Log an exception

With this example we are going to demonstrate how to log an exception. In order to do so, we will use a DateFormat and parse a String pattern to create a new Date. In short, to log the ParseException that occurs you should:

  • Create a new SimpleDateFormat with a specific String pattern.
  • Invoke the setLenient(boolean lenient) API method of the DateFormat, setting the lenient to false. Thus, the inputs of the DateFormat parser must match this object’s format, or else a ParseException will be thrown.
  • Invoke the log(Level level, String msg, Throwable thrown) API method to log a message, with the associated Throwable information.

Let’s take a look at the code snippet that follows:

package com.javacodegeeks.snippets.core;

import java.util.logging.Logger;
import java.util.logging.Level;
import java.util.Date;
import java.text.DateFormat;
import java.text.SimpleDateFormat;
import java.text.ParseException;
public class LogException {
    private static Logger logger = Logger.getLogger(LogException.class.getName());
    public static void main(String[] args) {

  DateFormat df = new SimpleDateFormat("dd/MM/yyyy");


  try {

// Set wrong date

Date date = df.parse("11/08/1984");

System.out.println("Date = " + date);

  } catch (ParseException e) {

  	// Create log message 

if (logger.isLoggable(Level.SEVERE)) {

    logger.log(Level.SEVERE, "Error parsing date", e);




Αυγ 12, 2012 1:30:09 ΜΜ com.javacodegeeks.snippets.core.LogException main
SEVERE: Error parsing date
java.text.ParseException: Unparseable date: "11/08/1984"
	at java.text.DateFormat.parse(Unknown Source)
at com.javacodegeeks.snippets.core.LogException.main(LogException.java:22)

This was an example of how to log an exception in Java.

Ilias Tsagklis

Ilias is a software developer turned online entrepreneur. He is co-founder and Executive Editor at Java Code Geeks.
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