当前位置: 代码迷 >> Java Web开发 >> easyui post值的有关问题
  详细解决方案

easyui post值的有关问题

热度:345   发布时间:2016-04-13 22:45:20.0
easyui post值的问题
我是在datagrid中combobox进行级联的,当第一个combobox选择值后,在第二个combobox中将这个值post出去,以便取取得第二个combobox的值,代码如下:GET方法我知道什么做
 { field: 'no', title: '编号', width: 50, editor: 'numberbox' },
                  {
                      field: 'DEPATMENT', title: '名称', width: 60,
                      formatter: function (value, row) {
                          return row.DEPATMENT;
                      },
                      editor: {
                          type: 'combobox',
                          options: {
                              valueField: 'DEPATMENT',
                              textField: 'DEPATMENT',
                              method: 'POST',
                              url: '../ashprama/test2.ashx',
                              onBeforeLoad: function (param) {  
                                  param.action = 'depatment';
                              },  

                              required: true,
                              onSelect: function (record) {
                                  var rows = $('#tt').datagrid("getSelected");
                                  var rowIndex = $('#tt').datagrid('getRowIndex', rows);
                                  var target = $('#tt').datagrid('getEditor', { index: rowIndex, field: 'user_name' }).target;
                                  target.combobox('clear');
                                  target.param.depatmentid=record.DEPATMENT//这里什么写才能将参数post过去
                                  target.combobox('reload', '../ashprama/test3.ashx');
                              }
                          }
                      }
                  },

                  {
                      field: 'user_name', title: '备注', width: 80, formatter: function (value, row) {
                          return row.bt;
                      },
                      editor: {
                          type: 'combobox',
                          options: {
                              valueField: 'user_name',
                              textField: 'user_name',
                              required: false,
                          }
                      }

                  },

由于没有分了,只能给20分了
------解决思路----------------------
都是url把参数带过去的啊
为什么要post?你想刷新页面啊?用ajax的话也没有url直接带参数写起来方便啊
  相关解决方案