There's no doc about playing a lot of audios at same time.

**Your Godot version:**
3.2.4 RC1

**Issue description:**
Altought the docs talk about how WAV file format can play thousands of it without any performance issue, the docs doesnt says how this can be archieved, I finally manage to found a tutorial from a user who did it right, but that is not begginer friendly and I couldnt understand it well too.

Maybe there's a better approach? In any case this should be documented, playing a lot of sounds in a little space of time is not simple!

**Source Content**

https://www.youtube.com/watch?v=DsAIrA9UE4E
https://www.youtube.com/watch?v=3CFWlQzMthY

**Source Code**

```gdscript
extends Node

var dic : Dictionary = {}

func play_sfx(audio_clip : AudioStream, priority : int = 0):
for child in $sfx.get_children():
if child.playing == false:
child.stream = audio_clip
child.play()
dic[child.name] = priority
break

if child.get_index() == $sfx.get_child_count() - 1:
var priority_player = check_priority_and_find_oldest(dic, priority) #finds player with same/lowest priority and oldest player
if priority_player != null:
$sfx.get_node(priority_player).stream = audio_clip
$sfx.get_node(priority_player).play()
else:
print("priority player is null")
pass

#playes at most 3 sounds at the same time, in a lot of cases bad, because you get less sound feedback.
#execept when you want to restrict the amount of sounds, for example crashes/destructions/debris
func check_priority(_dic : Dictionary, _priority):
var prio_list : Array = []

for key in _dic:
if _priority > _dic[key]:
prio_list.append(key)#append key(sfx_player.name) to the array

#get the lowest priority from prio_list
var last_prio = null
for key in prio_list:
if last_prio == null:
last_prio = key
continue
if _dic[key] < _dic[last_prio]:
last_prio = key
return last_prio
pass

#playes new sounds all the time, bad if you have important sound, wich you liked to play
func find_oldest_player():
var last_child = null

for child in $sfx.get_children():
if last_child == null:
last_child = child
continue
#find player wich played the longest
if child.get_playback_position() > last_child.get_playback_position():
last_child = child

return last_child.name
pass

#good for all types of situations, important sound get played most of the time and sounds doesn't get get
#swallowed up most of the time
func check_priority_and_find_oldest(_dic, _priority): #1,3,1 == 1
var prio_list : Array = []
for key in _dic:
if _priority >= _dic[key]:
prio_list.append(key) #append key(sfx_player.name) with same/lower priority to an array

#find oldest
if prio_list.empty():
return null
var oldest_player = prio_list[0]
for i in range(1, prio_list.size() -1):
if $sfx.get_node(oldest_player).get_playback_position() < $sfx.get_node(prio_list[i]).get_playback_position():
oldest_player = prio_list[i]

return oldest_player
pass

###### Here is a little coding challenge #######
#func lowest_priority_and_oldest(_dic, _priority):
#make prio_list
#append same/lower priority to prio_list
#get the lowest priority
#get all player with the same lowest priority
#if there is more then 1 player
#get the oldest player from the lowest priority players

func play_music(music_clip : AudioStream):
$music/music_player.stream = music_clip
$music/music_player.play()
pass

```