javascript - 如何使用一个id将多个值传递到模式中

Question

我在JQXgrid中创建了按钮渲染器，当单击按钮时，它将数据传递到 Controller ， Controller 发送到模型，然后返回来自mysql的数据结果。

这是我的 View 代码部分按钮渲染器:

var button_renderer = function (row, columnfield, value, defaulthtml, columnproperties) {
            var kode_keramik = $('#jqxgrid').jqxGrid('getcelltext', row, "kode_keramik");
            button = '<a href="#modal_details" class="btn btn-xs btn-success view_details" id="'+ kode_keramik +'" >Proceed</a>';
            return button;
            };

这是我将数据传递到 Controller 的 View 代码部分:

$(document).on('click', ".view_details", function() {

            //alert("aaa");
            var url = "<?php echo base_url().'getGlazeMM/ajax_get_item_list'?>";
            kode_keramik = this.id;

            $.post(url, {kode_keramik: kode_keramik} ,function(data) {
                $('.modal-body').empty();
                $('.modal-body').append(data);
                $('#modal_details').modal();
            });

        });

这是我的 Controller :

public function ajax_get_item_list(){

      $data['post'] = $_POST;
      $kode_keramik = $_POST['kode_keramik'];
      //$buyer = $_POST['buyer'];

      $this->load->model('get_glaze');

      $data['item_list'] = $this->get_glaze->action_ajax_get_item_list( $data['post'] );

      if ($data['item_list']){

         echo "<table class='table table-bordered'>
                  <tr>
                     <th>Inspect Date</th>
                     <th>Item Code</th>
                     <th>Type</th>
                     <th>Hasil KW1</th>
                     <th>Total Inspek</th>
                     <th>Aktual Yield</th>
                     <th>Buyer</th>
                  </tr>";

         foreach ($data['item_list'] as $key => $value) {

            echo "<tr>";
            echo "<td>".$value['inspect_date']."</td>";
            echo "<td>".$value['item_code']."</td>";
            echo "<td>".$value['sell_type']."</td>";
            echo "<td>".$value['hasil_kw1']."</td>";
            echo "<td>".$value['total_inspek']."</td>";
            echo "<td>".$value['aktual_yield']." %</td>";
            echo "<td>".$kode_keramik."</td>";
            echo "</tr>";

         }

         echo "</table>";

      } else {
         echo "Data tidak ditemukan";
      }

   }

最大的问题是如何从 view_details"id="'+ kode_keramik +'" + SECOND VALUE 传递多个数据

var button_renderer = function (row, columnfield, value, defaulthtml, columnproperties) {
            var kode_keramik = $('#jqxgrid').jqxGrid('getcelltext', row, "kode_keramik");
            button = '<a href="#modal_details" class="btn btn-xs btn-success view_details" id="'+ kode_keramik +'" >Proceed</a>';
            return button;
            }; 

进入:

var url = "<?php echo base_url().'getGlazeMM/ajax_get_item_list'?>";
            kode_keramik = this.id;
            ***SECOND VALUE;***

            $.post(url, {kode_keramik: kode_keramik, ***SECOND VALUE***} ,function(data) {

直到 Controller :

public function ajax_get_item_list(){

      $data['post'] = $_POST;
      $kode_keramik = $_POST['kode_keramik'];
      $***SECOND VALUE*** = $_POST['***SECOND VALUE***'];

Answer 1

您可以将数据格式化为一个 JSON 对象，并将其放入自定义 html 属性中。

示例

请小心单引号和双引号转义。

var button_renderer = function (row, columnfield, value, defaulthtml, columnproperties) {
  var kode_keramik = $('#jqxgrid').jqxGrid('getcelltext', row, "kode_keramik");
  button = "<a href='#modal_details' class='btn btn-xs btn-success view_details' data-custom='{\"kode_keramik\": \"" + kode_keramik + "\", \"second\": \"value\"}'>Proceed</a>";
  return button;
};

使用jQuery检索我们的对象

$(document).on('click', ".view_details", function() {
  //alert("aaa");
  var url = "<?php echo base_url().'getGlazeMM/ajax_get_item_list'?>";

  var obj = $(this).data('custom');   // get object using jQuery

  $.post(url, obj ,function(data) {
      $('.modal-body').empty();
      $('.modal-body').append(data);
      $('#modal_details').modal();
  });

});

它将自动解析对象。