c# webapi POST 参数解决方法

HttpWebRequest POST请求webapi：如果参数是简单类型，比如字符串（注意，拼接的字符串要HttpUtility.UrlEncode才行，否则服务端会丢失特殊字符&后面的数据）

要点：如下代码统一设置为：ContentType = "application/x-www-form-urlencoded";

服务端代码1：URL格式为 POSTapi/Values

public string Post([FromBody] string value)

则客户端Post的数据：拼接的字符串必须以 =开头，否则服务端无法取得value。例如：=rfwreewr2332322232 或者 {'':value}

服务端代码2：URL格式为 POST api/Values?value={value}

public string Post(string value)

则客户端Post的数据：需要url里拼接出KeyValue这样的数据

服务端代码3：URL格式为 POST api/Values

public string Post()

则客户端Post的数据：无要求。例如：key=rfwreewr2332322232。

服务端：可以用HttpContext.Current.Request.InputStream或者HttpContext.Current.Request.Form[0]都可以获取

如果post的参数类型比较复杂，则需要自定义类

要点：如下代码统一设置为：ContentType = "application/json";

服务端代码1：URL格式为 POST api/Values

public string Post([FromBody] Model value)或者 public string Post(Model value)

则客户端Post的数据：{\"id\":\"test1\",\"name\":\"value\"}。服务端会自动映射到对象。

提交代码如下：

HttpWebRequest wReq = (HttpWebRequest)WebRequest.Create("http://localhost:37831/api/Values");
wReq.Method = "Post";
//wReq.ContentType = "application/json";
//wReq.ContentType = "application/x-www-form-urlencoded";
wReq.ContentType = "application/json";

//byte[] data = Encoding.Default.GetBytes(HttpUtility.UrlEncode("key=rfwreewr2332322232&261=3&261=4"));
byte[] data = Encoding.Default.GetBytes("{\"id\":\"test1\",\"name\":\"value\"}");
wReq.ContentLength = data.Length;
Stream reqStream = wReq.GetRequestStream();
reqStream.Write(data, 0, data.Length);
reqStream.Close();
using (StreamReader sr = new StreamReader(wReq.GetResponse().GetResponseStream()))
{
string result = sr.ReadToEnd();

}

服务段代码如下：

// POST api/values
//public string Post()
//{
//    FileInfo fi = new FileInfo(AppDomain.CurrentDomain.BaseDirectory + "log.txt");
//    StreamWriter sw = fi.CreateText();
//    StreamReader sr = new StreamReader(HttpContext.Current.Request.InputStream);
//    sw.WriteLine("1:" + sr.ReadToEnd()+"--"+ HttpContext.Current.Request.Form[0]);
//    sw.Flush();
//    sw.Close();
//    return "{\"test\":\"1\"}";
//}

// POST api/values
public HttpResponseMessage PostStudent(Student student)
{
FileInfo fi = new FileInfo(AppDomain.CurrentDomain.BaseDirectory + "log.txt");
StreamWriter sw = fi.AppendText();

sw.WriteLine("2:" + student.id);
sw.Flush();
sw.Close();
HttpResponseMessage result = new HttpResponseMessage { Content = new StringContent("{\"test\":\"2\"}", Encoding.GetEncoding("UTF-8"), "application/json") };
return result;
}

这篇文章里有的方法也不错：http://www.cnblogs.com/rohelm/p/3207430.html