我用WPF c#编程。我有如下路径:
C:\Program Files\hello.txt
我想从中提取hello。
路径是从数据库中检索的字符串。目前，我正在使用以下代码通过'\'分割路径，然后再次通过'.'分割:
string path = "C:\\Program Files\\hello.txt";
string[] pathArr = path.Split('\\');
string[] fileArr = pathArr.Last().Split('.');
string fileName = fileArr.Last().ToString();
这是可行的，但我相信应该有更简单、更聪明的解决方案。任何想法?
当前回答
try
System.IO.Path.GetFileNameWithoutExtension(path);
demo
string fileName = @"C:\mydir\myfile.ext";
string path = @"C:\mydir\";
string result;
result = Path.GetFileNameWithoutExtension(fileName);
Console.WriteLine("GetFileNameWithoutExtension('{0}') returns '{1}'",
fileName, result);
result = Path.GetFileName(path);
Console.WriteLine("GetFileName('{0}') returns '{1}'",
path, result);
// This code produces output similar to the following:
//
// GetFileNameWithoutExtension('C:\mydir\myfile.ext') returns 'myfile'
// GetFileName('C:\mydir\') returns ''
https://msdn.microsoft.com/en-gb/library/system.io.path.getfilenamewithoutextension%28v=vs.80%29.aspx
其他回答
你可以像下面这样使用Path API:
var filenNme = Path.GetFileNameWithoutExtension([File Path]);
更多信息:路径。GetFileNameWithoutExtension
string Location = "C:\\Program Files\\hello.txt";
string FileName = Location.Substring(Location.LastIndexOf('\\') +
1);
string filepath = "C:\\Program Files\\example.txt";
FileVersionInfo myFileVersionInfo = FileVersionInfo.GetVersionInfo(filepath);
FileInfo fi = new FileInfo(filepath);
Console.WriteLine(fi.Name);
//input to the "fi" is a full path to the file from "filepath"
//This code will return the fileName from the given path
//output
//example.txt
try
System.IO.Path.GetFileNameWithoutExtension(path);
demo
string fileName = @"C:\mydir\myfile.ext";
string path = @"C:\mydir\";
string result;
result = Path.GetFileNameWithoutExtension(fileName);
Console.WriteLine("GetFileNameWithoutExtension('{0}') returns '{1}'",
fileName, result);
result = Path.GetFileName(path);
Console.WriteLine("GetFileName('{0}') returns '{1}'",
path, result);
// This code produces output similar to the following:
//
// GetFileNameWithoutExtension('C:\mydir\myfile.ext') returns 'myfile'
// GetFileName('C:\mydir\') returns ''
https://msdn.microsoft.com/en-gb/library/system.io.path.getfilenamewithoutextension%28v=vs.80%29.aspx
var fileNameWithoutExtension = Path.GetFileNameWithoutExtension(path);
路径。GetFileNameWithoutExtension
