The default value of working directory in visual studio property sheet is $(ProjectDir). If we have some dependencies (e.g., config files or libs & dlls), we have to put them in the working directory.
Since the default $(ProjectDir) is different from the output directory (namely $(TargetDir), folder where the EXE file resides), we have to copy the dependencies to the output directory if we run the EXE file by double clicking it.
So I'd like to change my working directory to $(TargetDir), in case I have to copy my dependencies twice. The question is: Why is the default working directory set to $(ProjectDir) but not $(TargetDir)? Is there any good reason for visual studio to do so?
I want VS to create subfolders in release directory with some linked files from project. Something like this:
Project/
-bin/
--Relase/
---program.exe
---Resources/
----input.txt
---Config/
----default.conf
How can I do this while having files "input.txt" and "default.conf" in solution hierarchy?
Heh, it's easy: just place needed items inside project in solution explorer and in properties window change property "Copy to output directory" to "Copy always" or "Copy if newer" .
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
In the context of a Visual Studio 2008 Setup Project, what are "Content Files". In other words, when creating a setup project and defining the File System settings and choosing: Add Project Output > Content Files, what files will be added?
For example, what has to be true about a file or its location for it to be considered a Content File for a given project's output?
The build action property of the file will be labeled "content". Here is a link to more information about File Properties.