Flutter Audio播放器出现意外错误

Question

我面临一个问题，它说。这个表达式的类型是"void"，所以不能使用它的值。我不知道如何解决这个

await audioPlayer.play(url) is This expression has a type of 'void' so its value can't be used.

playMusic(String url) async {
if(isPlaying && currentSong != url){
audioPlayer.pause();
int result = await audioPlayer.play(url);
if (result == 1){
setState(() {
currentSong = url;
});
}
}else if(!isPlaying){
// int result = 1;
int result = await audioPlayer.play(url);
if (result == 1){
setState(() {
isPlaying = true;
});
}
}

}

Answer 1

play方法获取一个源，并返回Future<void>。因此，你无法得到任何结果，只能用await来完成未来。

await audioPlayer.play(UrlSource(url));