本文介绍了.Net核心中Json()的小写属性名称的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

具有类ClientViewModel:

Have class ClientViewModel:

public class Result
{
  public long PKID
  {
      get;
      set;
  }

   public string Name
   {
      get;
      set;
   }
}

并具有post方法来返回此模型:

And have post method to return this model:

[HttpPost]
    public JsonResult Search(string orderBy, List<Order> order, Search search,
        int start, int length, string searchTerm, bool isAsc, bool hideVoidedAndDeclined, bool recent = false)
    {
        var take = length;
        var skip = start;
        var m = _clientsService.Search(searchTerm, orderBy, hideVoidedAndDeclined, isAsc, take, skip, recent);


        return Json(m);
    }

但是当尝试在json中返回对象时,它看起来是:

but when try to return object in json it looks as:

name:"Nick22"
pkid: 5

但是我需要保存寄存器

Name:"Nick22"
PKID: 5

推荐答案

应该更改合同解析器.但是,由于您的大小写并不总是相同,因此使用

You should change the contractresolver. But since your casing is not always the same, you might be better off using

services.AddJsonOptions(options => { options.SerializerSettings.ContractResolver = new DefaultContractResolver(); });

services.AddJsonOptions(options => { options.SerializerSettings.ContractResolver = new DefaultContractResolver(); });

这篇关于.Net核心中Json()的小写属性名称的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!

07-23 07:24