此调用作为同步调用。我想等待响应。我怎样才能使这个调用同步？staticasyncTaskUpload(stringkey,stringurl,stringsourceFile,stringtargetFormat){使用(HttpClientHandlerhandler=newHttpClientHandler{Credentials=newNetworkCredential(key,"")})使用(HttpClientclient=newHttpClient{(handler)varrequest=newMultipartFormDataContent();request.Add(newStringContent(targetFormat),"target_format");request.Add(newStreamContent(File.OpenRead(sourceFile)),"source_file",newFileInfo(sourceFile).Name);使用(HttpResponseMessageresponse=awaitclient.PostAsync(url,request).ConfigureAwait(false))使用(HttpContentcontent=response.Content){stringdata=awaitcontent.ReadAsStringAsync().ConfigureAwait(false);returnJsonObject.Parse(data);}}}感谢任何帮助！将awaitcontent.ReadAsStringAsync().ConfigureAwait(false)更改为content.ReadAsStringAsync().ResultReadAsStringAsync返回一个Task对象。该行末尾的“.Result”告诉编译器返回内部字符串。应该这样做：以上就是C#学习教程：同步http客户端：等待响应分享所有内容，如果对大家有用还需要详细了解C#学习教程，希望大家点赞更多关注——staticasyncTaskUpload(stringkey,stringurl,stringsourceFile,stringtargetFormat){using(HttpClientHandlerhandler=newHttpClientHandler{Credentials=newNetworkCredential(key,"")})using(HttpClientclient=newHttpClient(handler))){varrequest=newMultipart(FormData);request.Add(newStringContent(targetFormat),"target_format");request.Add(newStreamContent(File.OpenRead(sourceFile)),"source_file",newFileInfo(sourceFile).Name);使用(HttpResponseMessageresponse=awaitclient.PostAsync(url,request))使用(HttpContentcontent=response.Content){stringdata=awaitcontent.ReadAsStringAsync();返回JsonObject.Parse(数据);}}}本文收集自网络，不代表立场。如涉及侵权，请点击右侧联系管理员删除。如需转载请注明出处：
