问题描述

我在类似的javascript中创建了一个函数:

function addNewManufacturer() {
    var name = $("#id-manuf-name").val();
    var address = $("#id-manuf-address").val();
    var phone = $("#id-manuf-phone").val();
    
    var sendInfo = {
        Name: name,
        Address: address,
        Phone: phone
    };
    
    $.ajax({
        type: "POST",
        url: "/Home/Add",
        dataType: "json",
        success: function (msg) {
            if (msg) {
                alert("Somebody" + name + " was added in list !");
                location.reload(true);
            } else {
                alert("Cannot add to list !");
            }
        },
        data: sendInfo
    });
}

i调用jquery.json-2.3.min.js脚本文件，我使用它为toJSON(array)方法.

在控制器中，我有这个Add动作

[HttpPost]
public ActionResult Add(PersonSheets sendInfo) {
    bool success = _addSomethingInList.AddNewSomething( sendInfo );

    return this.Json( new {
         msg = success
    });
      
}

但sendInfo作为方法参数变为null.

模型:

public struct PersonSheets
{
    public int Id;
    public string Name;
    public string Address;
    public string Phone;
}

public class PersonModel
{
    private List<PersonSheets> _list;
    public PersonModel() {
         _list= GetFakeData();
    }

    public bool AddNewSomething(PersonSheets info) {
         if ( (info as object) == null ) {
            throw new ArgumentException( "Person list cannot be empty", "info" );
         }

         PersonSheets item= new PersonSheets();
         item.Id = GetMaximumIdValueFromList( _list) + 1;
         item.Name = info.Name;
         item.Address = info.Address;
         item.Phone = info.Phone;
             
         _list.Add(item);

         return true;
    }
}

如何在用帖子发送数据时在操作方法中进行操作方法?

我不知道如何使用.
此外，可以通过JSON发送响应(到Ajax)?

推荐答案

创建模型

public class Person
{
    public string Name { get; set; }
    public string Address { get; set; }
    public string Phone { get; set; }
}

控制器如下

public ActionResult PersonTest()
    {
        return View();
    }

    [HttpPost]
    public ActionResult PersonSubmit(Vh.Web.Models.Person person)
    {
        System.Threading.Thread.Sleep(2000);  /*simulating slow connection*/

        /*Do something with object person*/


        return Json(new {msg="Successfully added "+person.Name });
    }

javascript

<script type="text/javascript">
    function send() {
        var person = {
            name: $("#id-name").val(),
            address:$("#id-address").val(),
            phone:$("#id-phone").val()
        }

        $('#target').html('sending..');

        $.ajax({
            url: '/test/PersonSubmit',
            type: 'post',
            dataType: 'json',
            contentType: 'application/json',
            success: function (data) {
                $('#target').html(data.msg);
            },
            data: JSON.stringify(person)
        });
    }
</script>

其他推荐答案

var SendInfo= { SendInfo: [... your elements ...]};

        $.ajax({
            type: 'post',
            url: 'Your-URI',
            data: JSON.stringify(SendInfo),
            contentType: "application/json; charset=utf-8",
            traditional: true,
            success: function (data) {
                ...
            }
        });

和行动

public ActionResult AddDomain(IEnumerable<PersonSheets> SendInfo){
...

您可以像这样绑定你的数组

var SendInfo = [];

$(this).parents('table').find('input:checked').each(function () {
    var domain = {
        name: $("#id-manuf-name").val(),
        address: $("#id-manuf-address").val(),
        phone: $("#id-manuf-phone").val(),
    }

    SendInfo.push(domain);
});

其他推荐答案

使用JSON.stringify(<data>).

更改代码:data: sendInfo到data: JSON.stringify(sendInfo).
希望这可以帮助你.

以上所述是小编给大家介绍的通过POST（ajax）发送JSON数据，并从控制器（MVC）接收JSON响应。,希望对大家有所帮助，如果大家有任何疑问请给我留言，小编会及时回复大家的。在此也非常感谢大家对77isp云服务器技术网的支持！