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.
Related
I am trying to copy my build to a particular drop location. But instead of copying the some important files and directories it is copying all the solution file and directories there like .csproj files and .csfiles and also all the folders present there.
I only want to copy dlls(projects dlls and references dlls of the project) and some data Folders and files to drop location so that I can use them to execute my test cases. But it ends up copying the whole solution.
Below are the setting I am using to copy and publish build artifacts.
Can Anyone tell me how can it be done?
Note: when I tried to put $(build.sourcesdirectory)\bin in the copy root it gave me an error that "Cannot bind argument to parameter 'Path' because it is null".
I also tried putting only **.dll in my contents but then it copies nothing. The folder created in drop location comes empty.
You're telling it to publish **/*test*.dll, from the root of the build agent. That minmatch pattern is saying "find everything in every subfolder that matches test.dll", so the results you're getting aren't surprising.
What you probably want to do is specify the path to the binary output location of your test projects. Without seeing the structure of your code or what MSBuild variables you're providing/overriding, it's impossible to give more guidance. Look at the documentation on pre-defined variables and that should give you some ideas.
In general, you want to set your output folder (via MSBuild arguments) to $(Build.ArtifactStagingDirectory), so only build outputs go there. Then you can use that as your Copy Root.
Similar to the question here: Copy one file in target directory on deploy from visual studio team services, I am trying to copy the contents of the app.publish directory of a ClickOnce application via VSTS build/release.
This is the Copy Publish Artifact step of the Build
This obviously outputs everything in the bin folder to the drop folder on the server share.
This is the Copy Files step of the Release
This copies only the files in the app.publish folder but it builds out the entire hierarchy to the folder app.publish folder; however, it builds out the full folder structure. i.e. bin\Release\app.publish or bin\Debug\app.publish.
I like the idea of being able to publish and store different configurations (debug/release) but I don't like that I'm having to specify the InstallUrl in the msbuild arguments in order to pick up debug/release in the folder path. Even if I didn't try to maintain multiple configurations, the configuration would still be in the path.
So my question is: Is there a way to use the copy files task to just copy the app.publish folder and/or the contents to another location, without it creating the full hierarchy?
You need to specify the copy root in Source Folder of Copy Files, in your example, the copy root should be like:
Source Folder: $(System.DefaultWorkingDirectory)\$(Build.DefinitionName)\bin\$(BuildConfiguration)\
Contents: **\app.publish\**
In this way, you should only have app.publish in your Target Folder.
Configure the "Copy Root" of "Copy Publish Artifacts" step in your build definition to the path of build configuration folder and "Content" to "**".
For example, if the path of your build output is:
C:\a\1\s\SolutionFolder\ProjectFolder\bin\Relase\app.publish...
You can set the Copy Root to:
$(build.sourcesdirectory)\SolutionFolder\ProjectFolder\bin\$(BuildConfiguration)
And Content set to:
**
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 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.