如何将jqgrid的值保存在列表数组中

本文关键字：存在列表数组保存 jqgrid | 更新日期: 2023-09-27 18:28:12

我已经成功实现了jqgrid，我对将所选数据发布到控制器有一些问题

如何保持对所选行的跟踪意味着如何将数据保存在列表数组中

如何返回数据表以再次显示结果，如所选行数和所选行值的总和等。请提供帮助。我正在发布我的jquery代码以及控制器代码。提前谢谢。

视图：

onSelectRow: function (id, status)
    {
        alert('polo');
        var rowData = jQuery(this).getRowData(id);
        FirstName = rowData['FirstName'];
        LastName = rowData['LastName'];
        Salary = rowData['Salary'];
        Gender = rowData['Gender'];
        $.ajax({
            url: '/TodoList/notCk_Pk',
            data:{'FirstName':FirstName,'LastName':LastName,'Salary':Salary,'Gender':Gender},
        type: "post"
        })
        $('#example').dataTable({
            "bProcessing": true,
            "bServerSide": true,
            "sAjaxSource": "/TodoList/notCk_Pk",
            "sAjaxDataProp": "",                
            "columns": [
                       { "data": "FirstName" },
                       { "data": "LastName" },
                       { "data": "Salary" },
                       { "data": "Gender" },
            ]
        });
}

控制器代码：

public ActionResult notCk_Pk(String FirstName,String LastName,int Salary,String Gender)
    {                            
        l.Add(FirstName);
        l1.Add(LastName);
        i = Salary + i;
        l2.Add(Gender);
        string ConnectionString = ConfigurationManager.ConnectionStrings["EmployeeContext"].ConnectionString;
        using (SqlConnection connection = new SqlConnection("data source=.; database=Srivatsava; integrated security=SSPI"))
        {
            connection.Open();
            SqlCommand com = new SqlCommand("insertinto", connection);
            com.CommandType = CommandType.StoredProcedure;
            com.Parameters.AddWithValue("@FirstName", FirstName);
            com.Parameters.AddWithValue("@LastName",LastName);
            com.Parameters.AddWithValue("@Salary", Salary);
            com.Parameters.AddWithValue("@Gender", Gender);
            com.ExecuteNonQuery();
            connection.Close();
        }
        var todoListsResults = l.Count();
        var t1=l1.Count();
        var t3=l2.Count();
        var aaData=new{
                       todoListsResults,
                       t1,
                       i,
                       t3
                   };
        return Json(aaData, JsonRequestBehavior.AllowGet);
        //Console.WriteLine(FirstName + "" + LastName + "" + Salary + "" + Gender);
        //return Content("<script language='javascript' type='text/javascript'>alert("+FirstName+");</script>");
    }

每次我选择一行时，它都会将I变量初始化为"0"，即使我将其设置为全局变量。

如何将jqgrid的值保存在列表数组中

您不需要使用列表数组。您可以使用$("#ListGrid").jqGrid('getGridParam', 'selarrrow')获取所有选定的行。下面的代码片段可能会对您有所帮助。

onSelectRow: function(id, status) {
     var currentRow = $(this).getRowData(id);
     FirstName = currentRow['FirstName'];
     LastName = currentRow['LastName'];
     Salary = currentRow['Salary'];
     Gender = currentRow['Gender'];
    $.ajax({
        url: '/TodoList/notCk_Pk',
        data:{'FirstName':FirstName,'LastName':LastName,'Salary':Salary,'Gender':Gender},
        type: "post"
    })
    var rows = $("#ListGrid").jqGrid('getGridParam', 'selarrrow');
    var totalRow = rows.length;
    var totalAmount = 0;
    $.each(rows, function() {
      var rowData = $("#ListGrid").getRowData(this);
      totalAmount += rowData["Salary"] * 1;
    });
    dataTable.row($('#example').find('tbody tr')).remove().draw();
    dataTable.row.add([totalRow, totalAmount]).draw();
}

如何将jqgrid的值保存在列表数组中

演示