Is there a simple way to copy a file in a post-build event for a Visual Studio project, but only if the file exists? (i.e. don't fail the build if the file doesn't exist)
I've tried some options using xcopy. But I feel so stupid - I can't seem to get my head around what switches I might need with xcopy.
Use "IF" command:
IF EXIST file.txt xcopy file.txt [destination_folder]\ /Y
Also without the IF EXIST but using the /U option of XCOPY
xcopy source_file_name dest_folder /u /y
If you prefer to use "Copy to Output Directory" in the Properties panel (provided by Visual Studio when you right-click on a project-file and select "Properties"), you can do so with a little bit of text-editing. This will work for all versions of Visual Studio that use MSBuild (i.e. Visual Studio 2010 and onward).
First, choose an appropriate value for "Copy to Output Directory", such as "Copy always" or "Copy if newer". Save your changes to the project (e.g. "Save all" from the "File" menu).
Then, edit the project file in a text editor, and inside the opening tag of the newly-added <CopyToOutputDirectory> element, add an attribute like Condition="Exists('$(MSBuildProjectDirectory)\FILENAME')", ,where FILENAME is the path of the file being copied (i.e. as referred to in the parent element).
The GUI won't edit this setting properly in the future (i.e. it'll display "Do not copy" as the value if the file doesn't exist), but at least it'll work.
Related
I want to open a specific solution (.sln) in visual studio directly from the command line.
Using the command, I tried
devenv "full-path-to-sln-file.sln"
but it didn't work.
How do I do this?
You can just use explorer.exe !!
In your command line type in:
explorer mysolution.sln
Use the full path to the devenv.exe executable and the full path to the sln solution file, both wrapped in quotes and with a space in between. If your solution file is in a network path, make sure that it does not require authentication before accessing the destination folder.
C:\Users\YourWindowsUser>"D:\Visual Studio\Common7\IDE\devenv.exe" "\\networkDirectory\profiles\Desktop\VisualStudioSolutions\Project999.sln"
I wrote an open source application called OpenVSSolution to do exactly this.
Essentially:
Put this helper exe somewhere in your PATH
The exe scans the current directory for a .sln file
The exe opens devenv passing in the sln file
The explanation is on here:
https://davemateer.com/coding/2018/11/14/Open-visual-studio-from-command-line.html
I find this incredibly useful and is how I open all solutions.
You can open a solution file with Visual Studio directly if you are within a folder that contains the solution file, looking for it recursively (Powershell 7.0) by doing the following:
Open powershell, type
echo $profile
Open up the location of your profile, save the below into it:
function vs
{
Get-ChildItem *.sln -Recurse | Invoke-Item
}
Then just type vs into a folder and it will go through the sub directories looking for the solution file and open one if found
Try WhatsNew. From the readme:
Why fish around for Visual Studio solution files using Windows Explorer when you can find and launch them from your terminal window with just three little letters? Run sln (an alias for Open-Solution) to recursively search your current working directory for a solution file and launch it, if one is found.
for macOs I just use:
open projectname.csproj
it will open it in visual studio 2019 for me
Can a Windows batch file determine its invoked filename when invoked through a shortcut?
For example, I create real.bat, and create its shortcut named phony.bat (.lnk?)
And invoke phony by double-click on it.
Can this batch file detect the name phony.bat instead of real.bat?
Of course I can just copy it to another name, but when I edit one of them, I have to manually sync the content to all files.
The question is related to Can a Windows batch file determine its own file name?, but different.
As in your you mentioned that you've created the shortcut I assume you can create the with any properties you want.
So right click on your lnk file and change the the target line to:
C:\Windows\System32\cmd.exe /c "set "lnk_call=1"&"C:\PATH\TO\your.bat" "
This will change the icon of the link so to set back to batch file cog click on change icon and find the bat file icon in :
%SystemRoot%\System32\SHELL32.dll
Finally in your bat put this line:
if defined lnk_call echo triggered from lnk file
the lnk_call now can be used to determine if your file is called from double clicking on a .lnk file. I don't think it is possible to detect this from a shortcut that anyone else created.
Oh yeah, I found hardlink useful in this case:
mklink /h <link-name> <source-file>
I can create many hardlinks with different name, and they all points to the same file, so I can freely edit any one of them without manually sync their content.
I'm trying to change the start action for debugging in a VS2010 (C-)project to run a .bat file.
I found on MSDNA a How to: Change the Start Action for Application Debugging on how to do that. Unfortunately, I can't find the mentioned "start action", although I'm using VS2010 (no express version). I have - when choosing the "Local Windows Debugger" - a "Command" field I can edit. I tried to enter there the full path to my .bat file, but that doesn't work, because the "file is unrecognized or unsupported binary format".
Do you know a) why I don't have "start action", b) why the bat file in the "command" field won't run?
Edit This is what it looks like:
just created "console application" in vs2010 and found "Start Action" in place. What kind of project do you have?
edit:
a) "Start Action" is available only for .net projects
b) may be there is a problem with your .bat file. I just created an empty cpp project and set simple .bat file as Command. Everything works as expected.
Here is my .bat file
#echo on
pause
Try to use simple .bat and post the result here
I realize this is going to be an exotic question, but I just can't find the answer.
I'm trying to fix up and enhance an old visual basic-based application.
My problem is: where is the output directory with the compiled binaries?
Thanks.
If some clarification is needed, please ask.
In the .VBP, if there is a line specifying the path like this
Path32="C:\"
Then the resulting EXE will be built at that location. Otherwise, it will be built in the same directory as the .VBP file. The path can be relational as well and may not be a fully qualified path.
I think you want the /outdir switch. This overrides the Path32 setting in the Project file.
VB6[.EXE] [[{/run | /r}] | [/runexit] | [{/make | /m}] projectname]
[/out filename] [/outdir path] [/d const=value{[:constN=valueN]}]
[/mdi | /sdi] [{/cmd argument | /c argument}]
Options:
/run or /r projectname Tells Visual Basic to compile projectname and run it,
using the arguments stored in the Command Line
Arguments field of the Make tab of the Project
Properties dialog box.
/runexit projectname Tells Visual Basic to compile projectname and run it.
Visual Basic will exit when the project returns to
design mode.
/make or /m projectname Tells Visual Basic to compile projectname and make an
executable file from it, using the existing settings
stored in the project file.
/out filename Specifies a file to receive errors when you build using
/m or /runexit. If you do not use /out, command line
bild errors are displayed in a message box.
/outdir path Specifies a directory path to place all output files in
when using /make. This path must already exist.
/d or /D const=value... Tells Visual Basic which values to use for conditional
compilation constants when making an .EXE or ActiveX
component with the /make switch. Separate multiple
constants with colons.
/cmd or /c argument Specifies a command string to be passed to the Command$
function. When used, it must be the last switch on the
command line.
/mdi or /sdi Changes the Visual Basic environment to either Single
Document Interface (SDI) or Multiple Document Interface
(MDI) mode. Visual Basic remains in this mode until
you change it.
/? Displays a list of valid command line switches.
You may use a group name in place of projectname in any of the above switches.
Works fine here.
There is no equivalent 'bin/lib' output directory for VB6. When you compile a VB6 project, the default is to compile to the same folder as the project (vbp file). It is also possible to compile to any other folder available to the user.
What I want to do is very simple, yet can't find the way to do it.
In my projects I have several .bat files (I know, I'm old) the perform some tasks like running SqlMetal, etc.
I just want to right-click the file and select "run" or something so the bat is executed. How do I do it?
Note: I know I can use Tools -> External Tools, but it's not what I'm looking for
If each project has a few batch files associated with it then why not include them in the project and add a new external tool and link it with a custom menu item to run the tool? This will give you a list of batch files in each project and a context menu command to run them. The only downside is that it isn't file type specific (it will let you run any file as a batch file).
Here's how to do it...
Create an external tool called "Run batch file"
Set the Command to: CMD.EXE
Set the Arguments to: /c "$(ItemPath)"
Set the Initial directory to: $(ItemDir)
Check the "use output window" checkbox and then Apply to create the command
Note where the new command appeared in the list of commands. The external commands are numbered from 1 starting below the divider bar. #1 is usually "Create GUID"
Now go to Tools -> Customize and select the commands tab.
Select the Context menu radio button and select "Project and Solution Context menus | Item" from the dropdown.
Now use "Add Command..." to add a new command
In the Categories list select "Tools"
From the commands select the "External Command #" that corresponds to the position of the "Run Batch file" custom command you noted the number of in step 5 above.
Move it to the correct position in the list add keyboard shortcuts etc.
Close the dialog.
Now right click on the batch file and you should see a "Run batch file" menu item. This will execute the batch file and show its output in the VS Output window.
Right click the batch file in the Solution Explorer
Select “Open With…” from the context menu
Click “Add…”
In the “Program name” textbox, enter powershell.exe
In the “Friendly name” textbox enter “PowerShell”
Select “Set As Default”
Click OK
via
Check out the Open Command Line Visual Studio extension.
As per the description, it features:
Opening of a command line at the root of the project.
Support for all consoles such as CMD, PowerShell, Bash etc.
Syntax highlighting, Intellisense and execution of .cmd and .bat files.
For files in your solution, a context-menu button shows up.
Alternatively, the keyboard shortcut Shift+Alt+5 can be used when editing a batch file. This makes it really easy and fast to execute any batch file - even ones that are not part of your project.
Right-click a file in the Solution Explorer window, Open With, click the Add button to add your .bat files. Use %1 in your .bat to get the path to the selected file.
Couldn't get any of the above to quite work, but this did (VS 2019/2022):
Right click a .bat file in Solution Explorer and Open With...
Add...
Program: CMD.exe
Arguments: /c %1
Friendly name: Command
OK and Set as Default for double clicking of batch files to run from Solution Explorer as the default (or right click Open With... and select Command if you don't want it the default).
Merely a suggestion: This is a scenario which can be done with the VsCommandBuddy extension. The following configuration is an example:
{
"cmdname": "mk",
"title": "Run MK.BAT",
"description": "Run the MK bat file.",
"cwd": "$(SolutionDir)",
"filename": "cmd.exe",
"arguments": "/C mk.bat",
"async": false
}
You can assign shortkeys as well, so you can invoke your batch files with shortcut keys. Output comes in your output pane.