Question

我尝试使用以下代码获取xml文件：

HttpWebRequest webReq = (HttpWebRequest)WebRequest.Create(url);
HttpWebResponse response = (HttpWebResponse)webReq.GetResponse();
string xml = string.Empty;
using (StreamReader sr = new StreamReader(response.GetResponseStream()))
{
    xml = sr.ReadToEnd();
}

XmlDocument xmlDoc = new XmlDocument();
//xml = xml.Replace((char)(0x1F), ' ');
xmlDoc.LoadXml(xml);

但我得到例外如下：

' ', hexadecimal value 0x1F, is an invalid character. Line 1, position 1.

所以根据stackoverflow上的许多类似问题，我尝试添加这个注释行，但后来我得到了异常：

Data at the root level is invalid. Line 1, position 2.

怎么了？

Answer 1

假设正在应用于XML的压缩是GZip，您可以像这样解压缩XML：

HttpWebRequest webReq = (HttpWebRequest)WebRequest.Create(url);
HttpWebResponse response = (HttpWebResponse)webReq.GetResponse();
string xml = string.Empty;
using (GZipStream gzip = new GZipStream(response.GetResponseStream(), CompressionMode.Decompress))
using (StreamReader sr = new StreamReader(gzip))
{
  xml = sr.ReadToEnd();
}

XmlDocument xmlDoc = new XmlDocument();
//xml = xml.Replace((char)(0x1F), ' ');
xmlDoc.LoadXml(xml);

如果GZipStream不能解压缩XML，则必须用适当的解压缩流替换它。

Xml解析错误：十六进制值是无效字符

1 个答案: