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Getting path to the parent folder of the solution file using C#

I am a beginner in C#, and I have a folder from which I am reading a file.

I want to read a file which is located at the parent folder of the solution file. How do I do this?

string path = "";
StreamReader sr = new StreamReader(path);

So if my file XXX.sln is in C:\X0\A\XXX\ then read the .txt files in C:\X0\A\.

like image 886
User1204501 Avatar asked Sep 25 '13 09:09

User1204501


3 Answers

You may enjoy this more general solution which depends on finding the solution *.sln file by scanning all parent directories from current or selected one while covering the case of not finding the solution directory!

public static class VisualStudioProvider
{
    public static DirectoryInfo TryGetSolutionDirectoryInfo(string currentPath = null)
    {
        var directory = new DirectoryInfo(
            currentPath ?? Directory.GetCurrentDirectory());
        while (directory != null && !directory.GetFiles("*.sln").Any())
        {
            directory = directory.Parent;
        }
        return directory;
    }
}

Usage:

// get directory
var directory = VisualStudioProvider.TryGetSolutionDirectoryInfo();
// if directory found
if (directory != null)
{
    Console.WriteLine(directory.FullName);
}

In your case:

// resolve file path
var filePath = Path.Combine(
    VisualStudioProvider.TryGetSolutionDirectoryInfo()
    .Parent.FullName, 
    "filename.ext");
// usage file
StreamReader reader = new StreamReader(filePath);

Enjoy!

Now, a warning.. Your application should be solution-agnostic - unless this is a personal project for some solution processing tool I wouldn't mind. Understand that, your application once distributed to users will reside in a folder without the solution. Now, you can use an "anchor" file. E.g. search parent folders like I did and check for existence of an empty file app.anchor or mySuperSpecificFileNameToRead.ext ;P If you want me to write the method I can - just let me know.

Now, you may really enjoy! :D

like image 65
Demetris Leptos Avatar answered Oct 26 '22 07:10

Demetris Leptos


Try this:

string startupPath = Path.Combine(Directory.GetParent(System.IO.Directory.GetCurrentDirectory()).Parent.Parent.Parent.FullName,"abc.txt");

// Read the file as one string. 
string text = System.IO.File.ReadAllText(startupPath);
like image 27
Thilina H Avatar answered Oct 26 '22 07:10

Thilina H


It would be remiss, I feel, if your application relied on the location of a file based on the relationship between the file path and the solution path. Whilst your program may well be executing at Solution/Project/Bin/$(ConfigurationName)/$(TargetFileName), that works only when you are executing from within the confines of Visual Studio. Outside of Visual Studio, in other scenarios, this is not necessarily the case.

I see two options:

  1. Include the file as part of your project, and in its' properties, have it copied to the output folder. You can then access the file thusly:

    string filePath = Path.Combine(Path.GetDirectoryName(Assembly.GetExecutingAssembly().Location), "Yourfile.txt");
    

    Note, during deployment you'll have to ensure that this file is also deployed alongside your executable.

  2. Use command line arguments to specify the absolute path to the file on startup. This can be defaulted within Visual Studio (see Project Properties -> Debug Tab -> Command line arguments". e.g:

    filePath="C:\myDevFolder\myFile.txt"
    

    There's a number of ways and libraries concerning parsing the command line. Here's a Stack Overflow answer on parsing command line arguments.

like image 7
Moo-Juice Avatar answered Oct 26 '22 07:10

Moo-Juice