Question

我使用PostAsync()发送数据，代码执行完全正确。我发送数据的站点提供XML响应通知，通知成功或失败。 XML响应如下：成功帖子会返回1

<?xml version="1.0" encoding="utf-8" ?>
<result>
 <success>1</success>
 <uploadid/>
 </errors>
</result>

不成功会在<error>代码

之间返回原因

<?xml version="1.0" encoding="utf-8" ?>
<result>
 <success>0</success>
 <iploadid/>
 <errors>
    <error>File is not in correct format</error>
 </errors>
</result>

这是我用来执行PostAsync()我需要添加的 C＃语法，以便从页面读取XML响应？

private void SendAsyncData()
{
  string connectionString = "Data Source=hmod;Initial Catalog=salesdb;uid=userid;password=password;MultipleActiveResultSets=True";
  DataTable grds = new DataTable();
  using (var con = new SqlConnection(connectionString))
  using (var cmd = new SqlCommand("procedureone", con))
  using (var da = new SqlDataAdapter(cmd))
  {
    cmd.CommandType = CommandType.StoredProcedure;
    da.Fill(grds);
  }
  var client = new HttpClient();
  var q = string.Empty;
  foreach (DataRow row in grds.Rows)
  {
    var parameterSet = new List<KeyValuePair<string, string>>();
    for (int i = 0; i < grds.Columns.Count; ++i)
    {
        parameterSet.Add(new KeyValuePair<string, string>(grds.Columns[i].ColumnName, row[i].ToString()));

    }

    var result = client.PostAsync("http://weneedsampledata/ashx", new FormUrlEncodedContent(parameterSet)).Result;
    string resultContent = result.Content.ReadAsStringAsync().Result;
  }
}

修改
目前这个语法返回代码200（成功），即使接收数据的站点引发错误，我需要一种“读取”XML响应以确定数据是否实际成功的方法。我试图在下面添加以读取响应，但它给我一个编译错误：

'System.Net.Http.HttpResponseMessage'不包含'GetResponse'的定义，并且没有扩展方法'GetResponse'接受类型'System.Net.Http.HttpResponseMessage'的第一个参数可以找到（你错过了吗？） using指令或程序集引用？）

var response = result.GetResponse();
Stream receiveStream = response.GetResponseStream();
StreamReader readStream = new StreamReader(receiveStream, Encoding.UTF8);
var postresult = readStream.ReadToEnd();
var xml = System.Xml.Linq.XElement.Parse(postresult);
if (xml.Elements("success").FirstOrDefault().Value == "1")
{
  lblgoodbad.Text = "It was good.";
}
else
{
  var errors = xml.Elements("errors");
  foreach (var error in errors.Elements("error"))
  {
    lblgoodbad.Text = error.Value;
  }
}

编辑＃2
感谢@Hussey的帮助 - 我只是对我的语法进行了一次快速调整，它正在运行。我的完整语法如下所示：

    private void SendAsyncData()
{
  string connectionString = "Data Source=hmod;Initial Catalog=salesdb;uid=userid;password=password;MultipleActiveResultSets=True";
  DataTable grds = new DataTable();
  using (var con = new SqlConnection(connectionString))
  using (var cmd = new SqlCommand("procedureone", con))
  using (var da = new SqlDataAdapter(cmd))
  {
    cmd.CommandType = CommandType.StoredProcedure;
    da.Fill(grds);
  }
  var client = new HttpClient();
  var q = string.Empty;
  foreach (DataRow row in grds.Rows)
  {
    var parameterSet = new List<KeyValuePair<string, string>>();
    for (int i = 0; i < grds.Columns.Count; ++i)
    {
        parameterSet.Add(new KeyValuePair<string, string>(grds.Columns[i].ColumnName, row[i].ToString()));

    }

    var result = client.PostAsync("http://weneedsampledata/ashx", new FormUrlEncodedContent(parameterSet)).Result;
    string resultContent = result.Content.ReadAsStringAsync().Result;
    var xml = System.Xml.Linq.XElement.Parse(resultContent);
    if (xml.Elements("success").FirstOrDefault().Value == "1")
    {
      lblgoodbad.Text = "It was good.";
    }
    else
    {
      var errors = xml.Elements("errors");
      foreach (var error in errors.Elements("error"))
      {
        lblgoodbad.Text = error.Value;
      }
    }
  }
}

Answer 1

您能否请添加此行并尝试：

这一行之后：

string resultContent = result.Content.ReadAsStringAsync().Result;

添加：

XElement incomingXml = XElement.Parse(resultContent);

或者您也可以使用ReadAsStreamAsync尝试此选项以将流读入XmlDocument：

string resultContent = result.Content.ReadAsStreamAsync().Result;

XDocument incomingXml = XDocument.Load(resultContent);

从PostAsync读取XML响应（）

1 个答案: