I am trying to create a batch file which runs an exe from a specific path.
Eg: I have my exe in E drive. Exact path is E:\kk.exe. I want to run this kk.exe from D:\bin folder.
I use the following command in my batch file:
start "D:\bin" "E:\kk.exe"
So far no luck. Any help would be appreciated.
start "" /d "d:\bin" "e:\kk.exe"
start command has a peculiar behaviour: the first quoted argument is the title of the window. That is the reason for the initial "" (you can include the title you want). The rest of the line is the starting folder (/d, what will be the current active folder for the started process) and the command to execute.
cd /d "D:\bin"
start "window name" "E:\kk.exe"
If I've decoded your meaning correctly, you wish to run kk.exe while your current directory is d:\bin. This will create an independent process to run that program.
Note: the syntax of "start" is such that it's advisable to assign a window title (the first quoted parameter) - if you don't ant a title, leave the text out and use en empty quoted string.
However, if you just want to execute e:\kk.exe then
cd /d "D:\bin"
"E:\kk.exe"
Related
I'm writing a .bat file to launch two programs, not having success so far. I've followed the guide here.
#echo off
cd "D:\CRCR 0.4.0\"
start "" "D:\CRCR 0.4.0\Chernobyl Relay Chat Rebirth.exe"
cd "D:\MO2\"
start "" "D:\MO2\ModOrganizer.exe moshortcut://:Anomaly Launcher"
exit
Whenever I try to launch the batch file, it tells me "Windows cannot find 'D:\MO2\ModOrganizer.exe "moshortcut://:Anomaly Launcher'. Make sure you've typed the name correctly, then try again."
I've tried removing 'start' from those lines, removing quotations from the cd lines, etc. How can I get this working?
You probably want to pass moshortcut://:Anomaly Launcher as a parameter.
In that case you should also separate the string:
start "" "D:\MO2\ModOrganizer.exe" "moshortcut://:Anomaly Launcher"
Otherwise Windows will interpret the path as D:\MO2\ModOrganizer.exe moshortcut://:Anomaly Launcher not D:\MO2\ModOrganizer.exe and assume that the entire string is part of the path.
Currently this is how I have written the code in the batch file:
C:\ cd C:\abc\xyz\build-scripts-master
call setEnv.cmd
cmd ant do-clean
cmd ant do-dist
This is not working. It just executes the setEnv and breaks out. It does not run the remaining commands
Manually this is how it works:
I first go to the folder C:\abc\xyz\build-scripts-master through the Command Prompt
Then I type in setEnv, which is a windows command script, and hit return.
Then I type in ant do-clean
And then ant do-dist
I want to automate this process and hence was trying to achieve this using batch file.
Try the following:
#CD /D "C:\abc\xyz\build-scripts-master"
#Call setEnv.cmd
#Call ant.bat do-clean
#Call ant.bat do-dist
The latter two lines assume that ant.bat is located somewhere in the current working directory or %PATH%
It is not imperative that the directory path is doublequoted in this case, just good practice.You could continue not to use the .bat extension with ant. I've included it just to make it clear that it is a batch file, and should be Called in the same way as the setEnv batch file.
it didn't run the bat files because you didn't specify the files' location in the code. As it stands right now, the script expects the .bats to exist in the working directory or it has been placed in the folder. The only way it will run the files arbitrarily is if you had placed your working location in the system variables or set the Path to the folders location. I don't know if the cmd and call are needed. I have never used them in my scripts.
I try this code :
start /d "D:\test\CONTOH\DATA\QGIS2\bin\" qgis.bat
based on :
Bat file to run a .exe at the command prompt
But, I want to be relative path, something like this :
start /d %~dp0\DATA\QGIS2\bin\qgis.bat
based on :
relative path in BAT script
but, nothing happen. So, can someone give me information, what is wrong?
Open a command prompt window and run start /?. This outputs the help for the command START which should be read on using this command to get knowledge about its options.
Run in command prompt window call /? and read the output help pages to understand %~dp0. The drive and path of the batch file (argument 0) always ends with a backslash. Therefore don't add an extra backslash on concatenating it with another string.
And the first double quoted string is interpreted by START as title for the new console window displayed in title bar of the new window. Therefore better specify always a title string which can be also an empty string like "" in case of starting a GUI application on which no console window is opened at all.
start "Running QGIS2" /D "%~dp0DATA\QGIS2\bin" qgis.bat
Also possible is using this command line in batch file:
start "Running QGIS2" /D"%~dp0DATA\QGIS2\bin" qgis.bat
Here start in directory as defined with /D"%~dp0DATA\QGIS2\bin" is 100% correct specified according to help of command START as one parameter string.
But Windows command interpreter accepts also the first variant with just option /D without any folder path and next parameter string "%~dp0DATA\QGIS2\bin" after a separating space is the folder path for start in directory.
The first variant with just /D as one parameter string and "%~dp0DATA\QGIS2\bin" as one more parameter string is easier to read in comparison to second variant with /D"%~dp0DATA\QGIS2\bin" being just one parameter string.
I want to create a batch file to launch my executable file after it has made some changes to itself.
My batch file is:
START /D "C:\Users\me\AppData\Roaming\Test\Test.exe"
When I run it though I just get a brief console flash and Test.exe doesn't start up.
I've verified the EXE is there in the directory.
I've launched the exe manually to verify it is working as well.
My batch file resides in
C:\Users\admin\AppData\Roaming\run.bat"
There are two issues:
The /D option solely defines the starting or working directory, but not the program to execute.
The start command considers the first quoted argument as the title of the new window. To avoid confusion with other arguments, always provide a window title (that may also be empty).
There are two solutions, which are actually not exactly equivalent:
Remove the /D option, so the current working directory is used:
start "" "C:\Users\me\AppData\Roaming\Test\Test.exe"
Keep the /D option and explicitly provide the new working directory to be used:
start "" /D "C:\Users\me\AppData\Roaming\Test" "Test.exe"
try changing to this
start /d "C:\Users\me\AppData\Roaming\Test" Test.exe
You will see the console flash and your program should startup.
Update
Thanks for #SomethingDark 's suggestion to use the following code.
start "" C:\Users\me\AppData\Roaming\Test\Test.exe
However, the above code will not work if your filename contains space.
Try with the following command.Add it to your batch script.Notice that you have to add double quotes after start keyword if there is/are whitespaces in the path string.
start "" "C:\Users\me\AppData\Roaming\Test\Test.exe"
Enclose any directory names which are longer than one-word into quotation marks. So the following path:
start C:\Program Files\MySQL\MySQL Workbench 8.0 CE\MySQL.exe
Should become something like this:
start C:\"Program Files"\MySQL\"MySQL Workbench 8.0 CE"\MySQL.exe
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.