我正在使用默认的 UITableViewCell 。我想从本地服务器在单元格上设置图像...cell.imageView.image

Question

BLImageCache *sharedImageCache = [BLImageCache sharedInstance];
NSDictionary *dictionary = [NSDictionary dictionaryWithObjectsAndKeys:indexPath ,@"IndexPath", nil] ;
NSString *urlString;
urlString = [[mNewsArray objectAtIndex:indexPath.row] objectForKey:@"imgUrl"];
urlString =[urlString stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding];
UIImage *image=[sharedImageCache imageForURL:urlString];
if(image==nil)
{
    urlString = [urlString stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding];

   [sharedImageCache loadImageFromURL:[NSURL URLWithString:urlString] sender:self context:(NSDictionary*)dictionary];
}
else{
    cell.imageView.image =image;
    cell.imageView.contentMode = UIViewContentModeScaleAspectFit;       
}

应用程序崩溃了，任何人都可以告诉我我做错了什么，是否存在任何与内存相关的问题...

-(void)didLoadImage:(UIImage *)inImage contextInfo:(void *)inContext
{
    if(inImage )
    {
        NSIndexPath *indexPath= [(id)inContext objectForKey:@"IndexPath"];

        UITableViewCell *cell=[mTableView cellForRowAtIndexPath:indexPath];

        inImage=[inImage resizeImagewithwidth:69 height:67];

        cell.imageView.image=inImage;
    }

}   

Answer 1

回溯会有所帮助。哪里崩溃了？

我立即看到的问题是，您在远程加载时转义了 urlString 两次。另外，如果 urlString 碰巧为零，当您尝试将其转换为创建 NSURL 时，您会遇到崩溃。