WebApi传递JSON参数

开发过程中经常进行JSON的传递,在WebApi中传递JSON字串时,会发现服务器端接收到不参数值,看下面代码

服务端:

public void Post([FromBody]string value)
        {
            LoggerHelper.Info("Post:{0}", value);
        }

客户端:

HttpClient client = new HttpClient();
            client.DefaultRequestHeaders.Accept.Add(new MediaTypeWithQualityHeaderValue("application/json"));
            string url = "http://api.oa.com/api/Test/Post";

            var json = "{ \"Name\": \"Test\" }";
            var httpContent = new StringContent(json, Encoding.UTF8);
            httpContent.Headers.ContentType = new MediaTypeHeaderValue("application/json");
            var response = client.PostAsJsonAsync(url, httpContent).Result;
            if (!response.IsSuccessStatusCode)
            {
                Response.Write(string.Format("{0} ({1})", (int)response.StatusCode, response.ReasonPhrase));
            }

运行客户端,查看服务端的日志,结果为“Post:”,调用成功,但参数接收失败。

查了些资料,显示WebApi不支持JSON字串做为简单参数传递,既然如此就将JSON字串做为复杂类型进行传,对代码稍做调整,服务端接收JObject参数:

public void Post([FromBody]JObject value)
        {
            
            LoggerHelper.Info("Post:{0}", value.ToString());
        }

客户端:

HttpClient client = new HttpClient();
            client.DefaultRequestHeaders.Accept.Add(new MediaTypeWithQualityHeaderValue("application/json"));
            string url = "http://api.oa.com/api/Test/Post";

            var json = "{ \"Name\": \"Test\" }";
            var jObject = JObject.Parse(json);
            var response = client.PostAsJsonAsync(url, jObject).Result;
            if (!response.IsSuccessStatusCode)
            {
                Response.Write(string.Format("{0} ({1})", (int)response.StatusCode, response.ReasonPhrase));
            }

运行客户端,再次查看服务端的日志,结果为:

Post:{
"Name": "Test"
},参数传递成功

posted @ 2015-01-25 11:43  flysoul  阅读(32876)  评论(2编辑  收藏