代码之家 › 专栏 › 技术社区 › Sumit raj

将Post-API调用参数获取为空

postman asp.net-web-api2 c#

Sumit raj · 技术社区 · 6 年前

我正试着用很大的参数来做一个后调用。但电话不通。

发行 :当参数的大小为大时,不会进行调用。但如果参数较小,则一切正常,这意味着API正在工作。我试着研究 POST 打电话但没能成功。

我所做的 我做了一个示例api项目,并尝试通过postman和project进行测试。

邮递员电话 这是工作正常,如预期。

在项目中

如果有人要指出 rquest.ContentLength = 0; . 这是因为param只被附加到url。我不想把它附加上去,但是我不知道怎么做,还用谷歌搜索了一下。对于较小的长度,这是可行的,但如果字符串长度达到3000,则失败。

API调用发起程序

UriBuilder uriBuilder = new UriBuilder(restURL);
var query = HttpUtility.ParseQueryString(uriBuilder.Query);
query["json"] = new JavaScriptSerializer().Serialize(reportParameterDictionary);
uriBuilder.Query = query.ToString();
restURL = uriBuilder.ToString();
HttpWebRequest request = (HttpWebRequest)WebRequest.Create(restURL);
request.Method = "Post";
request.ContentType = "application/json";
rquest.ContentLength = 0;
using (HttpWebResponse response = request.GetResponse() as HttpWebResponse)
{
  StreamReader reader = new StreamReader(response.GetResponseStream());
  string responseData = reader.ReadToEnd();
  response.Close();
  dynamic obj = JsonConvert.DeserializeObject(responseData);
  if (obj != null)
  {
   printresult = obj.Success;
  }
}

接收端

[Route("PrintReport")]
public IHttpActionResult PrintReport(string json)

2 回复 | 直到 6 年前

artista_14 6 年前

// Create a request using a URL that can receive a post.   
WebRequest request = WebRequest.Create ("http://www.contoso.com/PostAccepter.aspx");  
// Set the Method property of the request to POST.  
request.Method = "POST";  
// Create POST data and convert it to a byte array.  
string postData = "This is a test that posts this string to a Web server.";  
byte[] byteArray = Encoding.UTF8.GetBytes (postData);  
// Set the ContentType property of the WebRequest.  
request.ContentType = "application/x-www-form-urlencoded";  
// Set the ContentLength property of the WebRequest.  
request.ContentLength = byteArray.Length;  
// Get the request stream.  
Stream dataStream = request.GetRequestStream ();  
// Write the data to the request stream.  
dataStream.Write (byteArray, 0, byteArray.Length);  
// Close the Stream object.  
dataStream.Close ();  
// Get the response.  
WebResponse response = request.GetResponse ();  
// Display the status.  
Console.WriteLine (((HttpWebResponse)response).StatusDescription);  
// Get the stream containing content returned by the server.  
dataStream = response.GetResponseStream ();  
// Open the stream using a StreamReader for easy access.  
StreamReader reader = new StreamReader (dataStream);  
// Read the content.  
string responseFromServer = reader.ReadToEnd ();  
// Display the content.  
Console.WriteLine (responseFromServer);  
// Clean up the streams.  
reader.Close ();  
dataStream.Close ();  
response.Close ();

跟着这个 Link 完整的解释。

Tanato 6 年前

你试过用 HttpClient 而不是 HttpWebRequest ?

var client = new HttpClient();
var content = new StringContent("YOUR LONG STRING", 
                             System.Text.Encoding.Unicode, 
                             "application/json");            
var task = client.PostAsync(restURL, content);
var str = await task.Result.Content.ReadAsStringAsync();