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我想出了这个解决方案：

  for(var i=0; i < max; i++){
   
   var note=arrays[i].pitch;
   var time=arrays[i].timing;            
   if(i===0){synth.triggerAttackRelease(note,time,0)}
   if(i===1){synth.triggerAttackRelease(note,time,arrays[0].timing)}
   if(i===2){synth.triggerAttackRelease(note,time,arrays[1].timing+arrays[0].timing)}
   if(i===3){synth.triggerAttackRelease(note,time,arrays[2].timing+arrays[1].timing+arrays[0].timing)}
   if(i===4){synth.triggerAttackRelease(note,time,arrays[3].timing+arrays[2].timing+arrays[1].timing+arrays[0].timing)}
   if(i===5){synth.triggerAttackRelease(note,time,arrays[4].timing+arrays[3].timing+arrays[2].timing+arrays[1].timing+arrays[0].timing)}
   if(i===6){synth.triggerAttackRelease(note,time,arrays[5].timing+arrays[4].timing+arrays[3].timing+arrays[2].timing+arrays[1].timing+arrays[0].timing)}
   if(i===7){synth.triggerAttackRelease(note,time,arrays[6].timing+arrays[5].timing+arrays[4].timing+arrays[3].timing+arrays[2].timing+arrays[1].timing+arrays[0].timing)}
}

但是代码太多，只能弹7个音符左右；

是否有一个简单的快捷方式，只需几行？

4

1 回答 1

1

如果您可以在数组中声明音符及其长度（您似乎正在这样做）。然后，您可以循环播放音符并播放它们，累积时间延迟以计算每个音符从起点开始的正确延迟。

例如

const synth = new Tone.Synth().toDestination();

const notes = [
    { pitch: "C4", timing: 0 },
    { pitch: "D4", timing: 1 },
    { pitch: "E4", timing: 1 },
    { pitch: "F4", timing: 1 },
    { pitch: "G4", timing: 1 }
];

function play() {
    let delay = Tone.now();
    for(let i = 0; i < notes.length; i++) {
        delay += notes[i].timing;
        synth.triggerAttackRelease(notes[i].pitch, '8n', delay);  
    }
}

play() // call this when someone interacts with your program.
于 2021-12-03T04:55:58.450 回答