JSF page refresh

To refresh a JSF page the following piece of code is useful

FacesContext context = FacesContext.getCurrentInstance();
String viewId = context.getViewRoot().getViewId();
ViewHandler handler = context.getApplication().getViewHandler();
UIViewRoot root = handler.createView(context, viewId);
root.setViewId(viewId);
context.setViewRoot(root);


If your page is not getting refreshed due to browser caching

 FacesContext facesContext = event.getFacesContext();
HttpServletResponse response = (HttpServletResponse) facesContext.getExternalContext().getResponse();
response.addHeader("Pragma", "no-cache");
response.addHeader("Cache-Control", "no-cache");
response.setHeader("Cache-Control", "no-store");
response.addHeader("Cache-Control", "must-revalidate");

Data Structures

Few good links for Data Structures

• C API : http://code.google.com/p/c-generic-library/
  
• google collections : http://code.google.com/p/google-collections/
• java listenable collections : http://code.google.com/p/jalico/
• JRDF : http://code.google.com/p/jrdf/
• J5Collections : http://code.google.com/p/j5collections/
• Stanford video tutorial : http://in.youtube.com/view_play_list?p=84A56BC7F4A1F852
• Book : http://www.brpreiss.com/books/opus5/html/page257.html

More links

• all slide handout from the textbook http://ww0.java4.datastructures.net/handouts/
• comparison of link implementation in java (ArrayList, Vector, LikedList) http://www.onjava.com/pub/a/onjava/2001/05/30/optimization.html?page=1
• time your execution program example http://www.rgagnon.com/javadetails/java-0132.html *in Java 5, you can also use nano Timing method.
• Queue in Java 5 http://www.devx.com/Java/Article/21983
• Josephus Problem http://en.wikipedia.org/wiki/Josephus_problem
• Pigeonhole principle http://en.wikipedia.org/wiki/Pigeonhole_principle
• Iterator Example http://www.particle.kth.se/~lindsey/JavaCourse/Book/Part1/Java/Chapter10/iterator.html or http://www.java-examples.com/iterate-through-elements-java-arraylist-using-iterator-example
• Tree in Java 5 http://it-essence.xs4all.nl/roller/technology/entry/three_tree_traversals_in_java
• Prime in Hash http://www.interactivecode.com/programming-coding-1/hashing-why-mod-prime-2101/
• Last 2 slides of ch9.2 page17-18ch09.2.pdf

How to empty a StringBuffer

We know that StringBuffer is reusable , We may want to empty a String Buffer ( lets say at end of a loop )
so how do you do it ?

The easiest way i came across use the method setLength()

public void setLength(int newLength)

You may have the case like
StringBuffer sb = new StringBuffer("HelloWorld")
// after many iterations and manipulations
sb.setlength(0);
// reuse sb

Please note you cant say
StringBuffer sb = "";
You will get a type mismatch error saying that "Cannot assign a value of type String to StringBuffer"
Makes sense as StringBuffer is not String , and they both inherit from java.lang.Object

Java ques

There are three types of empty string, null, "" and " ". Here is how to check for each flavour:

if ( s == null ) echo ( "was null" );
else if ( s.length() == 0 ) echo ( "was empty" );
else if ( s.trim().length () == 0 ) echo ( "was blank or other whitespace" );

We cant use the following
String s="";
if(s.equals(""))
sop("string equal");

Java tricky ques

Q: How can I check a string has no numbers?

A: The simplest way to check that a String does not contain any numbers is to use the regular expression class Pattern in the java.util.regex package. The method below uses the regular expression [^0-9]+ to check that none of the characters in the input string is a number. The square brackets define a character class. The negation modifier, ^, followed by the number range 0-9 means "not a number". The + quantifier asks the regular expression to match the character class one or more times.

Java String tricky ques

String s1 = "hello world";
String s2 = "hello world";

if(s1==s2)
System.out.println("Yes");
else
System.out.println("No");

// output of first is Yes
the reason being that in above code you are comparing two references

public static void point1()
{
String s = new String("Friday");
String s1 = new String("Friday");
if (s == s1)
System.out.println("Equal A");
if (s.equals("Friday"))
System.out.println("Equal B");
}
ANS : Equal B
Because they are two different objects on heap , refering diff memeory addresses
hence their reference is not same.

more explanation here.

dont develop for a living you will be a better developer

Hi
Totally agree with this article