Question

我有一个modal（bootstrap弹出窗口）包含一个表。现在，当我调用ajax并获得成功响应时，我希望得到弹出窗口的响应并将其显示在表格中，一切正确但我的图像没有显示，有什么问题？

的Ajax：

function detailOrder($id){
    var url= "{{ URL::to('admin/detail_order')}}"+"/"+$id;
    $.ajax({
        type:"get",
        url:url,
        data:"",
        success:function(response){
            console.log(response);
            $('#order').modal('show');
            var trHTML = '';
            $.each(response, function (i, detail_orders) {
                for (i = 0; i < response.order[0].detail_orders.length; i++) {
                    for(j = 0;j<response.order[0].detail_orders[i].product.length;j++){
                        trHTML +=
                        '<tr><td>'
                        +'<img class="table_image" id="pic">'
                        + '</td><td>'
                        + response.order[0].detail_orders[i].product[j].product_name
                        + '</td><td>'
                        + response.order[0].detail_orders[i].quantity 
                        + '</td><td>'
                        + response.order[0].detail_orders[i].product[j].price
                        + '</td><td>'
                        + response.order[0].detail_orders[i].total_amount
                        + '</td></tr>';
                        $("#pic").attr("src","{{ URL::to('admin/storage')}}"+"/"+response.order[0].detail_orders[i].product[j].product_image); 
                    }
                    $('#tBody').append(trHTML);
                }
            });
        },
        error:function(){
            console.log("fail");
        }
    })
}

Answer 1

试试这个

success:function(response){
   var html = ''; //initialize your html
   for(i=0;i<response.length;i++) //loop depends on the length of response
   {
      html+= //concatinate the html data in this variable
        '<tr>'+
            '<td>'+response[i].id+'</td>'+
        '</tr>';
   }
   $('#tBody').html(html); //pass the html data to your body
}

如何从ajax获取laravel 5.2中modal（bootstrap）刀片的数据

1 个答案: