Question

I’m having a hard time understanding some of the logic behind the UITableView. I am populating my table from a MPMediaItemCollection as a queue of songs. What I am trying to achieve is to have a now playing indicator as the accessory view of the cell at the index path that matches the currently playing song.

I originally tried this with the following:

if (indexPath.row == [mutArray indexOfObject:[mainViewController.musicPlayer.nowPlayingItem valueForProperty:MPMediaItemPropertyTitle]]) {        
    UIImageView *playButtonView = [[UIImageView alloc] initWithImage:[UIImage imageNamed:@"PlayButton.png"]];
    [cell setAccessoryView:playButtonView];
}

This works fine for the now playing item, which is always actually objectAtIndex:0. But what I don’t understand is why my table seems to define this index once every height of combined visible cells.

Let me try to explain this more clearly. Lets say that my table is 600px tall, and its content height is 1800px. This causes the indicator to be added roughly every 600px down the content height.

Now my first thought was that this was something wrong with the code for judging the index based off the name of the song, so I tried changing it to:

if (indexPath.row == 0)

But this produces the same result!

This screenshot should help explain what I’m talking about.

So, is there anything I can do to make the table treat indexPath0 as only the first cell in reference to the entire table instead of in reference to the currently visible cells?

Answer 1

You have to state explicitly also when the accessory should not be there:

if (indexPath.row==0) {
   UIImageView *playButtonView = [[UIImageView alloc] initWithImage:
     [UIImage imageNamed:@"PlayButton.png"]];
   [cell setAccessoryView:playButtonView];
} 
else {
   [cell setAccessoryView:nil];
}

The reason is that when cell 0 gets dequeued (i.e. reused) on a different row it still has the accessory view in it.

Actually, this is a nice case study for understanding how dequeueing table view cells actually works ;-).