[英]How to read zipfile as StreamContent from httpResponseMessage?
[英]StreamContent from HttpResponseMessage into XML
我正在从具有此代码的WebApi控制器调用现有的get方法(我无法对其进行修改)
[HttpGet]
public HttpResponseMessage Get()
{
XmlDataDocument xmldoc = new XmlDataDocument();
FileStream fs = new FileStream("d:\\document.xml", FileMode.Open, FileAccess.Read);
xmldoc.Load(fs);
string str = xmldoc.DocumentElement.InnerXml;
return new HttpResponseMessage() { Content = new StringContent(str, Encoding.UTF8, "application/xml") };
}
我一直在尝试阅读这样的信息
HttpClient client = new HttpClient();
HttpResponseMessage response = client.GetAsync("http://localhost/api/info");
HttpContent content = rm.Content;
我得到了StreamContent,但是我现在想做的是读取此内容,然后尝试将其反序列化为Xml Document以读取节点。
如何从HttpContent的流内容中获取此信息?
string response;
using(var http = new HttpClient())
{
response = await http.GetStringAsync("http://localhost/api/info");
}
var xml = new XmlDataDocument();
xml.LoadXml(response);
您可以使用GetStringAsync获取字符串,而不是HttpContent对象。 您还错过了GetAsync中的等待状态。
注意:代码尚未经过测试
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