在数据库中。它在

中

-(void)imagePickerController:(UIImagePickerController *)picker didFinishPickingMediaWithInfo:(NSDictionary *)info
    NSURL *imageUrl = [info  valueForKey:UIImagePickerControllerReferenceURL];

    NSURL *imageUrl = [info  valueForKey:UIImagePickerControllerReferenceURL];
    ALAssetsLibrary *assetLibrary = [[ALAssetsLibrary alloc]init];
    ALAssetsLibraryAssetForURLResultBlock resultblock = ^(ALAsset *myasset)
    {
        NSLog(@"asset");
        CGImageRef iref = [myasset thumbnail];
        if(iref)
        {
            NSLog(@"ifref");
            UIImage *thethumbnail = [UIImage imageWithCGImage:iref];
            NSLog(@"the thumbnail %@",thethumbnail);
            [[self photo]setImage:thethumbnail];
            lpdel.imageurl = UIImageJPEGRepresentation (thethumbnail, 1);
        }
    };

    ALAssetsLibraryAccessFailureBlock failureblock = ^(NSError *myerror)
    {
        NSLog(@"failed");
        NSLog(@"cant get image -- %@",[myerror localizedDescription]);
    };

    if (imageUrl) {
        NSLog(@"if url");
        [assetLibrary assetForURL:imageUrl
                      resultBlock:resultblock
                     failureBlock:failureblock];

        //lpdel.imageurl =imageUrl;

        NSLog(@"the image string    %@",lpdel.imageurl);
        NSLog(@"lpdel.image%@",lpdel.imageurl);
        [picker dismissModalViewControllerAnimated:YES];
        [imageUrl release];
    }
    [picker dismissModalViewControllerAnimated:YES];
    [picker release]
}

我将imageURL存储在数据库中。然后在另一个视图控制器中，我试图使用相同的URL使用下面的代码

来检索图像

  ALAssetsLibraryAssetForURLResultBlock resultblock = ^(ALAsset *myasset)
    {
        NSLog(@"asset");
        CGImageRef iref = [myasset thumbnail];
        if(iref)
        {
            NSLog(@"ifref");
            thethumbnail = [UIImage imageWithCGImage:iref];
            NSLog(@"the thumbnail to upload  %@",thethumbnail);
            [self uploadImage:UIImageJPEGRepresentation (thethumbnail, 1)];
            //[[self photo]setImage:thethumbnail];
        }
    };

    ALAssetsLibraryAccessFailureBlock failureblock = ^(NSError *myerror)
    {
        NSLog(@"failed");
        NSLog(@"cant get image -- %@",[myerror localizedDescription]);
    };

    if (image) 
    {
        NSLog(@"if url");
        ALAssetsLibrary *assetLibrary = [[[ALAssetsLibrary alloc]init]autorelease];
        NSLog(@"url");
        [assetLibrary assetForURL:image
                      resultBlock:resultblock
                     failureBlock:failureblock];
    }
NSLog(@"the thumbnail %@",thethumbnail);

此时应用程序崩溃，没有任何错误消息。任何人都可以帮助我

Answer 1

我找到了解决方案。这是最后一个nslog的问题。但我需要知道原因。使用类范围声明变量thethumbnail。是否必须对块做任何事情

iphone图像ALAsset问题

1 个答案: