如何使用 codeigniter 在 ajax 数据表中进行特定响应？

Question

这是我在视图文件中的数据表代码 (Codeigniter 3)

function trade_Datatable()
{
    $.ajax({
        url:'<?php echo base_url('user/get_trade_manage'); ?>',
        type:'POST',

        success:function(data)
        {
            var dataTable=$('#trade_management').DataTable({
            'searching' : false,
            'ordering'  : false,
            'processing': true,
            'serverSide': true,
            'ajax':'<?php echo base_url('user/get_trade_manage'); ?>',
            
            "rowCallback": function( row,response) {

                $('td', row).css('background-color', '#00b0f0');
                
                },

            });
        }

    });
}
trade_Datatable();

控制器代码

......
$output['draw']=$draw;
$output['recordsTotal']=$counts;
$output['recordsFiltered']=$counts;
$output['data']=$data;
$output['colors']=$bgcolor;
echo json_encode($output);

我需要回显 $bgcolor 变量 #00b0f0 以应用样式 dynamically.Anyone 建议最佳答案？

Answer 1

用户为 ajax 数据表中的应用样式创建了行

$.ajax({
        url:'<?php echo base_url('user/get_trade_manage'); ?>',
        type:'POST',
        success:function(data)
        {
            var dataTable=$('#trade_management').DataTable({
            'searching' : false,
            'ordering'  : false,
            'processing': true,
            'serverSide': true,
            'ajax':'<?php echo base_url('user/get_trade_manage'); ?>',
    
            columns:[
                {data: 'sno'},
                {data: 'created'},
                {data: 'updatedon'},
                {data: 'tip_type'},
                {data: 'tip_for'},
                {data: 'script_name'},
                {data: 'tip_date'},
                {data: 'tip_position'},
                {data: 'tip_price'},
                {data: 'action'},
                {data: 'tip_exitat'},
                {data: 'photo'},
                {data: 'profit_loss'},  
                ],
    
                createdRow: function(row, rowColors, dataIndex)
                {
                    var colorRow = rowColors.colors;
                    $(row).css('background-color',colorRow);
                }
    
                });
            }
        });

Answer 2

应该在 createdRow 函数中传递三个参数以获取 json 值

var table = $('#emergency_appointment_list_test').DataTable({
                "ordering": true,
                "processing": true, 
                "serverSide": true, 
                "bDestroy": true,
                "order": [], 
                "language": {
                    "paginate": {
                        "previous": '<i class="fa fa-arrow-left" aria-hidden="true"></i>',
                        "next": '<i class="fa fa-arrow-right" aria-hidden="true"></i>'
                    }
                },
    
                "ajax": 
                {
                    "url": baseurl+"emergency_app_list_test",
                    "type": "POST",
                    "data": 
                    {
                        "_token": $('#token').val(),
                    },
                    
                    
                },
                "columns": [
                    { "data": "id" },
                    { "data": "applicant_name" },
                    { "data": "applicant_passport_no" },
                    { "data": "serviceapplied" },
                    { "data": "apptmnt_reason1" },
                    // { "data": "appointment_status" },
                    // { "data": "action" },
                    { "data": "view" },
                ],
                createdRow: function(row, rowColors, dataIndex)
                {
                    if(rowColors.emg_status == 1)
                    {
                        $(row).css('background-color','red');
                    }else
                    {
                        $(row).css('background-color','');
                    }
                }
                  
            });

如何使用 codeigniter 在 ajax 数据表中进行特定响应？

how to specific response in ajax datatable using codeigniter?

php

ajax

codeigniter