Better way to do skip to previous with AVQueuePlayer?

does the queue handle more than one skip forward smoothly? if so, you could constantly re-insert the previous video back into the queue at index n+2. when the user wishes to play the previous track, you would skip forward twice.

if playing from track A to F without any skips, the pattern would look like this:

A B C D E F
B C D E F

// re-insert A after next track
B C A D E F
C A D E F

// remove A then re-insert B
C D E F
C D B E F
D B E F

// remove B then re-insert C
D E F
D E C F
E C F

// remove C then re-insert D
E F
E F D
F D

// remove D then re-insert E
F
FE

using this pattern you could only smoothly skip backwards once, but it could be modified to allow more.

definitely not an ideal solution, but may work!

It seems like AVQueuePlayer removes the current item from the play queue when calling advanceToNextItem. Theoretically, there is no way to get this item back without rebuilding the queue.

What you could do is use a standard AVPlayer, have an array of AVPlayerItems, and an integer index which keeps the index of the current track.

Swift 3:

let player = AVPlayer()
let playerItems = [AVPlayerItem]() // your array of items
var currentTrack = 0

func previousTrack() {
    if currentTrack - 1 < 0 {
        currentTrack = (playerItems.count - 1) < 0 ? 0 : (playerItems.count - 1)
    } else {
        currentTrack -= 1
    }

    playTrack()
}

func nextTrack() {
    if currentTrack + 1 > playerItems.count {
        currentTrack = 0
    } else {
        currentTrack += 1;
    }

    playTrack()
}

func playTrack() {

    if playerItems.count > 0 {
        player.replaceCurrentItem(with: playerItems[currentTrack])
        player.play()
    }
}

Swift 2.x:

func previousTrack() {
    if currentTrack-- < 0 {
        currentTrack = (playerItems.count - 1) < 0 ? 0 : (playerItems.count - 1)
    } else {
        currentTrack--
    }

    playTrack()
}

func nextTrack() {
    if currentTrack++ > playerItems.count {
        currentTrack = 0
    } else {
        currentTrack++;
    }

    playTrack()
}

func playTrack() {

    if playerItems.count > 0 {
        player.replaceCurrentItemWithPlayerItem(playerItems[currentTrack])
        player.play()
    }
}