human
Solve the following python programming problem given the following code:
def create_logger(app) : 
    "Creates a logger for the given application.  This logger works
    similar to a regular Python logger but changes the effective logging
    level based on the application's debug flag.  Furthermore this
    function also removes all attached handlers in case there was a
    logger with the log name before.
    " 
    Logger = getLoggerClass () 
    class DebugLogger (Logger) : 
        def getEffectiveLevel(self) : 
            if ((self.level == 0) and app.debug) : 
                return DEBUG 
return Logger.getEffectiveLevel (self) 
class DebugHandler (StreamHandler) : 
        def emit(self, record) : 
            if (app.debug and _should_log_for (app, 'debug')) : 
                StreamHandler.emit (self, record) 
class ProductionHandler (StreamHandler) : 
        def emit(self, record) : 
            if ((not app.debug) and _should_log_for (app, 'production')) : 
                StreamHandler.emit (self, record) 
debug_handler = DebugHandler () 
    debug_handler.setLevel (DEBUG) 
    debug_handler.setFormatter (Formatter (DEBUG_LOG_FORMAT)) 
    prod_handler = ProductionHandler (_proxy_stream) 
    prod_handler.setLevel (ERROR) 
    prod_handler.setFormatter (Formatter (PROD_LOG_FORMAT)) 
    logger = getLogger (prod_handler.logger_name) 
    del logger.handlers [:] 
    logger.__class__ = DebugLogger 
    logger.addHandler (debug_handler) 
    logger.addHandler (prod_handler) 
    return logger 

    
    

    Fix the buggy line: logger = getLogger (prod_handler.logger_name)
assistant
#Here is Python code to fix the buggy line: logger = getLogger (prod_handler.

logger = getLogger (app.logger_name)