What is the best way to get the executing exe'

问题:

From program a.exe located in c:/dir I need to open text file c:/dir/text.txt. I don't know where a.exe could be located, but text.txt will always be in the same path. How to get the name of the currently executing assembly from within to program itself so that i can access the text file?

EDIT: what if a.exe is a Windows service? It doesn't have Application as it is not a Windows Applicaion.

Thanks in advance.

回答1:

I usually access the directory that contains my application's .exe with:

System.IO.Path.GetDirectoryName(System.Reflection.Assembly.GetEntryAssembly().Location);

回答2:

string exePath = Application.ExecutablePath;
string startupPath = Application.StartupPath;

EDIT - Without using application object:

string path = System.IO.Path.GetDirectoryName( 
      System.Reflection.Assembly.GetExecutingAssembly().GetName().CodeBase );

See here for more info:

http://msdn.microsoft.com/en-us/library/aa457089.aspx

回答3:

Application.ExecutablePath

Application.StartupPath

回答4:

Get the assembly you are interested in (eg. assigned to a System.Reflection.Assembly a variable):

• System.Reflection.Assembly.GetEntryAssembly(), or
• typeof(X).Assembly for a class X that's in the assembly you're interested in (for Windows Forms you could use typeof(Program))

Then get the path of where the file from which that assembly a was loaded from:

• System.IO.Path.GetDirectoryName(a.Location)

The Application object from a Windows Forms application is also a possibility, as explained in other answers.

回答5:

using peSHlr's answer worked well when testing in NUnit as well.

var thisType = typeof(MyCustomClass);

var codeLocation = Path.GetDirectoryName(thisType.Assembly.Location);

var codeLocationPath = Path.GetDirectoryName(codeLocation);

var appConfigPath = Path.Combine(codeLocationPath, "AppConfig");

回答6:

In VB.NET we can get it in following way:

Assembly.GetEntryAssembly.Location

回答7:

MessageBox.Show("This program is located in: " + Environment.CurrentDirectory);