CODEWITHSUNDEEP
phploggingsymfony1
Matt Sidesinger
Matt Sidesinger

Reputation: 2154

Symfony Logging

In Java, I use log4J that is a logging framework. 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
}

Reading some Symfony examples, I am unable to find a way to determine if DEBUG logging is activated in the current class context. Is there a way to do this?

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

Upvotes: 3

Views: 3829

Answers (5)

Ricardo
Ricardo

Reputation: 2497

you can use the symfony logger service if your are in a 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 yo can check MonologBundle http://symfony.com/doc/current/logging.html

Upvotes: 0

Valentin Hristov
Valentin Hristov

Reputation: 13

You can check is in debug mode with that:

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

This is information about logger service https://symfony.com/doc/2.8/components/console/logger.html and http://symfony.com/doc/current/cookbook/logging/index.html

Upvotes: 0

Geshan
Geshan

Reputation: 1141

You should check the usage for Monolog

Upvotes: 3

Fracsi
Fracsi

Reputation: 2314

You could also try: sfConfig::get('sf_logging_enabled'). It shows if logging is enabled. The levels can be configured in the factories.yml.

Upvotes: 0

ax.
ax.

Reputation: 60167

something like

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

should do.

Upvotes: 5

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