在C#中检索构建输出文件夹 [英] Retrieving build output folder in C#
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问题描述
我想得到项目构建输出文件夹的名称,例如,如果构建配置是Debug,那么它应该是bin / Debug或者如果构建配置是然后释放它应该是bin / Release。
我使用了以下代码
Hi,
I want to get the name of project build output folder e.g if Build configuration is Debug then it should be bin/Debug or if Build configuration is Release then it should be bin/Release.
I have used following code
Microsoft.Build.Evaluation.Project project =
new Microsoft.Build.Evaluation.Project(projectfilename);
string assemblyPath = project.GetPropertyValue("OutputPath");
但我总是得到assembleyPath =bin / Debug配置是什么。
我很感激提前回复。
我尝试过的事情:
我使用了以下代码
But I am always getting assembleyPath = "bin/Debug" what ever the configuration is.
I am thankful to any response in advance.
What I have tried:
I have used following code
Microsoft.Build.Evaluation.Project project =
new Microsoft.Build.Evaluation.Project(projectfilename);
string assemblyPath = project.GetPropertyValue("OutputPath");
推荐答案
可能最安全的方法是获取可执行文件路径:Application.ExecutablePath属性(System.Windows.Forms)| Microsoft Docs [ ^ ]
Probably the safest way is to fetch the executable path: Application.ExecutablePath Property (System.Windows.Forms) | Microsoft Docs[^]
您可以使用:
Assembly.GetExecutingAssembly().Location
但不适用于Windows服务,那么最好使用:
But not for a Windows service, then it's better to use:
string currentDir = Path.GetDirectoryName(AppDomain.CurrentDomain.BaseDirectory) ?? string.Empty;
我找到了一些示例,说明如何使用 project.GetPropertyValue():
Project.GetPropertyValue,Microsoft。 Build.Evaluation C#(CSharp)Code-Beispiele - HotExamples [ ^ ]
希望您觉得这个答案更有帮助:)
I found some examples that show how to use project.GetPropertyValue():
Project.GetPropertyValue, Microsoft.Build.Evaluation C# (CSharp) Code-Beispiele - HotExamples[^]
Hope you find this answer more helpful :)
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