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Setting the outputs field (correctly) was critical in order to prevent VS from always thinking the project is out of date and requiring to be rebuilt (I’m not certain if it needs to be prefixed with $(OutDir)\).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
Related
I have a C# Visual Studio solution with about 15 projects. When I build the solution I want all DLL and EXE files for each project to go to a common folder (called Deploy).
The way I was thinking about doing it was, for each project's Post-build Event Command Line section put the following commands:
IF NOT EXIST $(SolutionDir)Deploy (
'If directory does not exist, create it
MKDIR $(SolutionDir)Deploy
) ELSE (
'Delete directory to make sure it's "clean"
RMDIR /F /S /Q $(SolutionDir)Deploy
MKDIR $(SolutionDir)Deploy
)
'Copy executable and required DLLs to Deploy directory
COPY /Y $(TargetPath) $(SolutionDir)Deploy
COPY /Y $(TargetDir)*.dll $(SolutionDir)Deploy
The problem with doing it this way, however, is I have 15 projects and would have put this in each individual project's post build event section and also, every time I add a new project, I would have to remember to do the same for it.
I checked the solution file properties and did not see a way to set a solution-wide post build event to copy all the files so I did a few Google searches.
One page said to use a C++ Makefile project. I added this type of project to my solution and clicked on the project properties page and found that there is a section under Configuration Properties->NMake with the following:
Build Command Line
Rebuild All Command Line
Clean Command Line
Using the Makefile project's Command Line option poses a similar problem to above. Many different commands to copy each of my 15 project's output files such as:
COPY /Y $(SolutionDir)Project1\bin\$(ConfigurationName)\*.exe $(SolutionDir)Deploy
COPY /Y $(SolutionDir)Project2\bin\$(ConfigurationName)\*.exe $(SolutionDir)Deploy
...
COPY /Y $(SolutionDir)Project15\bin\$(ConfigurationName)\*.exe $(SolutionDir)Deploy
There is another apparent problem with doing it this way. As you can see I took advantage of the $(SolutionDir) and $(ConfigurationName) macros but I had to hard-code each project name.
I didn't see any macros like $(AllProjects), $(AllProjectDirs), etc.
Also, it appears that command line commands for Makefile projects are for building, not post-build events, so I gave up on this idea altogether.
I then tried using a Visual Studio Installer project. After adding the project to my solution I right-clicked the project and saw that there was an Add->Project Output... option. This brought up a dialog allowing me to add one of my other project's Primary Output. I repeated this for each of my other projects and rebuilt.
What resulted was an .MSI file in the output folder. I then opened installer project properties and changed the Package files option to As loose uncompressed files and rebuilt. The output folder now contained all my project's EXE and DLL files!
Most people would be satisfied at this point and move on but I am the kind of person who likes to find the best way to do things.
There was one thing I didn't like about using the installer project option, the fact that, besides copying the files from all my projects, it also created an MSI file (which I don't need) and I didn't see any option tell it not to create one.
Can anyone recommend a another/better way to accomplish my goal of copying all project output files to a single folder?
Thank you.
P.S. I was thinking I could just make a batch file to search and copy all the EXE and DLL files to the Deploy folder but I would have to run the batch file outside of the Visual Studio IDE and also hard-code the configuration folder (Debug or Deploy).
Can't you just change the Output Directory of the C++ projects? See How to: Change the Build Output Directory.
On the menu bar, choose Project, Properties.
Expand the Configuration Properties node, and choose General.
Change the Output Directory value to the new output directory.
If you want both options, you can also create multiple configurations for your VS projects and solutions, similar to the standard "Debug" and "Release" ones. Create a new configuration from one of the existing ones, then change the output directory and save. Now you can just switch the configuration at the solution level to build into another directory. See this link for detailed steps:
How to: Create and Edit Configurations
I have configured pre build events for a proj in visual studio.
All the files are copied in a particular folder as a part of this build command.
Recently i have added crx and xpi files in source folder.
These files alone are not getting copied to the destination folder.
Here is the build command:copy "$(SolutionDir)DLLRequired*.*" "$(TargetDir)"
here even though i have used "*.*", it skips the crx and xpi files.
Please help.
If these files are part of your project - you need to right click on them, select properties and in Copy to Output Directory select Copy always.
If these files are dynamic or are not part of your project - you should try doing this: XCOPY $(ProjectDir)\Libs\DLLRequired*.* $(TargetDir) /Y /R - this is using ProjectDir instead of SolutionDir and should always work.
Also if something does not work - check out Output windows for any error messages.
I have a few VS 2010 C# projects that are shared between several solutions. I would like these projects to build to the directory of the solutions they are open in. How do I do this?
I considered setting up different build configurations (Debug_Xsln, debug_Ysln, Release_Xsln...) but wasn't sure if there was a better way.
http://msdn.microsoft.com/en-us/library/42x5kfw4(v=VS.100).aspx
You can use a postbuild event with xcopy and the macro $(SolutionName) or $(SolutionDir) to copy the compiled files into the correct folder.
Go into properties for the project, build events tab, and in Post Build event command line enter something like this:
xcopy "$(ProjectDir)bin\$(ConfigurationName)\*.*" "$(SolutionDir)$(ProjectName)\bin\$(ConfigurationName)" /i /d /y
The benefit of this method is you can copy the build output of one project to multiple locations
OR
(as Ziplin discovered)
If you only have one location you want the build output to go, you can use the macros above to set the output path, like this:
$(SolutionDir)$(ProjectName)\bin\$(ConfigurationName)
just go to the project properties on the build tab and set your macroed location as the output path
I'm trying to publish my web solution to the file system, using the "Only files needed to run this appllication" option.
I have the following problems:
a) I have a connectionStrings.config file that should be included; I've marked this file as Content and Copy Always.
b) For testing purposes, I've added a textfile.txt that should not be included; I've marked this as Content and Do Not Copy.
Both of these settings fail...
It seems that the setting "Copy to output directory" is simply disregarded.
Any ideas why this happens, and what to do about it?
The "Copy to output directory" is for when you are building your application, not publishing it.
This property specifies the conditions under which the selected source file will be copied to the output directory. Select Do not copy if the file is never to be copied to the output directory. Select Copy always if the file is always to be copied to the output directory. Select Copy if newer if the file is to be copied only when it is newer than an existing file of the same name in the output directory.
Source
There's more information on publishing web sites and applications here
I had the same problem, that files marked as content, with "copy always" set, were not appearing in the output dir.
At the time i noticed this, i was tracking down why certain of my dlls were not getting deployed.
Sorting my project dependencies fixed both issues.
I just had this problem and for some reason choosing "Create application without a manifest" under the project's properties finally copied the linked content file to the build directory.
I re-started my visual studio and select the "copy always" option again. Rebuild and worked.
Good Luck.
I have a Visual Studio C++ project that relies on an external DLL file. How can I make Visual Studio copy this DLL file automatically into the output directory (debug/release) when I build the project?
Use a post-build action in your project, and add the commands to copy the offending DLL. The post-build action are written as a batch script.
The output directory can be referenced as $(OutDir). The project directory is available as $(ProjDir). Try to use relative pathes where applicable, so that you can copy or move your project folder without breaking the post-build action.
$(OutDir) turned out to be a relative path in VS2013, so I had to combine it with $(ProjectDir) to achieve the desired effect:
xcopy /y /d "$(ProjectDir)External\*.dll" "$(ProjectDir)$(OutDir)"
BTW, you can easily debug the scripts by adding 'echo ' at the beginning and observe the expanded text in the build output window.
The details in the comments section above did not work for me (VS 2013) when trying to copy the output dll from one C++ project to the release and debug folder of another C# project within the same solution.
I had to add the following post build-action (right click on the project that has a .dll output) then properties -> configuration properties -> build events -> post-build event -> command line
now I added these two lines to copy the output dll into the two folders:
xcopy /y $(TargetPath) $(SolutionDir)aeiscontroller\bin\Release
xcopy /y $(TargetPath) $(SolutionDir)aeiscontroller\bin\Debug
To do it with the GUI, first add the file(s) to the project: right-click the project, select "Add...", then "Existing Item", then browse to the file or files you want to add and click "Add". Next, tell Visual Studio to copy the file when you build: right-click the file you want to copy, select "Properties". You'll see a list of properties, including "Item Type". Change the "Item Type" to "Copy File". Hit OK and you're done.
Here's the file properties dialog:
Looking in the *.vcxproj file, the steps above add something like this:
<ItemGroup>
<CopyFileToFolders Include="libs\a.dll" />
<CopyFileToFolders Include="libs\a.dll" />
</ItemGroup>
I couldn't find any official documentation for <CopyFileToFolders>, but clearly it's supported or the GUI wouldn't use it. But, if you're doing it by hand and an undocumented item type makes you uncomfortable you can always use the well known but slightly more verbose <Content> type:
<ItemGroup>
<Content Include="libs\a.dll" >
<CopyToOutputDirectory>PreserveNewest</CopyToOutputDirectory>
</Content>
<Content Include="libs\b.dll" >
<CopyToOutputDirectory>PreserveNewest</CopyToOutputDirectory>
</Content>
</ItemGroup>
Add builtin COPY in project.csproj file:
<Project>
...
<Target Name="AfterBuild">
<Copy SourceFiles="$(ProjectDir)..\..\Lib\*.dll" DestinationFolder="$(OutDir)Debug\bin" SkipUnchangedFiles="false" />
<Copy SourceFiles="$(ProjectDir)..\..\Lib\*.dll" DestinationFolder="$(OutDir)Release\bin" SkipUnchangedFiles="false" />
</Target>
</Project>
(This answer only applies to C# not C++, sorry I misread the original question)
I've got through DLL hell like this before. My final solution was to store the unmanaged DLLs in the managed DLL as binary resources, and extract them to a temporary folder when the program launches and delete them when it gets disposed.
This should be part of the .NET or pinvoke infrastructure, since it is so useful.... It makes your managed DLL easy to manage, both using Xcopy or as a Project reference in a bigger Visual Studio solution. Once you do this, you don't have to worry about post-build events.
UPDATE:
I posted code here in another answer https://stackoverflow.com/a/11038376/364818
xcopy /y /d "$(ProjectDir)External\*.dll" "$(TargetDir)"
You can also refer to a relative path, the next example will find the DLL in a folder located one level above the project folder. If you have multiple projects that use the DLL in a single solution, this places the source of the DLL in a common area reachable when you set any of them as the Startup Project.
xcopy /y /d "$(ProjectDir)..\External\*.dll" "$(TargetDir)"
The /y option copies without confirmation.
The /d option checks to see if a file exists in the target and if it does only copies if the source has a newer timestamp than the target.
I found that in at least newer versions of Visual Studio, such as VS2109, $(ProjDir) is undefined and had to use $(ProjectDir) instead.
Leaving out a target folder in xcopy should default to the output directory. That is important to understand reason $(OutDir) alone is not helpful.
$(OutDir), at least in recent versions of Visual Studio, is defined as a relative path to the output folder, such as bin/x86/Debug. Using it alone as the target will create a new set of folders starting from the project output folder. Ex: … bin/x86/Debug/bin/x86/Debug.
Combining it with the project folder should get you to the proper place. Ex: $(ProjectDir)$(OutDir).
However $(TargetDir) will provide the output directory in one step.
Microsoft's list of MSBuild macros for current and previous versions of Visual Studio
I had a similar question. In my project, there were couple of external DLLs. So I created a new folder in the project called "lib" and copied all the external dlls to this folder.
Add a reference to these DLLs.
Go to Project References>dll properties and change the following properties
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