Javascript 为什么这行脚本不起作用？

当数据字被调用时，我想要一个声音和一个图像来显示和播放。我试过很多不同的方法，但都不明白

这里有一个方法

for(i = 0; i < ul.children.length; ++i){

listOfWords[ul.children[i].getAttribute("data-word")] = ul.children[i].getAttribute("data-pic", "data audio");

}
 console.log(listOfWords);

for（i=0；i

我也试过

for(i = 0; i < ul.children.length; ++i){

listOfWords[ul.children[i].getAttribute("data-word")] = ul.children[i].getAttribute("data-pic");
listOfWords[ul.children[i].getAttribute("data-word")] = ul.children[i].getAttribute("data-audio");


}
console.log(listOfWords);

for（i=0；i

但没有这样的运气

第二种方法是底部而不是顶部，当调用数据字时，我需要数据音频和数据pic

有人能帮忙吗

HTML

<ul style="display:none;" id="wordlist">
  <li data-word="mum" data-audio="http://www.wav-sounds.com/cartoon/daffyduck1.wav" data-pic="http://pixabay.com/static/uploads/photo/2012/04/13/00/06/head-31117_640.png"></li>
  <li data-word="cat" data-audio="http://www.wav-sounds.com/cartoon/porkypig1.wav" data-pic="http://pixabay.com/static/uploads/photo/2012/05/03/23/13/cat-46676_640.png"></li>
  <li data-word="dog" data-audio="http://www.wav-sounds.com/cartoon/porkypig1.wav" data-pic="http://pixabay.com/static/uploads/photo/2012/05/02/21/14/gray-46364_640.png"></li>
  <li data-word="bug" data-audio="http://www.wav-sounds.com/cartoon/porkypig1.wav" data-pic="http://pixabay.com/static/uploads/photo/2012/04/16/12/17/black-35741_640.png"></li>
  <li data-word="log" data-audio="http://www.wav-sounds.com/cartoon/daffyduck1.wav" data-pic="http://pixabay.com/static/uploads/photo/2012/04/13/11/18/fire-31929_640.png"></li>
  <li data-word="dad" data-audio="http://www.wav-sounds.com/cartoon/daffyduck1.wav" data-pic="http://pixabay.com/static/uploads/photo/2012/04/13/00/05/old-31110_640.png"></li>

类似这样的内容：

for(i = 0; i < ul.children.length; ++i){
   listOfWords[ul.children[i].getAttribute("data-word")] = {
         "pic" : ul.children[i].getAttribute("data-pic"),
         "audio" : ul.children[i].getAttribute("data-audio")
   };
}

使用jQuery：

var listOfWords = {};
$("#wordlist li").each(function() {
    var $item = $(this);
    listOfWords[ $item.attr("data-word") ] = {
        pic : $item.attr("data-pic"),
        audio : $item.attr("data-audio")
    };
});

使用jQuery，您可以使用来迭代列表项并检索数据属性值

var listOfWords = [];
$("#wordlist li").each(function(index, item) {
  // where liData is { word: "a", audio: "b", pic: "c" }
  var liData = $(item).data();
  listOfWords[liData.word] = liData;

  console.log("word: " + liData.word + ", audio: " + liData.audio + ", pic: " + liData.pic);
});

---

代码中没有jQuery。你想用jQuery获得答案吗？我想在你完成文字游戏后看看：-DIt不起作用，因为你没有按照我说的从

listowords

中检索

pic

和

audio

值，你只复制了将值放入其中的部分。当您想要与

rndWord

关联的音频时，您需要说

listOfWords[rndWord].audio

，同样，pic将是

listOfWords[rndWord].pic

。

var listOfWords = [];
$("#wordlist li").each(function(index, item) {
  // where liData is { word: "a", audio: "b", pic: "c" }
  var liData = $(item).data();
  listOfWords[liData.word] = liData;

  console.log("word: " + liData.word + ", audio: " + liData.audio + ", pic: " + liData.pic);
});