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

• Create a newSimpleDateFormat with a specific String pattern.
• Invoke thesetLenient(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 aParseException will be thrown.
• Invoke thelog(Level level, String msg, Throwable thrown) API method to log a message, with the associatedThrowable 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");  df.setLenient(false);   try {// Set wrong dateDate 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);}  }    }}

Output:

Αυγ 12, 2012 1:30:09 ΜΜ com.javacodegeeks.snippets.core.LogException mainSEVERE: Error parsing datejava.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.

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