我一直在尝试使用ajax的成功函数创建一个复选框, 表中的所有数据都已正确创建,但是我的输入复选框未出现在表中,但是如果我更改为按钮,则该按钮有效,只是该复选框不起作用。 我该如何解决? 谢谢。
function createTableByJqueryEach(data)
{
var eTable=`<table class="table">
<thead class="thead-light"><tr>
<th scope="col">Nome da Instituição</th>
<th scope="col">Categoria</th>
<th scope="col">UF</th>
<th scope="col">Responsável</th>
<th scope="col">E-mail</th>
<th scope="col">Nota</th>
<th scope="col">IAD Conh.</th>
<th scope="col">IAD Apli.</th>
<th scope="col">Enviar Ubc</th>
</tr>
</thead><tbody>`
$.each(data,function(index, row){
eTable += "<tr>";
$.each(row,function(key,value){
if(value['ubc'] !== undefined){
eTable += "<tr><td>"+value['ubc']['name']+"</td><td>" + value['ubc']['partner']['title'] + "</td><td>"+value['ubc']['uf']+"</td><td>"+
value['ubc']['coordinator']+"</td><td>"+value['ubc']['email']+"</td><td>"+ parseFloat(value['nf'].toFixed(3)) +"</td><td>"+
parseFloat(value['ap_grade'].toFixed(1))+"</td><td>"+parseFloat(value['kn_grade'].toFixed(1))+"</td><td><input type="checkbox"></td>";
}
});
eTable += "</tr>";
});
eTable +="</tbody></table>";
$('#eachTable').html(eTable);
}
$('#taxo_id').click(function(){
var radio_option = $("input[type='radio']:checked");
var ubcs = JSON.parse('{{ubcs}}')
$.ajax({
url:"/api/v1/oportuniy/diogoreturn/",
type:"GET",
data: {
option:radio_option.val(),
oportunity: '{{oportunity.id}}',
filter:'{{filter_oportunity.id}}',
ubcs:ubcs
},
dataType:'json',
success: function(data){
createTableByJqueryEach(data);
},
error: function(data){
swal({
title:'Atenção!',
text:data['responseJSON']['error'],
icon: "warning",
buttons: true,
dangerMode: true,
});
}
});
});