当 Content-Encoding 为空时，为什么 HttpWebResponse 尝试使用 GZip 解压缩流？

发布于 2025-01-07 17:52:48 字数 1019 浏览 1 评论 0原文

我已经查看了很多围绕这个主题的问答，并达到了使用以下代码来从给定 URI 获取字节的程度：

var request = (HttpWebRequest)WebRequest.Create(uri);
request.AutomaticDecompression = DecompressionMethods.GZip | DecompressionMethods.Deflate;
var response = request.GetResponse();
var stream = response.GetResponseStream();
if (stream != null)
{
    var buffer = new byte[4097];
    var memoryStream = new MemoryStream();

    do
    {
        var count = stream.Read(buffer, 0, buffer.Length);
        memoryStream.Write(buffer, 0, count);

        if (count == 0)
            break;
    } while (true);

    return memoryStream.ToArray();
}

response.Close();

return null;

现在，对于某个 URI（指向一个文件），在调试时，我看到 Web 响应的“Content-Encoding”标头不等于任何内容（“”），但是当尝试从流中读取时，它会抛出异常：

System.IO.InvalidDataException：GZip 标头中的幻数不正确。确保您传递的是 GZip 流。

在开发工具中调试相同的 URI 时，我在响应标头中得到以下信息：

内容编码：gzip、deflate

所以我真的不知道会发生什么。

关于如何避免此异常并成功读取文件字节的任何线索和想法？

谢谢！

原文

I've looked at a lot of Q&As around this subject and got to the point where I use the following code in order to get the bytes from a given URI:

var request = (HttpWebRequest)WebRequest.Create(uri);
request.AutomaticDecompression = DecompressionMethods.GZip | DecompressionMethods.Deflate;
var response = request.GetResponse();
var stream = response.GetResponseStream();
if (stream != null)
{
    var buffer = new byte[4097];
    var memoryStream = new MemoryStream();

    do
    {
        var count = stream.Read(buffer, 0, buffer.Length);
        memoryStream.Write(buffer, 0, count);

        if (count == 0)
            break;
    } while (true);

    return memoryStream.ToArray();
}

response.Close();

return null;

Now, for a certain URI (which is pointing to a file), when debugging, I see that the 'Content-Encoding' header of the web response equals to nothing (""), but when trying to read from the stream, it throws an exception: