AutoIt Script to run .exe file - exe

I want to run an application which is located in the following directory:
C:\LCR 12\stu.exe
With AutoIt, what would be the code to run the above stu.exe file?

Like this:
Run("C:\LCR 12\stu.exe")
Hope this is what you were after.

You can do by using a variable as below:
Local $exeLocation = "C:\LCR 12\stu.exe"
Run($exeLocation)

Related

Execute Python script using Laravel?

I'm trying to execute a python script in a Laravel 5.8 project but I'm having problems with the Symfony/process class.
Basically, I want to run this python script that takes an excel form from the storage folder.
My first try was this
$process = new Process('C:\Python\python.exe C:\Users\"my path"\laravel\storage\app\images\cargaExcel.py');
$process->run();
if (!$process->isSuccessful()) {
throw new ProcessFailedException($process);
}
echo $process->getOutput();
And the error is
Fatal Python error: _Py_HashRandomization_Init: failed to get random numbers to initialize Python
I also tried with shell_exec(), and if the two files (the excel and the python script are in the public path - app/public) it works.
I think the problem is that python only executes on the app/public folder, so I don't know how to run this in another path.
Python output is telling me that:
Working directory: C:\Users\"my path"\laravel\public
Does anyone know how to run this?
You can pass the working directory as the second argument of the process class
So it can be:
$process = new Process('C:\Python\python.exe C:\Users\"my path"\laravel\storage\app\images\cargaExcel.py', "/my/working/path/");
Also you may pass command as array
$process = new Process(['C:\Python\python.exe', 'C:\Users\"my path"\laravel\storage\app\images\cargaExcel.py'], "/my/working/path/");

Using node-cmd module while handling Squirrel Events function

I'm building a desktop app for Windows using electron-packager and electron-squirrel-startup, I would like to execute some Windows cmd commands during the installation of my application. To do so I was planning to use node-cmd node module, but I doesn't really work inside the handleSquirrelEvents function. An example command like this:
function handleSquirrelEvent(application) {
const squirrelEvent = process.argv[1];
switch (squirrelEvent) {
case '--squirrel-install':
case '--squirrel-updated':
var cmd=require('node-cmd');
cmd.run('touch example.created.file');
}
};
Seems to work. A file example.created.file in my_app/node_module/node-cmd/example directory is created.
But any other code does not work. Even if I only change the name of the file to be "touched" nothing happens.
Ok, example.created.file already exists in this directory and I suspect that you can only use update.exe supported commands in case '--squirrel-updated' sections. So this will not work.

How to create a JSCS config file on windows

When I try to create a JSCS config file:
C:\Blog\BlogWeb>jscs --auto-configure "C:\Blog\BlogWeb\temp.jscs"
I get the following error:
safeContextKeyword option requires string or array value
What parameter am I supposed to pass? What is a safecontextkeyword?
New to NPM and JSCS, please excuse ignorance.
JSCS was complaining that I didn't have a config file, so I was trying to figure out how to create one.
JSCS looks for a config file in these places, unless you manually specify it with the --config option:
jscs it will consequentially search for jscsConfig option in package.json file then for .jscsrc (which is a just JSON with comments) and .jscs.json files in the current working directory then in nearest ancestor until it hits the system root.
I fixed this by:
Create a new file named .jscsrc. Windows Explorer may not let you do this, so may need to use the command line.
Copy the following into it. It doesn't matter if this is the preset you want to use or not. The command will overwrite it.
{
"preset": "jquery",
"requireCurlyBraces": null // or false
}
Verify that it works by running a command such as:
run the command
jscs --auto-configure .jscsrc

Can any one explain the usage of Shell::Source perl module or Shell::GetEnv module

I m a beginner in perl. I want to know how to use this module. I read somewhere about this module but not getting its usage.
Actually I've a file which contains some environment paths which needs to be set while running some test(say file name is SET_ENV_TOOL1.csh or SET_ENV_TOOL1.sh) with particular tools.(say TOOL1, TOOL2 etc)
SET_ENV_TOOL1.sh file conatins:
setenv UVM_HOME /u/tools/digital/uvm/uvm-1.1a
setenv VIPP_HOME /u/tools/digital/vipcat_11.30-s012-22-05-2012
setenv VIP_AXI_PATH ${VIPP_HOME}/vips/amba_axi/vr_axi/sv/ #etc.(almost 10-15 paths are need to be set like this)
Everytime while running test, tool might get changed and so environment paths also needs to set to run that tool.
I have to make a perl script which sets these paths before running test. That test will run a command and that command will use these environment paths.
Any help would be greatly appreciated. Thanks !!
Reading and changing environment variables is built-in to Perl, you do not need the modules you mentioned.
$ENV{UVM_HOME} = '/u/tools/digital/uvm/uvm-1.1a';
$ENV{VIPP_HOME} = '/u/tools/digital/vipcat_11.30-s012-22-05-2012';
$ENV{VIP_AXI_PATH} = "$ENV{VIPP_HOME}/vips/amba_axi/vr_axi/sv/";

How do you get the path of the running script in groovy?

I'm writing a groovy script that I want to be controlled via a properties file stored in the same folder. However, I want to be able to call this script from anywhere. When I run the script it always looks for the properties file based on where it is run from, not where the script is.
How can I access the path of the script file from within the script?
You are correct that new File(".").getCanonicalPath() does not work. That returns the working directory.
To get the script directory
scriptDir = new File(getClass().protectionDomain.codeSource.location.path).parent
To get the script file path
scriptFile = getClass().protectionDomain.codeSource.location.path
As of Groovy 2.3.0 the #SourceURI annotation can be used to populate a variable with the URI of the script's location. This URI can then be used to get the path to the script:
import groovy.transform.SourceURI
import java.nio.file.Path
import java.nio.file.Paths
#SourceURI
URI sourceUri
Path scriptLocation = Paths.get(sourceUri)
Note that this will only work if the URI is a file: URI (or another URI scheme type with an installed FileSystemProvider), otherwise a FileSystemNotFoundException will be thrown by the Paths.get(URI) call. In particular, certain Groovy runtimes such as groovyshell and nextflow return a data: URI, which will not typically match an installed FileSystemProvider.
This makes sense if you are running the Groovy code as a script, otherwise the whole idea gets a little confusing, IMO. The workaround is here: https://issues.apache.org/jira/browse/GROOVY-1642
Basically this involves changing startGroovy.sh to pass in the location of the Groovy script as an environment variable.
As long as this information is not provided directly by Groovy, it's possible to modify the groovy.(sh|bat) starter script to make this property available as system property:
For unix boxes just change $GROOVY_HOME/bin/groovy (the sh script) to do
export JAVA_OPTS="$JAVA_OPTS -Dscript.name=$0"
before calling startGroovy
For Windows:
In startGroovy.bat add the following 2 lines right after the line with
the :init label (just before the parameter slurping starts):
#rem get name of script to launch with full path
set GROOVY_SCRIPT_NAME=%~f1
A bit further down in the batch file after the line that says "set
JAVA_OPTS=%JAVA_OPTS% -Dgroovy.starter.conf="%STARTER_CONF%" add the
line
set JAVA_OPTS=%JAVA_OPTS% -Dscript.name="%GROOVY_SCRIPT_NAME%"
For gradle user
I have same issue when I'm starting to work with gradle. I want to compile my thrift by remote thrift compiler (custom by my company).
Below is how I solved my issue:
task compileThrift {
doLast {
def projectLocation = projectDir.getAbsolutePath(); // HERE is what you've been looking for.
ssh.run {
session(remotes.compilerServer) {
// Delete existing thrift file.
cleanGeneratedFiles()
new File("$projectLocation/thrift/").eachFile() { f ->
def fileName=f.getName()
if(f.absolutePath.endsWith(".thrift")){
put from: f, into: "$compilerLocation/$fileName"
}
}
execute "mkdir -p $compilerLocation/gen-java"
def compileResult = execute "bash $compilerLocation/genjar $serviceName", logging: 'stdout', pty: true
assert compileResult.contains('SUCCESSFUL')
get from: "$compilerLocation/$serviceName" + '.jar', into: "$projectLocation/libs/"
}
}
}
}
One more solution. It works perfect even you run the script using GrovyConsole
File getScriptFile(){
new File(this.class.classLoader.getResourceLoader().loadGroovySource(this.class.name).toURI())
}
println getScriptFile()
workaround: for us it was running in an ANT environment and storing some location parent (knowing the subpath) in the Java environment properties (System.setProperty( "dirAncestor", "/foo" )) we could access the dir ancestor via Groovy's properties.get('dirAncestor').
maybe this will help for some scenarios mentioned here.

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