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 have a project in Visual Studio and want all files generated during build under the build\ subdirectory.
I set all output options I could find but there is still a Debug\ folder created with the files: projectname.dll.recipe and subfolder projectname.xxxx.tlog which contains .tlog and .lastbuildstate files.
I know the .tlog files are from the MSBuild File Tracker, know idea what the .dll.recipe is.
How can I set the output directory for those files to $(OutDir)?
You can try the following steps:
Function 1
Open your project in VS IDE and then right-click on the project Properties-->Configuration Properties-->General--> change Intermediate Directory to ..\$(Configuration).
Then, rebuild your project to get what you want.
Function 2
1) add a file called Directory.Build.props in your solution folder like this:
2) add these content in the Directory.Build.props:
<Project>
<PropertyGroup>
<IntDir>..\$(Configuration)\</IntDir>
</PropertyGroup>
</Project>
3) close VS Instance, delete any Debug folder in your solution folder.
Then restart VS(enable the function of the Directory.Build.props) to open your solution to build again and the files will be under the $(OutDir) path.
I have some custom made XML files and a read me file that I've included in my project.
How do I make it so that these files are copied to my debug/release folders automatically when I build my project? I've tried setting the Copy to Output Directory property to "Copy Always" but this doesn't seem to do the trick.
I'm using VS2010
I've found the answer. The build action needs to be set to Content if you want to just directly copy the file to the output folder.
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
Here is the problem:
1. Create a TestProject in your Visual Studio solution.
2. Open the .testrunconfig file and under the 'deployment' menu item, select the 'Enable Deployment' checkbox.
3. Now click on 'Add Directory...' button and add a folder which has some files in it.
4. Run the test project (use a dummy test).
Ok, now go check the TestResults folder: You will see that all the files got directly copied (to the top level)- the folder itself is not copied (with the files under them). This messes up my paths during testing. Can anybody tell how to get the folder copied instead of just the files underneath ?
Thanks.
Use the [DeploymentItem] attribute on the test classes that use it. You can specify a directory:
[TestClass]
[DeploymentItem("blahblah\\myDirectory", "myDirectory")]
public class MyTest
{
}
Note:
DeploymentItem is very slow when starting the tests. It seems to copy 2 files per second.
You can specify the attribute on a test base class. But it does not always work if you have more than one test project.
You can probably specify it on a TestClass that has a method marked with [AssemblyInitialize]. Then you have only to provide it once. Not sure, you have to try.
The source directory is relative to the solution location. This is hardly documented.
Open the .testsettings file in notepad. Now, you should see that for every folder to copy
<DeploymentItem filename="FolderName\" />
Change this to
<DeploymentItem filename="FolderName\" outputDirectory="FolderName\" />
The other option you have is to create another folder beneath the original folder, and then that folder will be deployed to the out directory. For example you can have this structure:
TestFolder/
TestFolder/TestDeployment/
And then in the testrunconfig you still select the TestFolder folder and the TestDeployment folder will be deployed to the out directory.
I just had this problem too today. I solved it by adding a folder called "deployment_files" in the project that contained the required folder. Then I put the required folder into the "deployment_files" folder. THEN, I opened the LocalTestRun.testrunconfig file under the "Solution Items" folder in the Solution Explorer. Went to the "Deployment" panel in the testrunconfig property window. Added the "deployment_files" directory to the deployment and voila. The folder within that was copied to the test results Out folder.