使用AJAX(ASP.NET MVC)将数据插入数据库

时间:2017-03-21 10:24:08

标签: c# jquery asp.net ajax asp.net-mvc

我有问题1的表 - 问题10

这是表格语法

CREATE TABLE [dbo].[QuestionBlocks] (
[Block_ID]     INT            IDENTITY (1, 1) NOT NULL,
[Question1]    NVARCHAR (MAX) NULL,
[Question2]    NVARCHAR (MAX) NULL,
[Question3]    NVARCHAR (MAX) NULL,
[Question4]    NVARCHAR (MAX) NULL,
[Question5]    NVARCHAR (MAX) NULL,
[Question6]    NVARCHAR (MAX) NULL,
[Question7]    NVARCHAR (MAX) NULL,
[Question8]    NVARCHAR (MAX) NULL,
[Question9]    NVARCHAR (MAX) NULL,
[Question10]   NVARCHAR (MAX) NULL,

此外,我还有针对这些问题的DropDownLists

这是看起来像

enter image description here enter image description here

我需要点击按钮从DropdownLists获取数据并在数据库中写入Question1-Question10行。

这是我的控制器

public ActionResult Index()
    {
        ViewBag.Question1 = new  SelectList(db.Questions,"QuestionId","question");
        ViewBag.Question2 = new SelectList(db.Questions, "QuestionId", "question");
        ViewBag.Question3 = new SelectList(db.Questions, "QuestionId", "question");
        ViewBag.Question4 = new SelectList(db.Questions, "QuestionId", "question");
        ViewBag.Question5 = new SelectList(db.Questions, "QuestionId", "question");
        ViewBag.Question6 = new SelectList(db.Questions, "QuestionId", "question");
        ViewBag.Question7 = new SelectList(db.Questions, "QuestionId", "question");
        ViewBag.Question8 = new SelectList(db.Questions, "QuestionId", "question");
        ViewBag.Question9 = new SelectList(db.Questions, "QuestionId", "question");
        ViewBag.Question10 = new SelectList(db.Questions, "QuestionId", "question");

        return View(db.Questions.ToList());
    }

这是View

  <div class="title2" style="margin-top: 15px; margin-left: 15px; margin-bottom: 15px; padding-top: 10px">
                @Html.DropDownList("Question1", null, "Вопрос 1", htmlAttributes: new {@class = "form-control", @style = "height:40px;margin-bottom: 20px;",placeholder="lol"})
                @Html.DropDownList("Question2", null, "Вопрос 2", htmlAttributes: new {@class = "form-control", @style = "height:40px; margin-bottom: 20px;"})
                @Html.DropDownList("Question3", null, "Вопрос 3", htmlAttributes: new {@class = "form-control", @style = "height:40px; margin-bottom: 20px;"})
                @Html.DropDownList("Question4", null, "Вопрос 4", htmlAttributes: new {@class = "form-control", @style = "height:40px; margin-bottom: 20px;"})
                @Html.DropDownList("Question5", null, "Вопрос 5", htmlAttributes: new {@class = "form-control", @style = "height:40px; margin-bottom: 20px;"})
                @Html.DropDownList("Question6", null, "Вопрос 6", htmlAttributes: new {@class = "form-control", @style = "height:40px; margin-bottom: 20px;"})
                @Html.DropDownList("Question7", null, "Вопрос 7", htmlAttributes: new {@class = "form-control", @style = "height:40px; margin-bottom: 20px;"})
                @Html.DropDownList("Question8", null, "Вопрос 8", htmlAttributes: new {@class = "form-control", @style = "height:40px; margin-bottom: 20px;"})
                @Html.DropDownList("Question9", null, "Вопрос 9", htmlAttributes: new {@class = "form-control", @style = "height:40px; margin-bottom: 20px;"})
                @Html.DropDownList("Question10", null, "Вопрос 10", htmlAttributes: new {@class = "form-control", @style = "height:40px; margin-bottom: 20px;"})
            </div>

我认为AJAX可以做到这一点,但是我需要如何编写代码或者我可以写些关于如何执行此操作的内容?

感谢的

更新

感谢Prasanna Kumar J的回答

我还有一个问题

我编写函数并尝试通过按钮单击运行它 我在html中编写了这段代码

<input id="save" type="button" value="Save" onclick="save();"/>

这在JS

$(document).ready(function () {
    $('#save').click(function () {
        save();
    });
});

但是按钮上没有运行功能。哪里出错?

2 个答案:

答案 0 :(得分:0)

您正在使用ASP.NET MVC,因此在您的问题控制器中,您可以创建一个[HttpPost]方法来处理数据库更新。

如果您使用的是SQL Server,则可以使用System.Data.SqlClient命名空间中的类,并查看MSDN documentation with examples

如果您使用的是MySQL,则可以使用MySQL .NET Connector - 可用信息on their site,包括documentation

在数据库更新方法中,您可以使用UPDATEINSERT查询来更新数据库中的数据。要检索信息,请设置<form method="post" action="your_update_page"><input type="submit" />。这将发布输入字段中包含的信息(所需名称等同于update方法中的参数),以便后端控制器在触发时进行数据库更新。

答案 1 :(得分:0)

试试这个。 在客户端

function save()
{
 $.ajax({
            type: 'Post',
            dataType: 'Json',
            data: { question1: $('#question1').val(),
                    question2: $('#question2').val(),
                    question3: $('#question3').val(),
                    question4: $('#question4').val(),
                    question5: $('#question5').val(),
                    question6: $('#question6').val(),
                    question7: $('#question7').val(),
                    question8: $('#question8').val(),
                    question9: $('#question9').val(),
                    question10: $('#question10').val()
            },
            url: '@Url.Action("SaveAction", "SaveController")',
            success: function (da) {
                if (da.Result == "Success") {
                 alert('saved sucessfully')


                } else {

                    alert( 'Error'+ da.Message);
                }
            },
            error: function (da) {
                alert('Error');
            }
        });
}

在服务器端

  [httppost]
  public ActionResult SaveAction(string question1,string question2,string question3,.....,string question10)
  {
        //do something
 return Json(new { Result = "Success", Message = "Saved Successfully" }, JsonRequestBehavior.AllowGet);
  }