I'm using Visual Studio 2022.
There is no Copy to Output Direcotry for files in Solution Items folder. But, this option does exist for files within a project.
Is there a way to make this happen instead of adding those files to a project or adding "xcopy..." to "Post-build event" for each project's page property?
I've got a Class Library project and a Console project. Inside the Class Library project there's a folder called "ConfigSettings" containing a configuration file that's referenced by the App.config file. Its "Copy to Output Directory" is set to "Copy Always".
The Console project references this Class Library project. When the console project is built, the ConfigSettings folder is created in the correct place along with the external config file. However, every time I Debug the Console project, the external config file is deleted. The ConfigSettings folder however is left untouched. If I mark the file as read-only it is not deleted, and warning appears in Visual Studio's Error List:
Warning Unable to delete file "C:\<path>\<app>\bin\Debug\ConfigSettings\settings.config". Access to the path 'C:\<path>\<app>\bin\Debug\ConfigSettings\settings.config' is denied.
Is there a setting I can change that will prevent VS from removing this file when I Debug the project? I am using Visual Studio 2015.
This might not be the best solution possible, but I found here, if you do a "Clean solution" just before executing your solution, the files will remain.
I'm trying to create a Visual Studio plugin, it's a menu item that executes batch files. I have no idea how to include the batch files (or any other additional files) with the VSIX when publishing so that they are available to all users that install the extension.
In solution explorer right click on the batch file (in this case I called it BatchFile.cmd) and choose 'Properties'
In the properties window change:
Build Action: Content
Include in VSIX: True
When the solution is built in release mode it creates a VSIX file in the bin/Release folder. This is the package and it contains all the assets required. When the package is installed on another machine, the batch file is included in the install location and can be referenced using:
Path.GetDirectoryName(Assembly.GetExecutingAssembly().Location) + "BatchFile.cmd"
You can just include the batch file as content in your project, and use GetAssembly() to find the location of your adin dll at runtime
When I build a Visual Studio project, the executable is written to the output directory specified in the projects Property Page.
I have a project that has some extra files (e.g., .ini file) that are used by the program.
How can I configure the project to copy the file to the output directory so that when the program runs, it has a copy of the other file in its CWD?
I checked the Property Page of the file and there was nothing useful other than an option to exclude it from the build (which is disabled), and the custom-build-tool command is empty (plus it is a plain-text file that does not need any processing).
For copying a files to the output directory in Visual Studio 2003 you could use Post-Build event:
Right click on the project->Properties
Common Properties->Build Events
Set Post-Build Event Command Line to:
xcopy /y $(ProjectDir)my_file.ini $(ProjectDir)$(OutDir)
OK and build!
Please try select the file in Solution Explorer. Then you should be able to see its properties in Properties window (press F4 if it is not visible). You will find there two properties:
"Build Action" and
"Copy to Output Directory"
Set "Build Action" to "Content", and then - select an appropriate value for "Copy to Output Directory" setting.
File properties window with "Build Action" and "Copy to Output Directory" settings
If the way above doesn't work for you, please read this post "Copy to output directory issue with .inf file". And have a look at this one then "Visual Studio: default build action for non-default file-types"
While I was searching the file’s Property Page for a build-action field, I had a thought: set the custom build step to copy the file (manually). This turned out to be easier than I thought. I had figured it would require using cmd or other external executable (xcopy, robocopy, etc.), but that is not necessary.
I set the Custom Build Step as follows:
Command Line : copy $(InputFileName) $(OutDir)
Description : Copying foobar...
Outputs : $(InputFileName)
Setting the outputs field (correctly) was critical in order to prevent VS from always thinking the project is out of date and requiring to be rebuilt (I’m not certain if it needs to be prefixed with $(OutDir)\).
It is reflected in the Output window as such:
Copying foobar...
1 file(s) copied.
Compiling resources...
Linking...
For VS 2017 the command Dmitry Pavlov posted would be the following:
xcopy /y "$(ProjectDir)my_file.ini" "$(OutDir)"
Quotes are important in case there are spaces in the path to the project directory.
Expanding on Synetech's answer.
In VS2019 right click the file you want to copy in the Solution Explorer and select Properties. Then under General >> Item Type change to Copy File then hit Apply.
You now should have UI fields in the Properties Page for Destination etc.
In case this helps anyone, I needed to copy the output dll of the project i was building into another project.
xcopy /y "$(ProjectDir)$(OutDir)$(TargetName)$(TargetExt)"
"C:\Application\MyApplicationName\bin\x86\Debug"
/y = overwrite file if already exists
$(ProjectDir) = location on your machine where the project lives
$(OutDir) = is where your current build setup outputs the build
$(TargetName) = What the project being built is set to be called. Ex: XXX of XXX.dll
$(TargetExt) = the extension of the build Ex: .dll of XXX.dll
"C:/..../x86/Debug" is the location to copy to.
You need the extra $(OutDir). Otherwise, in the rebuild/clean step it will throw away your source.
CommandLine : copy "$(SolutionDir)last-script.js" "$(TargetDir)Debug"
Outputs : $(TargetDir)Debug\last-script.js
Improving Synetech
answer :
In VS 2013 C++ project Command Line : copy %(Identity) $(OutDir) Description : Copying foobar... Outputs : %(Identity)
It works , But it leads to circular dependency , i.e. it will be executed each time you demand increamental build, no meter it has been already copied.
To solve this , you can add that item at target folder, change path to $(OutDir), and use that in first added item as Output. Drawback - two items with similar name are in solution.
Also usefull xcopy with /d /y parameters in postbuild - copy only if target file date is older.
You could also after the unload the project (Right click on the project >> Unload Project) add the following inside an existent <ItemGroup> tag:
<Content Include="..\..\Config\db.config">
<Link>Config\db.config</Link>
<CopyToOutputDirectory>Always</CopyToOutputDirectory>
<SubType>Designer</SubType>
</Content>
In this case it will grab the db.config file from 2 folders up and put it in the output folder(by default bin/Debug) after creating a Config folder with the db.config file inside
I want to change the build system path for building my VC++ project in VS2005.
When I try to build the project, I'm getting an error that a specified header file cannot be opened. I have that header file in "Microsoft SDKs\Windows\v7.0\Include".
But the path present in 'Build system path' is "Microsoft SDKs\Windows\v7.0\Include" which i found in the Error dialog box when i tried to open that header file through the code.
So I want to change the path in build system path to "Microsoft SDKs\Windows\v7.0\Include".
How can i do this.? How to open Build system paths in VS2005.?
You can add include directives rather than change the build system path. Right click the project in Solution Explorer | properties
C/C++ | General
Select 'Additional include directives'
Here browse to the location of where you include files are.