[HttpPost]
    public JsonResult ContactAdd(ContactViewModel contactViewModel)
    {
        if (ModelState.IsValid)
        {
            var job = new Job { Contact = new Contact() };

            Mapper.Map(contactViewModel, job);
            Mapper.Map(contactViewModel, job.Contact);

            _db.Jobs.Add(job);

            _db.SaveChanges();

            //you do not even need this line of code,200 is the default for ASP.NET MVC as long as no exceptions were thrown
            //Response.StatusCode = (int)HttpStatusCode.OK;

            return Json(new { jobId = job.JobId });
        }
        else
        {
            Response.StatusCode = (int)HttpStatusCode.BadRequest;
            return Json(new { jobId = -1 });
        }
    }

200只是一个成功请求的正常HTTP报头。如果这就是你所需要的，只要让控制器返回新的EmptyResult();

    [HttpPost]
    public JsonResult ContactAdd(ContactViewModel contactViewModel)
    {
        if (ModelState.IsValid)
        {
            var job = new Job { Contact = new Contact() };

            Mapper.Map(contactViewModel, job);
            Mapper.Map(contactViewModel, job.Contact);

            _db.Jobs.Add(job);

            _db.SaveChanges();

            //you do not even need this line of code,200 is the default for ASP.NET MVC as long as no exceptions were thrown
            //Response.StatusCode = (int)HttpStatusCode.OK;

            return Json(new { jobId = job.JobId });
        }
        else
        {
            Response.StatusCode = (int)HttpStatusCode.BadRequest;
            return Json(new { jobId = -1 });
        }
    }

在。net Core中这样做的方法(在撰写本文时)如下:

public async Task<IActionResult> YourAction(YourModel model)
{
    if (ModelState.IsValid)
    {
        return StatusCode(200);
    }

    return StatusCode(400);
}

StatusCode方法返回一种类型的StatusCodeResult，它实现了IActionResult，因此可以用作您的操作的返回类型。

作为一种重构，你可以通过使用HTTP状态码枚举的类型转换来提高可读性，比如:

return StatusCode((int)HttpStatusCode.OK);

此外，还可以使用一些内置的结果类型。例如:

return Ok(); // returns a 200
return BadRequest(ModelState); // returns a 400 with the ModelState as JSON

参考状态代码结果 – https://learn.microsoft.com/en-us/dotnet/api/microsoft.aspnetcore.mvc.statuscoderesult?view=aspnetcore-2.1

在你的控制器中，你会像这样返回一个HttpStatusCodeResult…

[HttpPost]
public ActionResult SomeMethod(...your method parameters go here...)
{
   // todo: put your processing code here

   //If not using MVC5
   return new HttpStatusCodeResult(200);

   //If using MVC5
   return new HttpStatusCodeResult(HttpStatusCode.OK);  // OK = 200
}

您可以简单地将响应的状态代码设置为200，如下所示

public ActionResult SomeMethod(parameters...)
{
   //others code here
   ...      
   Response.StatusCode = 200;
   return YourObject;  
}