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首先对糟糕的格式感到抱歉。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。我已经粘贴了我的代码请帮助我。我从很多天都遇到了这个问题。

在这里，我使用 referenceURL 将图像路径存储在数据库中。它在

-(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，我尝试使用以下代码使用相同的 url 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);

应用程序此时崩溃，没有任何错误消息。谁能帮我解决这个问题

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1 回答 1

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我找到了解决方案。这是最后一个nslog的问题。但我需要知道为什么。变量 thethumbnail 是在类范围内声明的。它与块有什么关系吗

于 2011-02-21T12:29:19.923 回答