原始图像(.jpg)文件大小为49kb,但当我下载它后,文件大小为87kb且已损坏。但是对于文本文件,它是有效的。使用HttpWebRequest或其他System.Net类下载图片需要做什么?我将XAMPP用于本地主机。
//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!");
}发布于 2019-04-10 14:58:26
对于非文本的内容,不应该使用StreamReader和StreamWriter。当您使用它们时,将应用编码。正如this blog将证明的那样,编码不能很好地与任意二进制数据混合。
相反,您应该使用简单的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)可能也会出现问题。您可以通过简单的设置更改来解决此问题:
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);
}
}https://stackoverflow.com/questions/55606343
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