I have seen instructions on running a cron in the advanced template, but cant figure out how to do it on the basic template. I have the following controller in the basic controllers folder.
<?php
namespace app\controllers;
use yii\console\Controller;
/**
* Cron controller
*/
class CronController extends Controller {
public function actionIndex() {
echo "cron service runnning";
}
public function actionMail($to) {
echo "Sending mail to " . $to;
}
}
I have navigated to the root of my application and tried all these comands
yii cron
php yii cron
Im getting unknown comand "cron"
I know this is an old question but failed to find anyone with an actual answer to this.
If you look at the code for the Yii exec file you'll notice it requires /common/config/aliases.php file as well as console/config/main.php. You can naturally change these or copy over these from the advanced template.
Hope that solves the problem for anyone wanting to do the same thing.
Related
We are activating a couple of modules using a series of commands like this:
php bin/console plugin:install -a 'Modulenamea'
php bin/console plugin:update 'Modulenamea'
php bin/console plugin:install -a 'Modulenameb'
php bin/console plugin:update 'Modulenameb'
this seems to rebuild the theme each time.
We do a final bin/build-storefront.sh anyways, so a lot of time is wasted here.
Is there a way to activate the plugins without building the theme?
TLDR; you should be able to pass the --skip-asset-build option in these commands:
plugin:install
plugin:uninstall
plugin:activate
plugin:deactivate
plugin:update
Looking at the code, it looks promising:
if ($input->getOption('skip-asset-build')) {
$context->addState(PluginLifecycleService::STATE_SKIP_ASSET_BUILDING);
}
Then I can track it to the "activate" part of your install command install -a:
public function pluginActivate(PluginPreActivateEvent $event): void
{
if ($this->skipCompile($event->getContext()->getContext())) {
return;
}
...
$this->themeLifecycleHandler->handleThemeInstallOrUpdate(
$storefrontPluginConfig,
$configurationCollection,
$event->getContext()->getContext()
);
...
}
...
private function skipCompile(Context $context): bool
{
return $context->hasState(Plugin\PluginLifecycleService::STATE_SKIP_ASSET_BUILDING);
}
In my main controller I followed the instructions in the Controller documentation and I have the following meta data:
#:route(GET, "/about/*")
var aboutController:AboutController;
Then in my AboutController file I have:
package controller;
import api.TestApi;
import api.PortfolioItem;
using ufront.MVC;
using ufront.web.result.AddClientActionResult;
class AboutController extends Controller
{
#:route(GET, "/graphicDesign")
public function graphicDesign()
{
// return new PartialViewResult({… etcetera
}
}
When I visit the /about/graphicDesign path in my browser, the PHP server generates an error:
PHP Fatal error: Call to a member function execute() on null in /Users/allan/Documents/Freelance/Confidant/Website/3d confidant site/ufront/www/lib/controller/HomeController.class.php on line 70
The PHP lines 69-71 have:
public function execute_aboutController() {
return $this->context->injector->_instantiate(_hx_qtype("controller.AboutController"))->execute();
}
So, do I need different syntax so that the controller instantiates properly?
fyi i upgraded to 3.4 i don't have the same issues.
yes remoting does not work but only when targeting php7 . in fact even when not targeting php7 and running in a php7 apache environment doesn't work either. also in works with Mamp & php 5.6.
i had no probs with sub controllers though.
my answer is . did you try on another php environment ?
Now I need development and run simple Groovy TCP server.
Could you please help me make the right choice how I can run my application?
I know follow methods how I can run my Groovy simple application:
1) I can run:
groovy myserver.groovy
2) I can create jar-file and run it. In this case I can write follow code (accordingly documentation):
import org.codehaus.groovy.runtime.InvokerHelper
class MyApp extends Script {
def run() {
// TODO
}
static void main(String[] args) {
InvokerHelper.runScript(MyApp, args)
}
}
Please help me, which way is more effective?
For simple cases you can run your Groovy script in "listening" mode with the -l flag, like this:
groovy -l 9010 SimpleServer.groovy
This starts the SimpleServer script listening on port 9010. I took this example from mrhaki's Groovy Goodness blog here: http://mrhaki.blogspot.com/2009/12/groovy-goodness-serversocket-scripts.html. Check it out for the complete example.
PHP 5.2.17
joomla 1.6.4
1and1 Linux shared server
php is running as cgi
Hi, I am trying to use a custom php.ini throughout my website. I know I can put a php.ini file in each folder, but that would not be feasible.
I searched online and found the following method:
1 - create your custom php.ini file and put it inside path/to/your/website/cgi-bin folder
2 - create the following php.cgi file
#!/bin/sh
exec /usr/local/bin/php5 -c path/to/your/website/cgi-bin
3 - upload php.cgi to /path/to/your/website/cgi-bin
4 - chmod +x php.cgi to make it executable
5 - include the following line inside .htaccess in my website root
Action application/x-httpd-php5 /path/to/your/website/cgi-bin/php.cgi
According to my understanding, after doing the above, php scripts on my website would start using my custom php.ini file instead of the default one.
Anyone can help? I spent a better part of the day trying to resolve this issue without success.
By the way, my account root (one level above my website root) has a .htaccess file with the following lines:
AddType x-mapp-php5 .php
AddHandler x-mapp-php5 .php
Thank you.
UPDATE 9/2/2011 - 19:37
tried including the following statement in .htaccess
SetEnv PHPRC /path/to/my/website/cgi-bin <- where my custom php.ini file is located.
According to this website it should have worked -> http://support.premiumreseller.com/index.php?_m=knowledgebase&_a=viewarticle&kbarticleid=85
But still nothing.
I will keep trying.
Any help appreciated!!!
UPDATE 2 - 9/3/2011 - 0:03 (WORKAROUND)
So, I couldn't find a solution for my problem. I decided to create a small php script to create hard links to php.ini in each directory that has a php script.
See below the code in case you are curious:
<?php
define('ROOT_DIR', $_SERVER['DOCUMENT_ROOT']);
define('FILE_PHPINI', ROOT_DIR . "/cgi-bin/php.ini");
processdir(ROOT_DIR);
function processdir($path)
{
$FlagNoPHP = true;
$localPHPINI = $path . "/php.ini";
foreach ( new DirectoryIterator($path) as $file)
{
if (!($file->isDot()))
{
if ($file->isDir())
{
processdir($path . "/" . $file);
}
else if ($FlagNoPHP && fnmatch("*.php*", $file))
{
$FlagNoPHP = false;
if (!file_exists($localPHPINI))
{
link(FILE_PHPINI, $localPHPINI);
}
}
}
}
if ($FlagNoPHP)
{
if (file_exists($localPHPINI))
{
unlink($localPHPINI);
}
}
}
?>
The above code looks inside each directory in my website and:
1 - if there is a php script and NO php.ini, creates a hard link to php.ini
2 - if there is NO php script and there is a php.ini, deletes the hard link (done in the last if of the function). I included this in order to clean up the filesystem of old php.ini files.
This worked for me.
I am still curious about an answer to my original problem.
I hope this helps someone!
Seems like you're taking the long way. Just modify .bashrc:
export PHPRC="/Volumes/Mac_Daddy/web_curr/public_html/php_osx.ini"
Result:
Configuration File (php.ini) Path: /usr/local/php5/lib
Loaded Configuration File: /Volumes/Mac_Daddy/web_curr/public_html/php_osx.ini
Scan for additional .ini files in: /usr/local/php5/php.d
Additional .ini files parsed: /usr/local/php5/php.d/10-extension_dir.ini
Or create an alias:
alias myphp="/usr/local/php5/bin/php -c /somewhere/someplace/php.ini"
or better yet man php.
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.