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Question

In Java, I use log4J, which is the basis of logging. In Log4j, you can do something like this:

if (log.isDebug()) {
  // do some expensive operation that should only be displayed if DEBUG is turned on
}

After reading some Symfony examples, I cannot find a way to determine if DEBUG logging is activated in the context of the current class. Is there any way to do this?

if (--need something here--) {
  $this->logMessage('Expensive operation return value: '.expensiveFunction(), 'debug');
}

Answer 1

sort of

$this->getLogger()->getLogLevel() == sfLogger::DEBUG

.

Answer 2

You should check the use of Monolog

Answer 3

: sfConfig::get('sf_logging_enabled'). , . .yml.

Answer 4

, :

if ($this->get('kernel')->isDebug()) {
    ...
}

https://symfony.com/doc/2.8/components/console/logger.html http://symfony.com/doc/current/cookbook/logging/index.html

Answer 5

you can use the symfony registration service if you are in the controller, you can call this service using this code:

if ($this->get('kernel')->isDebug()) {
 $this->get('logger')->err('my custom message');
}

If you want to improve your system, you can check out MonologBundle http://symfony.com/doc/current/logging.html