Use*_*r57 2 javascript php ajax jquery
我想向下面的HTML显示ajax响应,其中最初的0可用插槽可以在ajax响应上更改.并基于数据响应dropdown_wrapper class <li>将自动出现.我如何让它充满活力?
<span class="selected_item">0 available Slots</span>
<div class="form_item_dropdown">
<div class="dropdown_wrapper">
<li class="timeSlot" id="time1">
<i class="icon fa fa-clock-o" aria-hidden="true">
</i>
<span class="item_title">10am - 4pm (value comes through Ajax)</span>
</li>
</div>
</div>
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Ajax代码:
$(".school_title").on("click", function (argument) {
var id = $(this).attr('id');
var info=$.get("{{url('school-slot')}}",{id:id});
info.done(function(data){
console.log(data.length);
/* $.each(data,function(index,subcatObj){
}); */
});
});
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以下是回复:
{id: 3, school_id: 3, break_time: "10am - 8pm"}
{id: 4, school_id: 3, break_time: "10am - 4:30pm"}
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只需更改foreach循环的内容即可.
var divContent = '';
$.each(data,function(index,subcatObj){
divContent += '<li class="timeSlot" id="time'+subcatObj.id+'"><i class="icon fa fa-clock-o" aria-hidden="true"></i><span class="item_title">'+subcatObj.break_time+'</span></li>';
});
$(".dropdown_wrapper").append(divContent);
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