我正在调试一些数据(JSON字符串),这些数据被发送到使用DeflateStream
在C#中压缩的PHP API:
using (var writer = new StreamWriter(new MemoryStream())){
ServicePointManager.SecurityProtocol = SecurityProtocolType.Ssl3 | SecurityProtocolType.Tls | SecurityProtocolType.Tls11 | SecurityProtocolType.Tls12;
request.Headers.Add("Content-Encoding", "gzip");
request.Headers.Add(HttpRequestHeader.AcceptEncoding, "gzip, deflate");
byteArray = Compress(new MemoryStream(Encoding.UTF8.GetBytes(incomingData)));
Random random = new Random();
int randomNumber = random.Next(0, 100);
File.WriteAllBytes(@"compressed-"+ randomNumber + ".gz", byteArray);
}
这是Compress()
的定义
private static byte[] Compress(Stream input)
{
using (var compressStream = new MemoryStream())
using (var compressor = new DeflateStream(compressStream, CompressionMode.Compress))
{
input.CopyTo(compressor);
compressor.Close();
return compressStream.ToArray();
}
}
有没有办法打开这个在Windows中创建的文件?我试过7zip和Winrar无济于事。如果我使用GZipStream
我可以打开它,但PHP与该格式不兼容。
它不会写头信息和元数据,所以你不能像拉链文件那样“打开”...你总是可以简单地执行反向(即用DeflateStream解压缩)来获取原始数据......