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.
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
Inside the project folder there is a "project's name" folder, Release and Debug folder. I'm right now working in debug mode and my files keep getting updated everytime i run it, and i copy it into the other 2 folders everytime to avoid confusion. Is this the right way to do it or is it enough if the files are in the main folder ?
If you have files in the project that need to be copied to the current output folder on build then select the appropriate options in the file's properties: select the file in solution explorer and then set the properties of that file:
Build Action: None
Copy to output directory: Copy Always, or Copy If Newer.
Thus VS will put the files in the right place on build. You do not need to maintain multiple copies.
If your program needs access to text files, images, etc. which are separated from the program's code, you can put these both into the debug or into the release folder.
Beware that if you modify these files into the debug folder, you must update the files into the release, and vice-versa. This is valid if you use to switch between the debug mode and the release mode. If you use only the debug mode as I do while I'm developing, you could put all the files in the debug folder only, and copy these files into the release folder only when you will end the development.
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 $(OutDir), and use that in first added item as Output. Drawback - two items with similar name are in solution.
Also usefull xcopy with /d /y parameters in postbuild - copy only if target file date is older.
You could also after the unload the project (Right click on the project >> Unload Project) add the following inside an existent <ItemGroup> tag:
<Content Include="..\..\Config\db.config">
<Link>Config\db.config</Link>
<CopyToOutputDirectory>Always</CopyToOutputDirectory>
<SubType>Designer</SubType>
</Content>
In this case it will grab the db.config file from 2 folders up and put it in the output folder(by default bin/Debug) after creating a Config folder with the db.config file inside
I am currently creating a console application which is reading from a folder which contains xml files, like so:
/Templates/Create.xml
/Template/Update.xml
But when I debug the application it is looking for the files in /bin/Debug/Templates because I use: System.IO.Path.GetDirectoryName(System.Reflection.Assembly.GetExecutingAssembly().Location);
So it is doing exactly what i want it to ..but is there any way to get Visual Studio to copy the folder and files automatically to the Debug/Release folder?
At the moment I have to keep the two folders updated.
I have done this before in a Windows application, but I can't seem to figure it out in a console
SOLUTION:
I added the following code in Pre-build:
XCOPY "$(ProjectDir)Templates" "$(TargetDir)Templates" /R /Y /I /T
..and it works like a charm
In visual studio you can have post build actions. In this you can write command to copy the xml to the Debug/Release folder depending upon your BuildMode.
Update:
In Project Properties you can find this.
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