WebJul 11, 2015 · Get Executing Assembly Folder Path using CodeBase public static string GetAssemblyPathByCodeBase() { string codeBase = … Web2 days ago · I have recently begun the task of converting a dynamo node i made in c# to an addin. But when i click the button to run the method it returns the command failure as shown in the image. ... it and i think it might have something to do with the Autodesk.Revit.DB.Document i use in my methods versus when executing the code in …
c# - How do I get the path of the assembly the code is in?
WebMay 8, 2009 · System.Reflection.Assembly.GetExecutingAssembly(). Location 1. Combine that with System.IO.Path.GetDirectoryName if all you want is the directory.. 1 As per Mr.Mindor's comment: System.Reflection.Assembly.GetExecutingAssembly().Location returns where the executing assembly is currently located, which may or may not be … WebSep 10, 2013 · They all give the same result. They certainly don’t. currentDir and dir both give you the current working directory – i.e. by default the directory your executable was run from (but it can be changed during the execution).. By contrast, appBaseDir and path get the directory which contains the executing assembly’s file. To illustrate how they differ, … shreve elementary school ohio
Assembly.GetExecutingAssembly Method …
WebFeb 15, 2024 · All of these return the current execution location on Windows (i.e. C:/SomePath/SomeProject/Name/api.dll) which I can use with Path.Combine to produce the path to the schema file. However, on linux, these all return /home/app/ which is not where the dll should be according to the Jenkins logs. This is leading to failures loading the … WebJan 14, 2015 · So here is a quick method that you can use to return the directory of the current executable that takes the URI into account: public static DirectoryInfo GetExecutingDirectory () { var location = new Uri (Assembly.GetEntryAssembly ().GetName ().CodeBase); return new FileInfo (location.AbsolutePath).Directory; } WebYou can adjust your file with his path var path = Assembly.GetAssembly (MyType.GetType ()).Location; var thisAssembly= Assembly.LoadFrom (path); var TypeName = ""; Type type = thisAssembly.GetType (TypeName); object instance = Activator.CreateInstance (type); Share Improve this answer Follow edited Jan 8, 2015 at 10:32 Bellash 7,379 6 51 84 shreve fire department