Question

当我单击编辑按钮时。可编辑的数据将传递到相关的文本框中进行编辑。但没有显示数据。到目前为止我尝试过的东西附在下面。

Response.Write（JsonConvert.SerializeObject（employees））;一个东西非静态字段需要参考

这是一个按钮

  {
     "sTitle": "Edit",
     "mData": "id",
     "render": function (mData, type, row, meta) {
      return '<button class="btn btn-xs btn-success" 
      onclick="get_category_details(' + mData + ')">Edit</button>';
   }

如果我单击“编辑”按钮，则ID将成功传递给get_category_details方法。然后发布到edit_return.aspx / doSome页面但未检索到数据。我认为出现错误Edit_retun.aspx。

 function get_category_details(id) {
            $.ajax({
                type: 'POST',
                url: 'edit_return.aspx/doSome',
                dataType: 'JSON',
                data: "{id: '" + id + "'}",
                contentType: "application/json; charset=utf-8",
                success: function (data) {                            

                    $("html, body").animate({ scrollTop: 0 }, "slow");
                    isNew = false
                    id = data.id     
                    $('#fname').val(data.fname);
                    $('#age').val(data.age);               

                },
                error: function (xhr, status, error) {
                    alert(xhr.responseText);            
                }

            });
        }

Edit_retun.aspx

public class Employee
        {

            public string id { get; set; }
            public string fname { get; set; }
            public string age { get; set; }
        }

 [WebMethod]
            public static string doSome(int id)
            {
                SqlConnection con = new SqlConnection("server=.; Initial Catalog = jds; Integrated Security= true;");
                string sql = "select * from record where id='" + id + "'";
                SqlCommand cmd = new SqlCommand(sql, con);
                con.Open();
                cmd.ExecuteNonQuery();
                DataTable dt = new DataTable();
                SqlDataAdapter da = new SqlDataAdapter(cmd);
                da.Fill(dt);
                List<Employee> employees = new List<Employee>();

                employees = dt.AsEnumerable()
                        .Select(x => new Employee()
                        {
                            id = x.Field<int>("id").ToString(),
                            fname = x.Field<string>("name"),
                            age = x.Field<int>("age").ToString(),
                        }).ToList();
                Response.Write(JsonConvert.SerializeObject(employees));
                return JsonConvert.SerializeObject(employees);
            }

HTML表单

  <form  id="frmProject" runat="server">
            <div>
                <label class="form-label">First Name</label>     
               <input type="text" id="fname" name="fname" class="form-control"  required />

            </div>
            <div class="form-group" align="left">
             <label class="form-label">Age</label>
                <input type="text" id="age" name="age" class="form-control"  required />
            </div>
            <div> 
               <input type="button" id="b1" value="add" class="form-control" onclick="addProject()" />

            </div      
        </form>

Answer 1

您不能在Response.Write方法中使用static。您的doSome方法必须是non-static方法。或者，您可以从您的方法中删除Response.write，因为我猜这里不需要。

Answer 2

 [WebMethod]
    public List<Employee> doSome(int id)
    {
        SqlConnection con = new SqlConnection("server=.; Initial Catalog = jds; Integrated Security= true;");
        string sql = "select * from record where id='" + id + "'";
        SqlCommand cmd = new SqlCommand(sql, con);
        con.Open();
        cmd.ExecuteNonQuery();
        DataTable dt = new DataTable();
        SqlDataAdapter da = new SqlDataAdapter(cmd);
        da.Fill(dt);
        List<Employee> employees = new List<Employee>();

        employees = dt.AsEnumerable()
                .Select(x => new Employee()
                {
                    id = x.Field<int>("id").ToString(),
                    fname = x.Field<string>("name"),
                    age = x.Field<int>("age").ToString(),
                }).ToList();
        return employees;
    }

Your List<Employee> will be serialize automatically no need to serialize.
In ajax success 
  $.ajax({

            success: function(data){
          data=data.d;
     // now data is JavaScript array of employee object 
},

        });

使用Asp.net ajax

2 个答案: