resources can not be added

eclipse j2ee error There are No resources that can be added or removed from the server

Eclipse showing There are No resources that can be added or removed from the server when adding project

If you are trying to test the java web application (j2ee) on your local machine with built in eclipse support and getting error, then it is related to project facet.

Here is how you can solve the error There are No resources that can be added or removed from the server coming from eclipse

1. right click on your web project in eclispe

2. Select properties
properties

3. Select Project facets
project-facets

4. Select Dynamic Web Module
select-dynamic-web-module

5. Click on apply and then click on OK

This will solve the problem.

See also: How to resolve class pass error when working with maven

testing k complementary pairs algorithm with junit

This is the test class for the java solution of the k-complementary problem listed here


package algorithm;

import static org.junit.Assert.assertArrayEquals;

import org.junit.Before;
import org.junit.Test;

/**
 * Testing KComplementary algorithm
 * 
 * @author Kaleb Woldearegay
 *
 */
public class KComplementaryTest {
	
	private KComplementary kComplementary;
	
	@Before
	public void initiate() {
		this.kComplementary = new KComplementary();
	}
	
	
	@Test
	public void test1() {
		Integer[][] expectedResult = new Integer[][]{{1,9},{5,5},{9,1}};
		int[] test = new int[]{1,5,9};
		
		assertArrayEquals(this.kComplementary.getKComplementaryPairs(10,  test), expectedResult);
	}
	
	@Test
	public void test2() {
		Integer[][] expectedResult = new Integer[][]{{5,7},{7,5}};
		int[] test = new int[]{3,5,7};
		
		assertArrayEquals(this.kComplementary.getKComplementaryPairs(12,  test), expectedResult);
	}
	
	@Test
	public void test3() {
		Integer[][] expectedResult = new Integer[][]{{-1,1},{0,0},{1,-1}};
		int[] test = new int[]{5,-1,0,-2,3, 1};
		
		assertArrayEquals(this.kComplementary.getKComplementaryPairs(0,  test), expectedResult);
	}

}

See more algorithm solutions by clicking Here

Find the pairs that makes K Complementary in the given array java solution

Find K complementary pairs from the given array

So the problem is, given an array of numbers, you are going to find the pairs whose sum is K.

Say, if you are given 7, 1, 5, 6, 9, 3, 11, -1 and K = 10 then the pairs would be

=> 1 and 9
=> 3 and 7
=> 11 and -1

Here is the O(n) solution for the K complementary problem in java

It turns out, this is not a pure O(n) solution. Can you spot why?


package algorithm;

import java.util.HashMap;
//import java.util.List;
import java.util.Map;

/**
 * Algorithm to find the pairs making the K complementary in O(n) complexity
 * 
 * @author http://gullele.com
 * @see KComplementaryTest.java - a JUnit test for this class
 */
public class KComplementary {
	
	private Map pairs;
	
	public static void main(String[] args) {
		KComplementary kcomp = new KComplementary();
		int[] numbers = new int[]{7, 1, 5, 6, 9, 3, 11, -1};
		Integer[][] pairs = kcomp.getKComplementaryPairs(10, numbers);
		
		for (Integer[] thePairs : pairs) {
			System.out.println(" Pairs are "+thePairs[0] + " and " + thePairs[1]);
		}
	}
	
	public KComplementary() {
		this.pairs = new HashMap();
	}
	
	/**
	 * An algorithm to find the pair from the given array that would sum up the given K
	 * 
	 * @note - the algorithm would be done in o(n). First it will run through the whole 
	 * numbers and creates a temporary list of pairs in HashMap with 
	 * (value, sum-value). 
	 * @param sum
	 * @param listOfIntegers
	 * @return
	 */
	public Integer[][] getKComplementaryPairs(int sum, int[] listOfIntegers) {
		
		/*
		 * I could have used the ArrayList assuming the number of pairs we are getting 
		 * is not known, but giving it a little bit of thought, the number of pairs 
		 * is known in advance, at least the maximum we can get is known.
		 * By not using the Array list, the algorithm would run O(n) rather than
		 * O(n**2) since ArrayList.add would would be O(n) by itself 
		 */
		//List complementaryPairs = new ArrayList();
		Integer[][] complementaryPairs = new Integer[listOfIntegers.length][2];
		//First fill up the pairs with the complementary numbers
		for (int number : listOfIntegers) { //O(n) complexity
			this.pairs.put(number, sum-number);
		}
		
		//then filter out the pairs that don't have corresponding complementary number
		int index = 0;
		for (int number : listOfIntegers) { //O(n) complexity 
			int complementary = sum - number;
			//check if this key exists in the pairs
			if ( this.pairs.containsKey(complementary) ) {
				//Had I used array List this would have been used
				//Integer[] comps = {number, complementary};
				//complementaryPairs.add(comps); //this is O(n)
				complementaryPairs[index][0] = number;
				complementaryPairs[index][1] = complementary;
				index ++;
			}
		}
		//System.out.println(index);
		//Now trim the array since we know the exact record
		Integer[][] trimmed = new Integer[index][2];
		
		index = 0;
		for (Integer[] item : complementaryPairs) { //O(n) complexity
			if (item[0] != null) {
				trimmed[index][0] = item[0];
				trimmed[index][1] = item[1];
			}
			index++;
		}
	
		//Total complexity O(n)+O(n)+O(n) ==> O(n)
		
		//had I used arrayList this would have been used
		//return complementaryPairs.toArray(new Integer[0][0]);
		return trimmed;
	}
}


The above solution runs and would do the job without a problem. But it is not linear solution as it claims to be.

I have got another solution that would explain why and comes up with a solution for it. Check it out here

You can see the jUnit for this class also at here

Using mvn for creating and running a boilerplate minimal java application

It is almost a must to use mvn or any other build tool when dealing with java.

With mvn you can start a minimal simple all included java application and you run that too.. Just follow this. I will assume you are on linux or mac environment for this and also I am assuming you already have maven installed
Create a new folder and say mavenrocks

mkdir mavenrocks

Then add the following

mvn archetype:generate

What it does is, it will use a plugin to create a boilerplate application that holds basic src files along with standard directory structure.
if you do the listing on the created folder you will see:

.
./pom.xml
./src
./src/main
./src/main/java
./src/main/java/com
./src/main/java/com/gullele
./src/main/java/com/gullele/App.java
./src/test
./src/test/java
./src/test/java/com
./src/test/java/com/gullele
./src/test/java/com/gullele/AppTest.java
./target
./target/classes
./target/classes/com
./target/classes/com/gullele
./target/classes/com/gullele/App.class
./target/maven-status
./target/maven-status/maven-compiler-plugin
./target/maven-status/maven-compiler-plugin/compile
./target/maven-status/maven-compiler-plugin/compile/default-compile
./target/maven-status/maven-compiler-plugin/compile/default-compile/createdFiles.lst
./target/maven-status/maven-compiler-plugin/compile/default-compile/inputFiles.lst

and to run it

mvn execute:java -Dexec.mainClass="com.gullele.App"

JSF App slow with JPA connection

If you are working on Java Persistence API JPA on tomcat or any other web server this would be happening if you have multiple threads going off for connections.

The rule of thumb shall be to have one EntityManagerFactory and get EntityManagers out of it. Hence we would have one factory but multiple products that would take care of closing and managing them selves.

What are the signs:

1. Do you instantiate Persistence.createEntityManagerFactory(“name”) from multiple places?
2. What do you see on Process when you run

ps -aux | grep tomcat

Do you see multiple instances

If either or both of the above have yes, then here is the solution.

The first thing have single instance of ManagerFactory


package com.enderase.persistence;

import javax.persistence.EntityManagerFactory;
import javax.persistence.Persistence;

/**
* Singlton implementation for EntityManagerFactory
*
* @author Kaleb Woldearegay<kaleb@gullele.com>
*/
public class HibernateUtil {
private static final EntityManagerFactory entityManagerFactory;

static {
try {
entityManagerFactory = Persistence.createEntityManagerFactory("jpa");
} catch (Throwable exception) {
//log your error here
throw new ExceptionInInitializerError(exception);
}
}

public static EntityManagerFactory getEntityManager() {
return entityManagerFactory;
}
}

Then make sure you are taking care of the instances of the EntityManagers that are created from the factory using

EntityManagerFactory entityManagerFactory = HibernateUtil.getEntityManager();
EntityManager em = entityManagerFactory.createEntityManager();

Make sure to close them appropriately after using them

This should pretty much take care of the problem

 

Check if there are three numbers a, b, c giving a total T from array A

I got this question while helping a friend on the course work. It is relatively simple question. But the way how it is approached can make a difference on efficiency.

The question is, given an array of numbers, you are to find if there are three numbers that would total the given number T.

If done in a very naive way, it can soar to o(n^3) like having three loops and checking the sum inside the third loop.. well.. this is a no no..

I have approached it in a log n ( for sorting) and n for (searching) approach..


package algorithm;

import java.util.Arrays;

/**
 * Given an array of integers, find if there are three numbers that would sum up to the 
 * number T
 * 
 * @author http://gullele.com
 *
 */
public class ThreeNumbersSummingT {
	public static void main(String[] args) {
		int[] test = new int[]{1,3,4,5,10,12, 18};
		ThreeNumbersSummingT summt = new ThreeNumbersSummingT();
		
		int[] response = summt.findThreeNumbers(test, 29);
		
		if (response.length > 1) {
			for(int num : response) {
				System.out.println(num);
			}
		} else {
			System.out.println(":( Couldn't find those three gems");
		}
				
	}
	
	public int[] findThreeNumbers(int[] nums, int t) {
		
		int[] indexes = new int[1];
		if (nums.length == 0 || nums.length <= 2) {
			return indexes;
		}
		
		//for primitive this would be quick sort so we have nlogn
		Arrays.sort(nums);
		
		int minIndex =0;
		int maxIndex = nums.length-1;
		int current = 1;
		while (minIndex != maxIndex) {
			if (nums[minIndex] + nums[maxIndex] + nums[current] == t) {
				int[] summingNumbers = new int[3];
				summingNumbers[0] = nums[minIndex];
				summingNumbers[1] = nums[current];
				summingNumbers[2] = nums[maxIndex];
				return summingNumbers;
			}
			
			int lookingFor = t-(nums[minIndex] + nums[maxIndex]);
			//if the number being sought is beyond the max, then jack up the min index
			if (lookingFor >= nums[maxIndex]) {
				minIndex++;
				current = minIndex + 1;
			} else if (nums[minIndex] + nums[maxIndex] + nums[current] < t) {
				current++;
			} else {
				maxIndex--;
				current = minIndex + 1;
			}
			
		}
		
		return indexes;
	}
}


Accessing one bean in another without using annotation

I would say we have to use annotations and Injection to get the bean inside another bean.
But in case you are interested to get it without using annotation you can use the following in the action/actionListener

TheBean theBean = (TheBean)FacesContext.getCurrentInstance() .getExternalContext().getRequestMap().get("theBean");

Where theBean is one we are interested to get it from being in the other bean.

That is it!

Adding session bean to to requested bean using annotation JSF

One major part on JSF would separation of concerns even for beans. As a rule of thumb beans related to model are session beans and those which have actions to be taken care of are requested one.

So, In this particular scenario we would have two beans. Basically we don’t want to include any logic inside the session bean, rather we would add session bean as a member variable to request bean.

Lets take a simple registration process.

The requested bean which will be responsible for actions would look like

package com.enderase.beans;

import java.io.BufferedWriter;
import java.io.FileWriter;
import java.io.Serializable;

import javax.faces.bean.ManagedBean;
import javax.faces.bean.ManagedProperty;
import javax.faces.bean.RequestScoped;

import com.enderase.model.Contractor;

@ManagedBean
@RequestScoped
public class ActionListeners implements Serializable{
	
	@ManagedProperty(value="#{contractorBean.contractor}")
	private Contractor contractor;
	
	public void setContractor(Contractor contractor){
		this.contractor = contractor;
	}
	
	public Contractor getContractor() {
		return this.contractor;
	}
	
	private static final long serialVersionUID = 1L;
	
	
	/**
	 * Action handler for Contractor save.
	 * @return String, next
	 */
	public String registerContractor() {
		Contractor contractor = this.contractor;
		if (contractor != null) {
			try {
				FileWriter fileWriter = new FileWriter("/tmp/name.note");
				BufferedWriter bufferedWriter = new BufferedWriter(fileWriter);
				bufferedWriter.write("Name "+contractor.getName()+" Email "+contractor.getEmail()
						+contractor.getState());
				bufferedWriter.close();
			} catch (Exception ex) {
				//log the exception here
			}
		}
		return "navigated";
	}
}

So the key thing here would be the @ManagedProperty part.
That would inject the session bean into the request bean without creating any instance of it.

*Don’t for get to add getter and setter for the session bean you are adding otherwise you would get an error.

The session bean would be a simple holder of model

package com.enderase.beans;

import javax.faces.bean.ManagedBean;
import javax.faces.bean.SessionScoped;

import com.enderase.model.Contractor;

@ManagedBean
@SessionScoped
public class ContractorBean {
	
	private Contractor contractor;
	
	public ContractorBean() {
		this.contractor = new Contractor();
	}
	public Contractor getContractor(){
		return this.contractor;
	}
	
	public void setContractor(Contractor contractor){
		this.contractor = contractor;
	}
}

Where the contractor would be a simple POJO file

asadmin-command-unknown-in-glassfish-ee-application

am a great user of tomcat when it comes to Java web application. I had fun with it. Being fast and allowing a bunch of things to be done by myself.. that being said, I am a regular user of glassfish as well. Specially the later version 3 looks awesome in a lot of ways..

I will try to use this blog to amend any hiccups whenever they appear and a bit of more tutorials as well.

The first one is the command line friend asadmin.

On the new version it will be found on

/glassfish-main-directory/glassfish3/glassfish/bin

Being on this directory if you issie

./asadmin

You will get the command line for it.
To add it to your path so that you can use it from any where in your terminal, just add it to your path

open your ~/.bash_profile or .bash_rc [create it if it doesnt exist and add the above directory at the end of it separated by appropriate directory separator.

That is it..

Mass/Multiple file upload in Java ServerFaces JSF

Without knowing if it is the best approach or not, I will post how I solved the multiple file upload problem in JSF as follows.
Here is the xhtml file that would take the files


<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">

<html xmlns="http://www.w3.org/1999/xhtml"
xmlns:h="http://java.sun.com/jsf/html"
xmlns:f="http://java.sun.com/jsf/core"
xmlns:t="http://myfaces.apache.org/tomahawk">
<h:head>
<script type="text/javascript" src="https://ajax.googleapis.com/ajax/libs/jquery/1.6.2/jquery.js">
</script>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8" />
<h:outputScript library="javascript" name="amharic.js"/>
<title>OH YEA, PUT your face HERE</title>
</h:head>
<h:body>

<h:form id="uploadForm" enctype="multipart/form-data">
<h:panelGrid columns="3">
<h:outputLabel for="file1" value="Select file" />
<t:inputFileUpload id="file1" value="#{myBean.uploadedFile}" required="true" />
<h:message for="file1" style="color: red;" />
<h:outputLabel for="file2" value="Select file" />
<t:inputFileUpload id="file2" value="#{myBean.uploadedFile}" required="false" />
<h:message for="file2" style="color: red;" />
<h:outputLabel for="file3" value="Select file" />
<t:inputFileUpload id="file3" value="#{myBean.uploadedFile}" required="false" />
<h:message for="file3" style="color: red;" />

<h:panelGroup />
<h:commandButton value="save" action="#{myBean.uploadFiles}" />
<h:message for="uploadForm" infoStyle="color: green;" errorStyle="color: red;" />
</h:panelGrid>
</h:form>
</h:body>
</html>

As you can see, the file would be dealing with a single backing bean property uploadedFile. You would tomahawk for the file upload one. There is also file upload in richfaces as well.

Now, let’s see what is the hood of MyBean.java. In the bean you would have to member variables for this purpose:

private List<UploadedFile> uploadedFiles;
private UploadedFile uploadedFile;

You can get the uploaded file from org.apache.myfaces.custom.fileupload.UploadedFile.
Then have a normal getter and setter for both.
The trick is in the setter of the uploadedFile:

public void setUploadedFile(UploadedFile uploadedFile){
this.uploadedFiles.add(uploadedFile);
}

When the file is requested, add it to the list of the uploadedFiles.
For the action which would would do the actual uploading of the file, I have used the snippet from http://balusc.blogspot.com/2008/02/uploading-files-with-jsf.html.

public String uploadFiles(){
for(UploadedFile uploadedFile : this.uploadedFiles){
if (uploadedFile !=null ){

// Prepare filename prefix and suffix for an unique filename in upload folder.
String prefix = FilenameUtils.getBaseName(uploadedFile.getName());
String suffix = FilenameUtils.getExtension(uploadedFile.getName());

// Prepare file and outputstream.
File file = null;
OutputStream output = null;
try {
// Create file with unique name in upload folder and write to it.
file = File.createTempFile(prefix + "_", "." + suffix, new File("Your_Path"));
output = new FileOutputStream(file);
IOUtils.copy(uploadedFile.getInputStream(), output);

// Show succes message.
FacesContext.getCurrentInstance().addMessage("uploadForm", new FacesMessage(
FacesMessage.SEVERITY_INFO, "File upload succeed!", null));
} catch (IOException e) {
// Cleanup.
if (file != null) file.delete();

// Show error message.
FacesContext.getCurrentInstance().addMessage("uploadForm", new FacesMessage(
FacesMessage.SEVERITY_ERROR, "File upload failed with I/O error.", null));

// Always log stacktraces (with a real logger).
e.printStackTrace();
} finally {
IOUtils.closeQuietly(output);
}
}
}
return "done";
}