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Programming-Idioms

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  • Perl

Idiom #145 Print log line with datetime

Print message msg, prepended by current date and time.

Explain what behavior is idiomatic: to stdout or stderr, and what the date format is.

use Time::Piece qw(localtime);
my $logline = sprintf "%s %s\n", localtime->strftime('%F %T'), $msg;

print $logline; # stdout
warn $logline; # stderr
Console.WriteLine($"[{DateTime.Now}] {msg}");

Default behavior of Console.WriteLine is stdout. DateTime should respect system culture defaults.
New implementation...
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programming-idioms.org 
my $logline = sprintf "%s %s\n", localtime->strftime('%F %T'), $msg;

print $logline; # stdout
warn $logline; # stderr
Console.WriteLine($"[{DateTime.Now}] {msg}");
log(msg);
print('${DateTime.now()} msg');
integer :: value(8)
msg = "asdf"
call date_and_time (values=value)
write (unit=*,fmt='(I4,"-",I2.2,"-",I2.2," ",I2.2,":",I2.2,".",I4.4,": ",A)') value(1:3), value(5:7), msg
log.Println(msg)
@Slf4j
class X {
    def m(String message) {
        log.debug(message)
    }
}

console.log(Date(), msg);
console.error(Date(), msg);
Logger LOGGER = Logger.getLogger(MyClass.class.getName());

LOGGER.info(msg);
out.printf("%tc:  `%s`%n", currentTimeMillis(), msg);
writeln(DateToStr(Now), #32, msg);
logging.basicConfig(stream=sys.stdout, level=logging.DEBUG, format="%(asctime)-15s %(message)s")
logger = logging.getLogger('NAME OF LOGGER')

logger.info(msg)
logger = Logger.new('logfile.log') # or STDOUT or STDERR
logger.info(msg)
eprintln!("[{}] {}", humantime::format_rfc3339_seconds(std::time::SystemTime::now()), msg);
My.Application.Log.WriteEntry(msg,TraceEventType.Verbose)