问题
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).
回答1:
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!
回答2:
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"
Hope that helps
回答3:
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Setting the outputs field (correctly) was critical in order to prevent VS from always thinking the project is out of date and requiring to be rebuilt (I’m not certain if it needs to be prefixed with $(OutDir)\).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...
回答4:
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.
回答5:
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.
回答6:
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
回答7:
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.
来源：https://stackoverflow.com/questions/10827024/copying-visual-studio-project-files-to-output-directory-during-build