DirectoryInfo Dir = new DirectoryInfo(Server.MapPath(strheadlinesid));
FileInfo[] FileList = Dir.GetFiles("*.txt", SearchOption.AllDirectories);
在* .txt的地方,我想提一些文件扩展名,我该怎么做。
我使用了另一种方法,但是当我使用FI作为超链接时,我有一个小问题,它给出了总路径。但我想只打印文件名而不是完整路径。
string supportedExtensions = "*.jpg,*.gif,*.png,*.bmp,*.jpe,*.jpeg,*.wmf,*.emf,*.xbm,*.ico,*.eps,*.tif,*.tiff,*.g01,*.g02,*.g03,*.g04,*.g05,*.g06,*.g07,*.g08";
foreach (string FI in Directory.GetFiles(Server.MapPath(strheadlinesid), "*.*", SearchOption.AllDirectories).Where(s => supportedExtensions.Contains(Path.GetExtension(s).ToLower())))
{
Response.Write("<td><a href= view5.aspx?file=" + strheadlinesid + "\\" + FI + " target=_self;> " +
FI + "</a></td>");
}
答案 0 :(得分:3)
尝试
string fileFilter = "*.wma,*.jpeg,*.txt";
string[] fileExt = fileFilter.Split(',');
FileInfo[] fileInfo = null;
DirectoryInfo dir = new DirectoryInfo("D:\\Test");
List<FileInfo[]> listFileInfo = new List<FileInfo[]>();
foreach (string strVar in fileExt)
{
fileInfo = dir.GetFiles(strVar, SearchOption.AllDirectories);
listFileInfo.Add(fileInfo);
}