原始图像(.jpg(文件大小为49kb,但是当我下载它时,文件大小为87kb并且已损坏。但是对于文本文件,它可以工作。使用 HttpWebRequest 或其他 System.Net 类下载图像需要做什么?我正在使用XAMPP作为本地主机。
//Usage: HttpDownload("http://www.localhost/files/imagine.jpg", "seo.jpg");
static async void HttpDownload(string remoteFileOrUri, string localFileName)
{
HttpWebRequest request = (HttpWebRequest)WebRequest.Create(new Uri(remoteFileOrUri));
HttpWebResponse response = (HttpWebResponse)await request.GetResponseAsync();
StreamReader rdr = new StreamReader(response.GetResponseStream());
StreamWriter sw = new StreamWriter(File.OpenWrite(localFileName));
sw.Write(rdr.ReadToEnd());
sw.Flush();
rdr.Close();
sw.Close();
Console.WriteLine("fin!");
}
你不应该将 StreamReader 和 StreamWriter 用于非文本的内容。使用它们时,将应用编码。编码不能很好地与任意二进制数据混合,正如本博客将证明的那样。
相反,您应该使用一个简单的FileStream
:
using (var output = File.OpenWrite(localFileName))
{
using (var responseStream = response.GetResponseStream())
{
await responseStream.CopyToAsync(output);
}
}
不幸的是,这可能不是您唯一的问题,因为GZip压缩(如果服务器正在使用它(也可能出现问题。您可以通过简单的设置更改来说明这一点:
HttpWebRequest request = (HttpWebRequest)WebRequest.Create(new Uri(remoteFileOrUri));
request.AutomaticDecompression = DecompressionMethods.GZip | DecompressionMethods.Deflate;
HttpWebResponse response = (HttpWebResponse)await request.GetResponseAsync();
using (var output = File.OpenWrite(localFileName))
{
using (var responseStream = response.GetResponseStream())
{
await responseStream.CopyToAsync(output);
}
}