使用AVQueuePlayer跳到上一个的更好方法吗? [英] Better way to do skip to previous with AVQueuePlayer?

问题描述

我在我的应用程序中使用了AVQueuePlayer.我有两个滑动手势，可以跳到下一个和上一个avplayeritems.现在要跳到下一个，我只是在avqueueplayer上调用了AdvanceToNextItem，效果很好.

I am using an AVQueuePlayer in my app. I have a two swipe gestures to skip to next and skip to previous avplayeritems. Right now to do skip to next I am just calling advanceToNextItem on the avqueueplayer which works well.

但是，我跳到上一个视频时，我要删除所有项目并将它们添加回去，所以当多次跳到上一个视频时，这确实很慢.我怎样才能像调用advanceToNextItem那样更快?

However the skip to previous I am removing all items and adding them back in with the previous video up front, this is really slow when skipping to previous multiple times. How can I make this faster just like calling advanceToNextItem?

我的代码如下:

func skipToPrevious() {
    queuePlayer.removeAllItems()
    // move the previous playerItem to the front of the list then add them all back in
    for playerItem in playerItems:
        queuePlayer.insertItem(playerItem, afterItem: nil)

}

推荐答案

在调用advanceToNextItem时，似乎AVQueuePlayer从播放队列中删除了当前项目.从理论上讲，不重建队列就无法取回该项目.

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.

您可以做的是使用标准的AVPlayer，具有AVPlayerItems的数组以及保留当前曲目索引的整数索引.

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.

迅速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:

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()
    }
}

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