我正在尝试编写一个尝试使用async和await发布webapi调用的webapi,我的当前问题是,只要我将await client.PostAsync(url, content);
挂起就挂起。
1.如何调试它挂起的原因?
2.有没有办法在没有异步和等待的情况下做到这一点?我想按顺序执行
public static async Task<string> testWCF2(string xmlConfig)
{
string submitOut;
using (var client = new System.Net.Http.HttpClient())
{
var url = "http://server:8100/api/SoftwareProductBuild";
var content = new StringContent(xmlConfig, Encoding.UTF8, "application/xml");
var response = await client.PostAsync(url, content);
if (response.IsSuccessStatusCode)
{
var responseBody = await response.Content.ReadAsStringAsync();
submitOut = responseBody;
}
else
{
submitOut = string.Format("Bad Response {0} \n", response.StatusCode.ToString());
submitOut = submitOut + response;
}
}
return submitOut;
}
public async Task<string> QlasrSubmit(List<XMLSiInfo> xmlConfigs)
{
string submitOut = "QLASR: ";
foreach (XMLSiInfo xmlConfig in xmlConfigs)
{
submitOut = submitOut + "\n" + await testWCF2(xmlConfig.xml);
}
return submitOut;
}
public async Task<string> QlasrPostcommit(string si, string sp, string variant = null)
{
.....
string submitStatus = await QlasrSubmit(siInfo);
.....
return submitStatus;
}
服务:
public async Task<string> QlasrPostcommit(string si, string sp, string variant = null)
{
return await DPR.QlasrPostcommit(si, sp, variant);
}
控制器:
[Route("api/DevPool/QlasrPostcommit")]
[HttpPost]
public ResponseObject QlasrPostcommit(string si, string sp, string variant = null)
{
ResponseObject response = new ResponseObject();
try
{
response.status = 200;
response.data = DPS.QlasrPostcommit(si, sp, variant);
return response;
}
catch (Exception e)
{
response.status = 200;
response.data = null;
response.message = e.Message;
return response;
}
}
答案 0 :(得分:2)
您应该使用我async
all the way中提到的your previous question:
[Route("api/DevPool/QlasrPostcommit")]
[HttpPost]
public async Task<ResponseObject> QlasrPostcommit(string si, string sp, string variant = null)
{
ResponseObject response = new ResponseObject();
try
{
response.status = 200;
response.data = await DPS.QlasrPostcommit(si, sp, variant);
return response;
}
catch (Exception e)
{
response.status = 200;
response.data = null;
response.message = e.Message;
return response;
}
}
在这种特殊情况下,您正在遇到deadlock because you're blocking on asynchronous code。
答案 1 :(得分:0)
我解决了它并且它完美地工作,没有deallock和等待结果!!
您已修复服务:
public string QlasrPostcommit(string si, string sp, string variant = null)
{
Task<string > task = Task.Run<string >(async () => await
DPR.QlasrPostcommit(si, sp, variant));
task.Result;
}
通用答案:
public TypeToReturn MyAsyncMethod(myParams...)
{
Task<TypeToReturn> task = Task.Run<TypeToReturn>(async () => await
MyAsyncMethod(myParams...));
task.Result;
}