C# 如何在运行单元测试时获取目录

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时间:2020-08-09 12:51:52  来源:igfitidea点击:

How to get Directory while running unit test

c#unit-testingdirectory

提问by AnonyMouse

Hi when running my unit test I'm wanting to get the directory my project is running in to retrieve a file.

嗨,在运行我的单元测试时,我想获取我的项目正在运行的目录以检索文件。

Say I have a Test project named MyProject. Test I run:

假设我有一个名为 MyProject 的测试项目。我运行的测试:

AppDomain.CurrentDomain.SetupInformation.ApplicationBase

and I receive "C:\\Source\\MyProject.Test\\bin\\Debug".

我收到了"C:\\Source\\MyProject.Test\\bin\\Debug"

This is close to what I'm after. I don't want the bin\\Debugpart.

这接近我所追求的。我不要那bin\\Debug部分。

Anyone know how instead I could get "C:\\Source\\MyProject.Test\\"?

有谁知道我怎么能得到"C:\\Source\\MyProject.Test\\"

采纳答案by Ilian Pinzon

I would do it differently.

我会以不同的方式去做。

I suggest making that file part of the solution/project. Then right-click -> Properties -> Copy To Output = Copy Always.

我建议将该文件作为解决方案/项目的一部分。然后右键单击 -> 属性 -> 复制到输出 = 始终复制。

That file will then be copied to whatever your output directory is (e.g. C:\Source\MyProject.Test\bin\Debug).

然后该文件将被复制到您的任何输出目录(例如 C:\Source\MyProject.Test\bin\Debug)。

Edit: Copy To Output = Copy if Newer is the better option

编辑:复制到输出 = 如果较新是更好的选择则复制

回答by David Z.

I'm not sure if this helps, but this looks to be briefly touched on in the following question.

我不确定这是否有帮助,但在以下问题中似乎会简要介绍这一点。

Visual Studio Solution Path environment variable

Visual Studio 解决方案路径环境变量

回答by AHM

I normally do it like that, and then I just add "..\..\"to the path to get up to the directory I want.

我通常是这样做的,然后我只需添加"..\..\"到路径即可到达我想要的目录。

So what you could do is this:

所以你可以做的是:

var path = AppDomain.CurrentDomain.SetupInformation.ApplicationBase + @"..\..\";

回答by manu_dilip_shah

Directory.GetParent(Directory.GetCurrentDirectory()).Parent.FullName;

This will give you the directory you need....

这将为您提供所需的目录....

as

作为

AppDomain.CurrentDomain.SetupInformation.ApplicationBase 

gives nothing but

什么都不给

Directory.GetCurrentDirectory().

Have alook at this link

看看这个链接

http://msdn.microsoft.com/en-us/library/system.appdomain.currentdomain.aspx

http://msdn.microsoft.com/en-us/library/system.appdomain.currentdomain.aspx

回答by Andre L.A.C Bittencourt

The best solution I found was to put the file as an embedded resource on the test project and get it from my unit test. With this solution I don′t need to care about file paths.

我找到的最佳解决方案是将文件作为嵌入资源放在测试项目中并从我的单元测试中获取。有了这个解决方案,我不需要关心文件路径。

回答by Darien Pardinas

Usually you retrieve your solution directory (or project directory, depending on your solution structure) like this:

通常你会像这样检索你的解决方案目录(或项目目录,取决于你的解决方案结构):

string solution_dir = Path.GetDirectoryName( Path.GetDirectoryName(
    TestContext.CurrentContext.TestDirectory ) );

This will give you the parent directory of the "TestResults" folder created by testing projects.

这将为您提供由测试项目创建的“TestResults”文件夹的父目录。

回答by user1207577

/// <summary>
/// Testing various directory sources in a Unit Test project
/// </summary>
/// <remarks>
/// I want to mimic the web app's App_Data folder in a Unit Test project:
/// A) Using Copy to Output Directory on each data file
/// D) Without having to set Copy to Output Directory on each data file
/// </remarks>
[TestMethod]
public void UT_PathsExist()
{
    // Gets bin\Release or bin\Debug depending on mode
    string baseA = AppDomain.CurrentDomain.SetupInformation.ApplicationBase;
    Console.WriteLine(string.Format("Dir A:{0}", baseA));
    Assert.IsTrue(System.IO.Directory.Exists(baseA));

    // Gets bin\Release or bin\Debug depending on mode
    string baseB = AppDomain.CurrentDomain.BaseDirectory;
    Console.WriteLine(string.Format("Dir B:{0}", baseB));
    Assert.IsTrue(System.IO.Directory.Exists(baseB));

    // Returns empty string (or exception if you use .ToString()
    string baseC = (string)AppDomain.CurrentDomain.GetData("DataDirectory");
    Console.WriteLine(string.Format("Dir C:{0}", baseC));
    Assert.IsFalse(System.IO.Directory.Exists(baseC));


    // Move up two levels
    string baseD = System.IO.Directory.GetParent(baseA).Parent.FullName;
    Console.WriteLine(string.Format("Dir D:{0}", baseD));
    Assert.IsTrue(System.IO.Directory.Exists(baseD));


    // You need to set the Copy to Output Directory on each data file
    var appPathA = System.IO.Path.Combine(baseA, "App_Data");
    Console.WriteLine(string.Format("Dir A/App_Data:{0}", appPathA));
    // C:/solution/UnitTestProject/bin/Debug/App_Data
    Assert.IsTrue(System.IO.Directory.Exists(appPathA));

    // You can work with data files in the project directory's App_Data folder (or any other test data folder) 
    var appPathD = System.IO.Path.Combine(baseD, "App_Data");
    Console.WriteLine(string.Format("Dir D/App_Data:{0}", appPathD));
    // C:/solution/UnitTestProject/App_Data
    Assert.IsTrue(System.IO.Directory.Exists(appPathD));
}

回答by noelicus

For NUnit this is what I do:

对于 NUnit,这就是我所做的:

// Get the executing directory of the tests 
string dir = NUnit.Framework.TestContext.CurrentContext.TestDirectory;

// Infer the project directory from there...2 levels up (depending on project type - for asp.net omit the latter Parent for a single level up)
dir = System.IO.Directory.GetParent(dir).Parent.FullName;

If required you can from there navigate back down to other directories if required:

如果需要,您可以从那里导航回其他目录(如果需要):

dir = Path.Combine(dir, "MySubDir");

回答by houss

In general you may use this, regardless if running a test or console app or web app:

通常,您可以使用它,无论是运行测试或控制台应用程序还是 Web 应用程序:

// returns the absolute path of assembly, file://C:/.../MyAssembly.dll
var codeBase = Assembly.GetExecutingAssembly().CodeBase;    
// returns the absolute path of assembly, i.e: C:\...\MyAssembly.dll
var location = Assembly.GetExecutingAssembly().Location;

If you are running NUnit, then:

如果您正在运行 NUnit,则:

// return the absolute path of directory, i.e. C:\...\
var testDirectory = TestContext.CurrentContext.TestDirectory;

回答by Ogglas

You can do it like this:

你可以这样做:

using System.IO;

Path.GetFullPath(Path.Combine(AppDomain.CurrentDomain.SetupInformation.ApplicationBase, @"..\..\"));