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如题，这是一个检测有没有买票并开启闸机的动画代码

const elevator = world.querySelector('#闸机的门-2')
elevator.fixed = true
elevator.collides = true
const lo = [38, 27.3, 76.9]
const hi = [36.6, 27, 76.9]
const anissss = elevator.animate([
    { position: lo, duration: 1 },
    { position: lo, duration: 3 },
    { position: hi, duration: 1 },
    { position: hi, duration: 3 },
], {
    duration: 16 * 2,
    iterations: 0,
    direction: Box3AnimationDirection.WRAP,
})
anissss.onReady(() => {
    world.say('开启成功')
})
anissss.onFinish(() => {
    world.say('')
})

world.querySelectorAll('#闸机-1').forEach((npcs) => {
    npcs.enableInteract = true
    npcs.interactRadius = 2;
    npcs.interactHint = '闸机';
    npcs.onInteract(async ({ entity }) => {
        const result = await entity.player.dialog({
            type: Box3DialogType.SELECT,
            content: `${entity.player.name}，请问你要`,
            title: '自动售票机',
            options: ['通过', '通你/妈过'],
        })
        if (result.value == '通过') {
            if (entity.sjjak == 1) {
                anissss.play({
                    duration: 16 * 15,
                    iterations: 1,
                    direction: Box3AnimationDirection.WRAP,
                })
            }else{
                world.say(`${entity.player.name} 你还没买票去买票去`)
            }
        }
    })
})

有没有人可以把它改成按下E键后，打开离着玩家最近的那个闸机，而不是只打开一个，求求了awa