谢谢你路过我真的是一个新手,我正在尝试(通过编写一些代码,希望那没什么问题)5个不同的玩家,并且我使用了以下代码:https://codepen.io/katzkode/pen/ZbxYYG
我正在尝试将以下循环分为5个不同的div(5个不同的播放器,而不是#audio-players,每个播放器都有一个新的div):
function createAudioPlayers() {
for (f in files) {
var playerString = "<div id=\"audioplayer-" + f + "\" class=\"audioplayer\"><button id=\"playbutton-" + f + "\" class=\"play playbutton\"></button><div id=\"timeline-" + f + "\" class=\"timeline\"><div id=\"playhead-" + f + "\" class=\"playhead\"></div></div></div>";
$("#audio-players").append(playerString);
}
}
这是从中提取文件的代码:
var files = ["interlude.mp3", // 0
"chooseyourweapon.mp3", // 1
"interlude.mp3", // 2
"scriptures.mp3",
"scriptures.mp3"// 3
];
而且我不确定该如何进行,我仍在学习。谢谢。
最佳答案
似乎您只需要循环数组。使用forEach
为便于理解
function createAudioPlayers() {
files.forEach(function(f) {
var playerString = "<div id=\"audioplayer-" + f + "\" class=\"audioplayer\"><button id=\"playbutton-" + f + "\" class=\"play playbutton\">" + f + "</button><div id=\"timeline-" + f + "\" class=\"timeline\"><div id=\"playhead-" + f + "\" class=\"playhead\"></div></div></div>";
$("#audio-players").append(playerString);
})
}
var files = ["interlude.mp3", // 0
"chooseyourweapon.mp3", // 1
"interlude.mp3", // 2
"scriptures.mp3",
"scriptures.mp3" // 3
];
createAudioPlayers()
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div id="audio-players"></div>