这是我想要显示从ajax数据中获取的值的模态,但它不起作用。我在span元素上有id="view_name"
。我不知道它不会出现什么问题
这是触发fun_view
功能的代码:
<td>
<a href="" data-id="" class="display" data-toggle="modal" data-target="#myModal" onclick="fun_view('{{ $dat->id }},{{ $dat->name }},{{ $dat->company_name }}, {{ $dat->mobile_phone }}, {{ $dat->timezone }}, {{ $dat->best_time_to_call }}')"><?php echo $dat->name;?></a>
</td>
<div class="modal-body">
<p><b>Name : </b><span id="view_name" class="text-success"></span></p>
<p><b>Company Name : </b><span id="view_company_name" class="text-success"></span></p>
<p><b>Mobile Phone : </b><span id="view_mobile" class="text-success"></span></p>
</div>
<script>
function fun_view(id,name,company_name,mobile_phone,timezone,best_time_to_call)
{
var view_url = $("#hidden_view").val();
$.ajax({
url: view_url,
type:"GET",
data:
{
"id": id,
"name": name,
"company_name": company_name,
"mobile_phone": mobile_phone,
"timezone": timezone,
"best_time_to_call": best_time_to_call
},
success: function(result){
//when I alert(result.name) it says undefined
$("#view_name").text(result.name);
$("#view_company_name").text(result.company_name);
$("#view_mobile").text(result.mobile_phone);
}
});
}
</script>
答案 0 :(得分:0)
添加数据类型只是为了确保ajax响应在JSON对象中 而且我想知道选择器$('#hidden_view')也许你有一些隐藏它的CSS。
$。AJAX({
url: view_url,
type:"GET",
datatype: 'JSON',
data: {"id":id, "name":name, "company_name":company_name, "mobile_phone":mobile_phone, "timezone":timezone, "best_time_to_call":best_time_to_call },
success: function(result){
//when i alert(result.name) it says undefined
$("#view_name").text(result.name);
$("#view_company_name").text(result.company_name);
$("#view_mobile").text(result.mobile_phone);
}
});
尝试以上代码
答案 1 :(得分:-1)
您可能忘记包含对JQuery脚本的引用。
<script
src="https://code.jquery.com/jquery-3.2.1.min.js"
integrity="sha256-hwg4gsxgFZhOsEEamdOYGBf13FyQuiTwlAQgxVSNgt4="
crossorigin="anonymous"></script>