Question

我正在尝试使用RESTSHARP在HP-ALM上发布实体。到目前为止，我成功通过身份验证并获得了一些GET响应。但是，不知何故，对于我发送的每个POST请求，我得到了这个回复：

qccore.general误差不支持的媒体类型

这是我做的许多试验之一（用于发布缺陷）。对这里有什么不妥的想法？

    private RestRequest createPOSTRequest()
    {
        RestRequest Request = m_client.CreateRequest(m_client.BaseUrl + 
        "rest/domains/{domain}/projects/{project}/{entity-type}", Method.POST);
        Request.AddUrlSegment("domain", m_client.domain);
        Request.AddUrlSegment("project", m_client.project);
        Request.AddUrlSegment("entity-type", "defects");

        Request.AddHeader("Content-Type", "application/xml");
        Request.AddHeader("Accept", "application/xml");

        Request.RequestFormat = DataFormat.Xml;

        m_xmlBody = = @"<?xml version='1.0' encoding='UTF-8'? encoding='UTF-8' standalone='yes'?>"+
                                "<Entity Type='defect'>"+
                                "<Fields>" +
                                "<Field Name='detected-by'>"+
                                "<Value>sa</Value>"+
                                "</Field>"+
                                "<Field Name='creation-time'>"+
                                "<Value>2010-03-02</Value>"+ 
                                "</Field>"+
                                "<Field Name='severity'>"+
                                "<Value>2-Medium</Value>"+ 
                                "</Field>"+
                                "<Field Name='name'>"+
                                "<Value>Defect Entity.</Value>"+ 
                                "</Field>"+
                                "</Fields>"+
                                "</Entity>";

        return Request;
    }

谢谢。

Answer 1

Content-Type无法与AddHeader一起正常使用。

解决方案here。

实现此目的的方法是使用AddBody（）和RestRequest.RequestFormat。一个例子：

var client = new RestClient();
// client.XmlSerializer = new XmlSerializer(); // default
// client.XmlSerializer = new SuperXmlSerializer(); // can override with any implementaiton of ISerializer

var request = new RestRequest();
request.RequestFormat = DataFormat.Xml;
request.AddBody(objectToSerialize);

通过RestSharp POST到ALM：不支持的媒体类型

1 个答案: