选择列表中有使用图像的方法吗？

Question

我正在使用awesomplete自动完成插件，我尝试添加图像我的选择列表一个月，但我做不到，难道没有任何方式使用图像与自动完成或选择？

我说的是这个：

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<link href="https://cdnjs.cloudflare.com/ajax/libs/awesomplete/1.1.1/awesomplete.min.css" rel="stylesheet"/>

<input class="awesomplete" list="mylist" />
<datalist id="mylist">
	<option>Ada</option>
	<option>Java</option>
	<option>JavaScript</option>
	<option>Brainfuck</option>
	<option>LOLCODE</option>
	<option>Node.js</option>
	<option>Ruby on Rails</option>
</datalist>
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.1.1/jquery.min.js"></script>
<script src="https://cdnjs.cloudflare.com/ajax/libs/awesomplete/1.1.1/awesomplete.min.js"></script>

​

我不想使用内联css，因为它不是跨浏览器的

Answer 1

也支持具有不同标签和价值的建议。标签将显示在自动完成器中，并将该值插入到输入中。

<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.2.0/jquery.min.js"></script>
<link href="https://cdnjs.cloudflare.com/ajax/libs/awesomplete/1.1.1/awesomplete.min.css" rel="stylesheet"/>

<input id="myinput" />


<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.1.1/jquery.min.js"></script>
<script src="https://cdnjs.cloudflare.com/ajax/libs/awesomplete/1.1.1/awesomplete.min.js"></script>
<script>
    var input = document.getElementById("myinput");

    // Show label but insert value into the input:
    new Awesomplete(input, {
        list: [
            { label: "<img src='https://encrypted-tbn0.gstatic.com/images?q=tbn:ANd9GcRUxdD-Q4nIx3uIg9jBCe1oT5a9MHuWY5_pW4FoZSU-nQd1Y_WJPQ'/> Faceboock", value: "https://www.facebook.com/" },
            { label: "<img src='https://hydra-media.cursecdn.com/dota2.gamepedia.com/thumb/2/25/Pounce_icon.png/16px-Pounce_icon.png?version=77c984fc4a9c8ca491ead081322fa738'/> Youtube", value: "https://www.youtube.com/" },
            { label: "China", value: "CN" },
            { label: "United States", value: "US" }
        ]
    });

    // You can search for a better version
    $(document).find('.awesomplete').on('click',function(e)
    {
        if($('#myinput').val())
            window.location = $('#myinput').val();
        //console.log($('#myinput').val());
    });
</script>