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ArgumentError: Error #2068: Invalid sound. at flash.media::Sound/play() at animacion2_copia_fla::M

Community Beginner ,
Jun 08, 2024 Jun 08, 2024

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import flash.media.Sound;
import flash.media.SoundChannel;
button_9.addEventListener(MouseEvent.CLICK, resumeSound);
var sonido1: Sound = new Sound(); 
var soundChannel: SoundChannel;
function resumeSound(event: MouseEvent): void {
if (!soundChannel) {
soundChannel = sonido1.play();
}}
kathe37922242rzrk_1-1717863983411.pngkathe37922242rzrk_2-1717864096855.png

Pueden ayudarme? nose porque sucede este error

se supone que la dar click en el boton deberia seguir reproduciendo el audio pero no lo hace nose que estoy haciendo mal,  Estoy trabajando en adobe animate 2023 

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correct answers 1 Correct answer

Community Expert , Jun 08, 2024 Jun 08, 2024

Hi.

Adding to k's answer, the sound declaration should be something like:

var sd:sonido1 = new sonido1();


In your case, you're using sonido1 as the variable name but you should use as the class name.

Regards,
JC

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Community Expert ,
Jun 08, 2024 Jun 08, 2024

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do you have a sound in your library that has a class assigned?  if so use it, not new Sound();

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Community Beginner ,
Jun 08, 2024 Jun 08, 2024

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si el sonido es: sonido1 

kathe37922242rzrk_0-1717865864546.png

pero no se reproduce al dar click en el boton 

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Community Beginner ,
Jun 08, 2024 Jun 08, 2024

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este es actionScript del sonido 

kathe37922242rzrk_0-1717866052635.png

cuando le doy a aceptar para guadar cambios me aparece este mesaje es normla? 

kathe37922242rzrk_1-1717866129053.png

 

 

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Community Expert ,
Jun 08, 2024 Jun 08, 2024

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Hi.

Adding to k's answer, the sound declaration should be something like:

var sd:sonido1 = new sonido1();


In your case, you're using sonido1 as the variable name but you should use as the class name.

Regards,
JC

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Community Beginner ,
Jun 08, 2024 Jun 08, 2024

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una ultima pregunta si quiero hacer que el sonido se reprodusca es correcto utilizar esta funcion

function resumeSound(event: MouseEvent): void {
if (!soundChannel) {
soundChannel = sonido1.play();}}

kathe37922242rzrk_0-1717867542933.png

 

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Community Expert ,
Jun 08, 2024 Jun 08, 2024

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Now you have to use the variable name. In the example I gave, It would be sd.play();.

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Community Beginner ,
Jun 08, 2024 Jun 08, 2024

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si, me di cuenta de eso muchas gracias, pero aun no se reproduce el audio por alguna razon :c, aunque ya no me sale ningun error 

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Community Expert ,
Jun 08, 2024 Jun 08, 2024

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browser's block sound unless there's user interaction, eg a mouse click.

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