I created a batch file to open VS Code to a specific folder. Here is the code -
cd z:\files\development\xampp\code\htdocs\Acme1
code .
Note that Z: maps to an external SSD.
The above works as expected.
Then I added more lines to try and open multiple instances of VS Code -
cd z:\files\development\xampp\code\htdocs\Acme1
code .
cd z:\files\development\xampp\code\htdocs\Acme2
code .
I expected this to open 2 separate instances of VS Code, one with the Acme1 folder open, the other with the Acme2 folder open. However, it only opens VS Code once, with the Acme1 folder open.
Per this answer, I modified my Windows 11 Path environment variable to include C:\Users\knot22\AppData\Local\Programs\Microsoft VS Code\bin. Alas, this made no difference in the behavior of the batch file, even after restarting the machine.
What needs to be changed to get the batch file to open both instances of VS Code?
This is how you open multiple vscode instances with batch
code your_first_path & code your_second_path & code your_third_path etc.
so in your case this is how you open both folders with vscode
code z:\files\development\xampp\code\htdocs\Acme1 & code z:\files\development\xampp\code\htdocs\Acme2
alternatively you can open multiple vscode instances this way
cmd /c code your_first_path
cmd /c code your_second_path
cmd /c code your_third_path etc.
Related
Simply put, I have a VBScript titled "tyrian_soundtest.vbs" that plays an .mp3 that is titled "tyrian_soundtest.mp3"
The VBScript code is below
Set Sound = CreateObject("WMPlayer.OCX.7")
Sound.URL = "tyrian_soundtest.mp3"
Sound.Controls.play
do while Sound.currentmedia.duration = 0
wscript.sleep 1
loop
wscript.sleep (int(Sound.currentmedia.duration)+1)*1000
When opened, it plays the .mp3. Simple enough.
The trouble comes in when I run a batch script titled "tyrian_soundtest.bat". Relative to it, the .vbs and .mp3 are in a folder called sfx. Here is what one iteration of that file contained.
#echo off
start %cd%\sfx\tyrian_soundtest.vbs
exit /b
The result is an error stating that Windows couldn't find the file path, likely due to it containing a space. Other attempts of the .bat were replacing line 2 with
start .\sfx\tyrian_soundtest.vbs
or
start "%cd%\sfx\tyrian_soundtest.vbs"
Any attempt I've made gives one of three results. Option 1: There is no error, but the audio simply never plays. Option 2: An error is thrown about the file directory not being found. Option 3: That file path opens up in a new cmd window, but the .vbs is never run.
Is there any way format the .bat to get the .vbs to run through the without an error being caused?
The main issue is that there is used in batch file and in the VBScript file the current directory path. The current directory on starting %SystemRoot%\System32\cmd.exe to process the batch file can be any directory.
The Windows Explorer sets the directory of the batch file as current directory on double clicking the batch file resulting in starting cmd.exe by explorer.exe to process the double clicked batch file of which full qualified file name is passed to cmd.exe after the option /c as additional argument. But if the batch file is stored on a network resource accessed using UNC path, the Windows command processor changes the current directory from the network resource to %SystemRoot% (the Windows directory) and the batch file fails to start Windows Script Host to process the VBS file. See also: CMD does not support UNC paths as current directories.
A batch file can be also started by right clicking on it and using Run as administrator. This can result in making the directory %SystemRoot%\System32 (the Windows system directory) the current directory. See also: Why does 'Run as administrator' change (sometimes) batch file's current directory?
That are just two of many examples where the current directory is different to the directory containing the batch file. So if a batch file references other files stored in same directory as the batch file itself or a subdirectory of the batch files directory, it is advisable to reference these files with the full path of the batch file instead of using a path relative to current directory.
Example:
The used files are stored in directory C:\Temp\Development & Test as follows:
sfx
tyrian_soundtest.vbs
tyrian_soundtest.mp3
tyrian_soundtest.bat
A user opens a command prompt window which usually results in making the directory referenced with %USERPROFILE% or with %HOMEDRIVE%%HOMEPATH% the current directory. A user executes next in the command prompt window:
"C:\Temp\Development & Test\tyrian_soundtest.bat"
So the current directory is definitely not the directory containing the batch file.
The batch file can be coded as follows to start nevertheless the Windows Script Host for processing the VBS file tyrian_soundtest.vbs and successfully play the MP3 file tyrian_soundtest.mp3.
#start "" /D "%~dp0sfx" %SystemRoot%\System32\wscript.exe //NoLogo "%~dp0sfx\tyrian_soundtest.vbs"
%~dp0 references the drive and path of argument 0 which is the full path of the currently processed batch file. The batch file path referenced with %~dp0 always ends with a backslash. For that reason the concatenation of %~dp0 with a file/folder name or wildcard pattern should be always done without using an additional backslash as that would result in two \ in series in complete argument string and the Windows file management would need to fix that small error by replacing \\ by just \ before passing the argument string to the file system. See also the Microsoft documentation about Naming Files, Paths, and Namespaces which explains which automatic corrections are usually applied on file/folder strings before passing them to the file system.
The internal command START of cmd.exe interprets the first double quoted argument string as title string for the console window as it can be seen on running in a command prompt window start /? and reading the output help. For that reason it is not enough to use just:
#start "%~dp0sfx\tyrian_soundtest.vbs"
That would result in starting one more command process with its own console window with the full qualified file name of the VBS file as title of the console window.
The Windows Script Host processing a VBS file by default on Windows exists in two versions:
%SystemRoot%\System32\cscript.exe is the console version.
%SystemRoot%\System32\wscript.exe is the Windows GUI version.
The usage of the console version cscript.exe results usually in opening a console window by the parent process if the parent process is not itself a console application running already with an opened console window like on execution of a batch file processed by %SystemRoot%\System32\cmd.exe being also a console application.
The usage of the Windows GUI version wscript.exe results in no opening of a window by default at all. The processed script file must contain commands to open a window if that is wanted at all.
The difference can be also seen on running from within a command prompt window cscript /? and next wscript /?. The first command results in printing the help for the command line options of Windows Script Host in already opened command prompt window while the second command results in showing a graphic window by wscript.exe displaying the same usage help.
The usage help of Windows Script Host explains also how each user can define which version of Windows Script Host is the default for executing scripts. So it is not advisable to specify just the VBS file name with full path on the command line with start and let cmd.exe look up in Windows registry which version of Windows Script Host to run to process the VBS file. It is better to explicitly run the version of Windows Script Host most suitable for playing the MP3 file which is in this case the Windows GUI version wscript.exe opening by default no window at all to play the MP3 file in background.
So it would be possible to use:
#start "" %SystemRoot%\System32\wscript.exe //NoLogo "%~dp0sfx\tyrian_soundtest.vbs"
There is an empty title string defined with "" as the started executable wscript.exe is a Windows GUI application for which no console window is opened at all by cmd.exe. So the title string does not really matter and can be an empty string.
But there is left one problem with that command line. The VB script references the MP3 file without path which means with a path relative to current directory. The current directory is %USERPROFILE% and not C:\Temp\Development & Test\sfx which contains the MP3 file tyrian_soundtest.mp3. So the VB script would fail to find the MP3 file to play.
There are two solutions to fix this:
The usage of the following command line in the batch file:
#start "" /D "%~dp0sfx" %SystemRoot%\System32\wscript.exe //NoLogo "%~dp0sfx\tyrian_soundtest.vbs"
The option /D of command START is used to explicitly set the subdirectory sfx of the batch file directory as start in respectively current directory for the process wscript.exe which cmd.exe starts using the Windows kernel library function CreateProcess with an appropriate created structure STARTUPINFO.
The VBS file references the MP3 file tyrian_soundtest.mp3 with its full path which the VB script file determines itself from its own full qualified file name. That can be achieved in the VB script file tyrian_soundtest.vbs by using in the second line:
Sound.URL = CreateObject("Scripting.FileSystemObject").GetParentFolderName(WScript.ScriptFullName) + "\tyrian_soundtest.mp3"
Best would be using both possible solutions as in this case the VB script file would work also on being executed by a different process than cmd.exe processing the batch file tyrian_soundtest.bat and deletion of directory %~dp0sfx is not possible as long as started wscript.exe is running because of this directory is the current directory of running Windows Script Host.
So the batch file as well as the VB script file work now both independent on which directory is the current directory as both reference the files with full path determined by the scripts themselves from their full qualified file names.
I need to set up a long list of temporary environment variables every time when I start a Python Flask app on Windows 10. Now I would like to create a batch file to run with all settings in one double click. The following lines run well if I copy them and paste to cmd prompt, but I couldn't run them in the batch file.
The execution of batch file always gets tripped and exited at the 2nd line venv\scripts\activate in the batch file, which has no issue at all if I copy and paste line by line at cmd.
cd C:\py\project
venv\scripts\activate
set env1=val1
set env2=val2
set FLASK_APP=some.py
flask run
One of the many (far too many) quirks of .bat files is that if you launch another .bat file, it doesn't know where to return to.
You need to explicitly call it:
call venv\scripts\activate
you can just use
start venv\Scripts\activate
This will open up a new terminal just like that... you can pass other commands using the && or & sign.
I created the p5.js project with p5 -manager tool. I successfully automated the whole thing with a batch file.
I should be able to open sketch.js with atom using additional batch script on the same file. But nothing happens and also no errors shows up. But same script on seperate file works just fine.
set /p PROJECT_NAME="Enter project Name: "
cd /d i:\My Work\p5
p5 generate --bundle %PROJECT_NAME%
atom "I:\My Work\p5\%PROJECT_NAME%\sketch.js"
the last line seem not to be working here.
but if I write:
set /p PROJECT_NAME="Enter project Name: "
atom "I:\My Work\p5\%PROJECT_NAME%\sketch.js"
on a seperate file and run it with valid folder name, it runs and opens atom with sketch.js file.
here PROJECT_NAME variable is the name of the project and the folder that is created with that name.
I think i guessed what it is. the p5 generate command runs which closes the command prompt before the next line runs. cause even the pause and cmd /d command are not doing anything if i put them after p5 generate line. Do you know how can i prevent it from closing and forcing it to run the next lines?
I've searched in many posts that already deal with this kind of issue but I'm still not able to make it work in my case.
I'm trying to create a link file to Launch a bat file with cmd.exe (because the bat file launch a service in spy/console mode so the console has to be shown).
The bat file is created before the install process of the service.
The link file is created in Windows Start Menu while installing process, so it can embed the path where the exe file of the service is installed.
As the root path will be linked to cmd.exe, I need to pass the path of the exe of the service as an argument of the bat file, itself being an argument of the cmd.exe file...
I've tried the syntax given as a solution in this threat :
correct quoting for cmd.exe for multiple arguments
Spaces in Program Path + parameters with spaces :
cmd /k ""c:\batch files\demo.cmd" "Parameter 1 with space" "Parameter2 with space""
So in my case, I've writen:
cmd /k ""C:\Program Files (x86)\Company\Licenses Server\Debug Mode.bat" "C:\Program Files (x86)\Company\Licenses Server\""
Here is the content of the "Debug Mode.bat" file :
#ECHO off
net stop company.service
START %~f1ServiceFile.exe /console spymode
net start company.service
#ECHO On
But when I launch the link file, I get the message:
Windows cannot find C:\Program...
There must be a "quote" problem but I've tried a lot of different syntaxes and I cannot figure out what goes wrong here...
I am writing a batch file on my Windows 8.1 machine. In one section of my batch file I need to start a command prompt in the "current working directory".
So far, this is what my batch file looks like:
#echo OFF
set WORKING=%cwd%
start cmd.exe /K pushd %WORKING%
exit
Let's say the batch file is located in the folder C:\Temp\Utilities. If I open an explorer window and double click the batch file to run it everything works great. A new command prompt is created in the directory C:\Temp\Utilities. However, if I right-click the batch file and select Run as administrator the working directory is no longer the location of the batch file, it's C:\Windows\System32.
Similarly, if I create a shortcut to the batch file in a different folder (for example. C:\Temp) and repeat the two steps above the results are the same. If I double click the shortcut and run it as a normal user the working directory is what I would expect. (Note, the working directory for the shortcut it's whatever is set for "Start in" of the shortcut properties, not the location of the batch file.) If I right click the shortcut and run it as administrator I again get a command prompt opened to the folder C:\Windows\System32.
I assume this is a "bug" or "feature" (if you want to call it that) in Windows 8.1 and it probably happens because execution environments for programs run as administrator are forced to run in the System32 folder? (I remember with Windows 7 this did not happen so it must be a new feature to Windows 8.)
I found one way to fix the issue and stop the command prompt from starting in C:\Windows\System32. I did this by modifying the following line in the batch file:
set WORKING=%~dp0
Doing it this way sets the working directory to the location of the batch file. With this change, no matter how I run the batch file or the shortcut (administrator or normal) the working directory ends up being the same, C:\Temp\Utilities.
The problem with this solution is I don't want the working directory to always be the location of the batch file. If the batch file is run directly then it's okay but if I run it from a shortcut I need the working directory to be whatever is set in the "Start in" property of that shortcut. For example, if the batch file is located in the folder D:\Temp\Utilities this is what I need to happen regardless of whether I run as administrator or not:
Shortcut Location Start In Property Command Prompt Working Directory
-------------------- ------------------- ------------------------------------------
C:\Temp <undefined> D:\Temp\Utilities
C:\Data\bin C:\Data\bin C:\Data\bin
C:\Data\bin D:\Temp\Utilities D:\Temp\Utilities
What this means is I can't always use %~dp0 to set the working directory in my batch file. What I need is some way for the batch file to know if it was run either directly or by a shortcut. If the batch file is run directly then the working directory is easy to get, it's just the value of %cwd%. If the batch file is run by using a shortcut, I don't know how to get the "Start in" property inside the batch file.
Does anyone know how I can do these two things inside my batch file:
1. Check whether it was run directly or by a shortcut.
2. If run by a shortcut, get the "Start in" property of the shortcut that started it.
Thank you,
Orangu
UPDATE
I found sort-of a "hackish" way to fix the issue. For the shortcut I edited the "Target" field and changed it to the following:
cmd.exe /k pushd "C:\Temp" && "D:\Temp\Utilities\batchfile.bat"
Now the working directory can be obtained by calling %CD% in the batch file and this works for both administrator and normal users. It does not, however, work for the case when I run the batch file directly. I still need to use %~dp0 in that case.
I don't like this solution, however, because it requires me to manually change all shortcuts I make and it also makes the icon look like a cmd prompt icon rather than a batch file.
Have you already considered to not use shortcuts at all?
You could e.g. create a batchfile_exec.bat containing your call
REM optionally do
REM cd /D working_directory
REM if you want to force a special working directory
D:\Temp\Utilities\batchfile.bat
and replace all the shortcuts with batchfile_exec.bat. If you double-click batchfile_exec.bat, the working directory will be the one containing batchfile_exec.bat.
I personally don't like Windows shortcuts that much, because they are hard to handle within a revision control system. As you also noticed, it is very time consuming if you want to modify a lot of them.
By the way: If batchfile.bat was designed/written to be always run from the direcory where it is located, you might also consider to modify batchfile.bat to force that behaviour:
setlocal
cd /D %0\..
REM your original content
endlocal
In %0 the path to the batchfile is stored.
The trick is to assume that %0 is a directory and then to change one level lower based on that directory. With /D also the drive letter is changed correctly.
The cd command doesn't care if %0 is really a directory. In fact %d doesn't even have to exist (%0\dummy\..\.. would also work).
The setlocal command is to have the working directory being restored when batchfile.bat has finished (this would be good if batchfile.bat was called form another batch file).
I noticed that the endlocal command is not really necessary in this context since it is applied implicitly when batchfile.bat finishes.