我有这样的代码:
$.ajax({
type: "POST",
url: "/api/slide",
cache: false,
contentType: "application/json; charset=utf-8",
data: '{"Title":"fghfdhgfdgfd"}',
dataType: "json",这是我的控制器:
public class SlideController : ApiController
{
// POST /api/Slide
public void Post(string Title)
{
}当我运行代码并调用/api/Slide时,标题没有数据并且为空。
如何将JSON发布到API控制器?
POST http://127.0.0.2:81/api/slide HTTP/1.1
Host: 127.0.0.2:81
Connection: keep-alive
Content-Length: 18
Origin: http://127.0.0.2:81
X-Requested-With: XMLHttpRequest
User-Agent: Mozilla/5.0 (Windows NT 6.1; WOW64) AppleWebKit/537.1 (KHTML, like Gecko) Chrome/21.0.1180.89 Safari/537.1
Content-Type: application/json; charset=UTF-8
Accept: application/json, text/javascript, */*; q=0.01
Referer: http://127.0.0.2:81/
Accept-Encoding: gzip,deflate,sdch
Accept-Language: en-US,en;q=0.8
Accept-Charset: ISO-8859-1,utf-8;q=0.7,*;q=0.3
Title=fghfdhgfdgfd发布于 2012-09-16 20:11:18
定义视图模型:
public class SlideViewModel
{
public string Title { get; set; }
}然后让您的控制器操作将此视图模型作为参数:
public class SlideController : ApiController
{
// POST /api/Slide
public void Post(SlideViewModel model)
{
...
}
}最后调用该操作:
$.ajax({
type: 'POST',
url: '/api/slide',
cache: false,
contentType: 'application/json; charset=utf-8',
data: JSON.stringify({ title: "fghfdhgfdgfd" }),
success: function() {
...
}
});原因是简单类型(如字符串)是从URI绑定的。我还邀请您阅读有关Web API中的模型绑定的following article。
发布于 2012-09-22 08:08:41
确保您尝试转换到的对象具有默认(空)构造函数。
经验法则:如果你想反序列化到一个对象,你需要让创建的对象变得简单。这些指导原则可以帮助您:
例如,此JSON字符串/对象:
{ Name: "John Doe", Phone: "123-456-7890", Pets: [ "dog", "cat", "snake" ] }可以从以下类转换为对象:
public class Person {
public string Name { get; set; }
public string Phone { get; set; }
public string[] Pets { get; set; }
}或者这个:
public class Person {
public string Name { get; set; }
public string Phone { get; set; }
public string[] Pets { get; set; }
public Person() {}
public Person(string name, string phone) {
Name = name;
Phone = phone;
}
}或者这个:
public class Person {
public string Name { get; set; }
public string Phone { get; set; }
public string[] Pets { get; set; }
public Person() {}
}但不是这个
public class Person {
public string Name { get; set; }
public string Phone { get; set; }
public string[] Pets { get; set; }
public Person(string name, string phone) {
Name = name;
Phone = phone;
}
}现在让ASP.NET MVC 4来做剩下的事情。
public class PersonController : ApiController
{
// .. other actions
public HttpResponseMessage PostPerson(Person person)
{
if ( null != person)
// CELEBRATE by doing something with your object
else
// BE SAD and throw and exception or pass an error message
}
// .. other actions
}如果您的类不能具有默认构造函数,或者如果您无权访问该类的源代码,则可以创建一个适配器类,该适配器类
的公共属性
使用上面不带默认构造函数的Person类,适配器可能如下所示
public class PersonAdapter {
public Person personAdaptee;
public string Name {
get { return personAdaptee.Name; }
set { personAdaptee.Name = value }
}
public string Phone {
get { return personModel.Phone; }
set { personModel.Phone = value; }
}
public string[] Pets {
get { return personAdaptee.Pets; }
set {personAdaptee.Pets = value }
}
public PersonAdapter() {
personAdaptee = new Person("", "", null);
}
}现在让ASP.NET MVC 4来做剩下的事情。
public class PersonController : ApiController
{
// .. other actions
public HttpResponseMessage PostPerson(PersonAdapter person)
{
if ( null != person)
// CELEBRATE by doing something with your object
else
// BE SAD and throw and exception or pass an error message
}
// .. other actions
}发布于 2012-09-16 18:25:50
试试这个:
$.ajax({
type: "POST",
url: "/api/slide",
data: { Title: "fghfdhgfdgfd" }
});这是由data属性周围的引号引起的:
即>>数据:{标题:"fghfdhgfdgfd“}
非标题数据:'{ >>:"fghfdhgfdgfd“}'
更新:
而且你的控制器看起来有点奇怪,尽管如果不看你的路由就很难判断,等等。
我希望看到更多这样的东西:
public class SlideController : ApiController
{
public HttpResponseMessage PostSlide(string Title)
{
// Do your insert slide stuff here....
string uri = Url.Link("DefaultApi", new { id = item.Id });
response.Headers.Location = new Uri(uri);
return response;
}
}显然,您还需要更新jQuery中的URL。
看看这里:
http://www.asp.net/web-api/overview/getting-started-with-aspnet-web-api/tutorial-your-first-web-api
另一个更新:
通常会创建一个CLR对象来匹配您的Json,并使用MVC模型绑定器直接绑定到该对象。如果你不想这样做,你可以绑定到一个对象并反序列化到一个字典中:
// POST api/values
public void Post(object json)
{
Dictionary<string, string> values = JsonConvert.DeserializeObject<Dictionary<string, string>>(json.ToString());
var x = values["Title"];
}https://stackoverflow.com/questions/12446024
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