Creating implementation for idiom 145 : Print log line with datetime

Idiom

# 145 Print log line with datetime

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Lead paragraph

Print message msg, prepended by current date and time.

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

New implementation

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Other implementations

Console.WriteLine($"[{DateTime.Now}] {msg}");

Demo

import std.experimental.logger;

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

import "log"

log.Println(msg)

Demo
Doc

import groovy.util.logging.Slf4j

@Slf4j
class X {
    def m(String message) {
        log.debug(message)
    }
}

console.error(Date(), msg);

console.log(Date(), msg);

import java.util.logging.Logger;

Logger LOGGER = Logger.getLogger(MyClass.class.getName());

LOGGER.info(msg);

long unixEpoch = System.currentTimeMillis();
String dateAndTime = String.format("%1$tY-%1$tm-%1$td %1$tI:%1$tM:%1$tS", unixEpoch);
System.out.printf("%s: %s%n", dateAndTime, msg);
System.err.printf("%s: %s%n", dateAndTime, msg);

uses sysutils;

writeln(DateToStr(Now), #32, msg);

use Time::Piece qw(localtime);

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

print $logline; # stdout
warn $logline; # stderr

import sys, logging

logging.basicConfig(stream=sys.stdout, level=logging.DEBUG, format="%(asctime)-15s %(message)s")
logger = logging.getLogger('NAME OF LOGGER')

logger.info(msg)

require 'logger'

logger = Logger.new('logfile.log') # or STDOUT or STDERR
logger.info(msg)

eprintln!("[{}] {}", humantime::format_rfc3339_seconds(std::time::SystemTime::now()), msg);

Demo
Doc

My.Application.Log.WriteEntry(msg,TraceEventType.Verbose)

Preview