我认为这个问题的答案会如你所愿:
如何使用 C# 识别文件的扩展名/类型?
这将允许您检测以下 mime 类型:http:
//msdn.microsoft.com/en-us/library/ms775147%28VS.85%29.aspx#Known_MimeTypes
从中复制/粘贴:
[DllImport("urlmon.dll", CharSet = CharSet.Unicode, ExactSpelling = true, SetLastError = false)]
static extern int FindMimeFromData(IntPtr pBC,
[MarshalAs(UnmanagedType.LPWStr)] string pwzUrl,
[MarshalAs(UnmanagedType.LPArray, ArraySubType=UnmanagedType.I1, SizeParamIndex=3)]
byte[] pBuffer,
int cbSize,
[MarshalAs(UnmanagedType.LPWStr)] string pwzMimeProposed,
int dwMimeFlags,
out IntPtr ppwzMimeOut,
int dwReserved);
示例用法:
/// <summary>
/// Ensures that file exists and retrieves the content type
/// </summary>
/// <param name="file"></param>
/// <returns>Returns for instance "images/jpeg" </returns>
public static string getMimeFromFile(string file)
{
IntPtr mimeout;
if (!System.IO.File.Exists(file))
throw new FileNotFoundException(file + " not found");
int MaxContent = (int)new FileInfo(file).Length;
if (MaxContent > 4096) MaxContent = 4096;
FileStream fs = File.OpenRead(file);
byte[] buf = new byte[MaxContent];
fs.Read(buf, 0, MaxContent);
fs.Close();
int result = FindMimeFromData(IntPtr.Zero, file, buf, MaxContent, null, 0, out mimeout, 0);
if (result != 0)
throw Marshal.GetExceptionForHR(result);
string mime = Marshal.PtrToStringUni(mimeout);
Marshal.FreeCoTaskMem(mimeout);
return mime;
}