Refreshing UITableViewCell after fetching image data by dispatch_async - ios

Question

I'm writing app that shows TableView with entries that contains images.
I'm trying to fetch image by executing this line of code inside cellForRowAtIndexPath method:
cell.detailTextLabel.text = [artistData objectForKey:generesKey];
dispatch_async(backgroundQueue, ^{
NSURL *url_img = [NSURL URLWithString:[artistData objectForKey:pictureKey]];
NSData* data = [NSData dataWithContentsOfURL:
url_img];
cell.imageView.image = [UIImage imageWithData:data];
[self performSelectorOnMainThread:#selector(refreshCell:) withObject:cell waitUntilDone:YES];
});
After setting image I perform selector that contains:
-(void)refreshCell:(UITableViewCell*)cell{
[cell setNeedsDisplay];
[self.view setNeedsDisplay];
[self.tableViewOutlet setNeedsDisplay];
}
And image is not shown but when I click on cell or scroll entire list, images are shown. Why my View is not refreshing? Did I missed something?

Answer 1

It should be enough to call setNeedsLayout() on a cell.
In swift 4 it looks like this:
DispatchQueue.global().async {
let data = try? Data(contentsOf: URL(string: imageUrl)!)
DispatchQueue.main.async {
cell.imageView?.image = UIImage(data: data!)
cell.setNeedsLayout()
}
}

Answer 2

You'll could always reload the cell by calling [self.tableView reloadRowsAtIndexPaths#[indexPath] withRowAnimation:UITableViewRowAnimationAutomatic];
In order to prevent the infinite loop once you successfully download an image you'll want to cache the results. How long you cache is up to you.
NSCache *imageCache = [[NSCache alloc] init];
imageCache.name = #"My Image Cache";
UIImage *image = [imageCache objectForKey:url_img];
if (image) {
cell.imageView.image = image;
} else {
// Do your dispatch async to fetch the image.
// Once you get the image do
[imageCache setObject:[UIImage imageWithData:data] forKey:url_img];
}
You'll want the imageCache to be a property at the ViewController level. Don't create one each time incellForRowAtIndexPath

Answer 3

It might be related to interacting with the UI from a background queue. Try this:
dispatch_async(backgroundQueue, ^{
NSURL *url_img = [NSURL URLWithString:[artistData objectForKey:pictureKey]];
NSData* data = [NSData dataWithContentsOfURL:url_img];
dispatch_async(dispatch_get_main_queue(), ^{
cell.imageView.image = [UIImage imageWithData:data];
});
});