如何从httpclient调用中获取内容主体? [英] How to get content body from a httpclient call?
问题描述
我一直在尝试弄清楚如何读取httpclient调用的内容,但似乎无法理解。我得到的响应状态是200,但是我不知道如何获得返回的实际Json,这就是我所需要的!
I've been trying to figure out how to read the contents of a httpclient call, and I can't seem to get it. The response status I get is 200, but I can't figure out how to get to the actual Json being returned, which is all I need!
以下是我的代码:
async Task<string> GetResponseString(string text)
{
var httpClient = new HttpClient();
var parameters = new Dictionary<string, string>();
parameters["text"] = text;
Task<HttpResponseMessage> response =
httpClient.PostAsync(BaseUri, new FormUrlEncodedContent(parameters));
return await response.Result.Content.ReadAsStringAsync();
}
而我只是从方法中调用它:
And I am getting it just calling it from a method:
Task<string> result = GetResponseString(text);
这就是我得到的
响应ID = 89,状态= RanToCompletion,方法= {null},结果= StatusCode:200,ReasonPhrase:确定,版本:1.1,内容:系统.Net.Http.StreamContent,标头:\r\n {\r\n连接:keep-alive\r\n日期:2014年10月27日星期一21:56:43 GMT\r\ ETag:\ 5a266b16b9dccea99d3e76bf8c1253e0\服务器:nginx / 0.7.65\r\n内容长度:125\r Contentn内容类型:application / json\ r\n} System.Threading.Tasks.Task< System.Net.Http.HttpResponseMessage>
更新:这是我当前根据Nathan的回答在下面使用的当前代码
Update: This is my current code per Nathan's response below
async Task<string> GetResponseString(string text)
{
var httpClient = new HttpClient();
var parameters = new Dictionary<string, string>();
parameters["text"] = text;
var response = await httpClient.PostAsync(BaseUri, new FormUrlEncodedContent(parameters));
var contents = await response.Content.ReadAsStringAsync();
return contents;
}
我用这种方法称呼它。...
And I call it from this method....
string AnalyzeSingle(string text)
{
try
{
Task<string> result = GetResponseString(text);
var model = JsonConvert.DeserializeObject<SentimentJsonModel>(result.Result);
if (Convert.ToInt16(model.pos) == 1)
{
_numRetries = 0;
return "positive";
}
if (Convert.ToInt16(model.neg) == 1)
{
_numRetries = 0;
return "negative";
}
if (Convert.ToInt16(model.mid) == 1)
{
_numRetries = 0;
return "neutral";
}
return "";
}
catch (Exception e)
{
if (_numRetries > 3)
{
LogThis(string.Format("Exception caught [{0}] .... skipping", e.Message));
_numRetries = 0;
return "";
}
_numRetries++;
return AnalyzeSingle(text);
}
}
它会一直运行,直到命中
var model = JsonConvert.DeserializeObject< SentimentJsonModel>(result.Result);
一次,它会继续运行而不会在另一个断点处停止。
And it keeps running forever, It hits the line
var model = JsonConvert.DeserializeObject<SentimentJsonModel>(result.Result);
Once, and it continues to go without stopping at another breakpoint.
当我暂停执行时,它说
Id =无法评估表达式,因为代码当前方法的优化。,Status =由于当前方法的代码已优化而无法评估表达式。,Method =由于当前方法的代码已优化而无法评估表达式。,Result =由于该方法的代码而无法评估表达式当前方法已优化。
Id = Cannot evaluate expression because the code of the current method is optimized., Status = Cannot evaluate expression because the code of the current method is optimized., Method = Cannot evaluate expression because the code of the current method is optimized., Result = Cannot evaluate expression because the code of the current method is optimized.
。我继续执行,但它永远运行。不确定问题出在哪里
.. I Continue execution, but it just runs forever. Not sure what the problem is
推荐答案
您使用await / async的方式充其量是很差的,这使得很难跟随。您正在将 await
与 Task’1.Result
混合在一起,这很令人困惑。但是,看起来您正在查看的是最终任务结果,而不是内容。
The way you are using await/async is poor at best, and it makes it hard to follow. You are mixing await
with Task'1.Result
, which is just confusing. However, it looks like you are looking at a final task result, rather than the contents.
我已经重写了您的函数和函数调用,这应该可以解决您的问题:
I've rewritten your function and function call, which should fix your issue:
async Task<string> GetResponseString(string text)
{
var httpClient = new HttpClient();
var parameters = new Dictionary<string, string>();
parameters["text"] = text;
var response = await httpClient.PostAsync(BaseUri, new FormUrlEncodedContent(parameters));
var contents = await response.Content.ReadAsStringAsync();
return contents;
}
最后一个函数调用:
Task<string> result = GetResponseString(text);
var finalResult = result.Result;
甚至更好:
var finalResult = await GetResponseString(text);
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