Developing a RESTful Java Web service in Eclipse Indigo using maven

RESTful web services have become the kind of de-facto in the present days, thanks to Roy Fielding's dissertation stating its advantages. Developing these is also easy to someone who is NOT very accustomed to SOAP-WSDL based services. However it is little difficult to someone who has been working in the arena of SOAP, although not tough.
I believe a great deal of difficulty in developing RESTful web services is because of the unavailability of a uniform IDE that makes life easier for a developer. Although Netbeans has a very good abstracted mechanism that lets users develop RESTful web services with no pain, I don't prefer working with it, nor would I recommend anyone to use it. The major reason for the same is the highest levels of abstraction it offers to the developer which makes it difficult for him/her to understand the flow.
In this article I will try to walk you through the steps involved in successfully developing a RESTful java web service in eclipse using maven.
This article uses Eclipse Indigo, maven2, Sonatype M2Eclipse, Apache Tomcat 7.
Step1: Go to Help->Install New Software->Add. Enter the name as "m2eclipse" (you can give any name you wish to) and the location  http://m2eclipse.sonatype.org/sites/m2e. Then you will find the software "Maven Integration for Eclipse". Install it.
Step2: Developing a Dynamic web project also requires installing the software, "Maven Integration for Eclipse WTP", which can be found in the repository, http://m2eclipse.sonatype.org/sites/m2e-extras. Proceed similarly to add this repository too.
Step3: Now choose the directory where the workspace needs to be stored and run the folowing command.

mvn -Declipse.workspace= eclipse:add-maven-repo

This will be the local maven repository.
Step4: Now New->Project->Maven Project->Select Artifact ID as maven-archetype-webapp->Give groupID (similar to package name), artifact ID (similar to project name) and click Finish.
Step5: Now a directory structure will be formed as follows.

Step6: This is the most important step. The directory structure formed by default has to be changed a little. As per the Maven directory structure specifications, all the source java files (services) need to be inside /src/main/java. Follow the following steps to change the directory structure.

• Remove src/main/resources under Java Resources
• Add java folder to src/main directory
• Right click on Java Resources->New->Source Folder and give the previously added java folder in the Folder name and click Finish as shown below.

• The final directory structure should look like this.

Step7: Now add the following lines to pom.xml.

  

 
  maven2-repository.java.net
  Java.net Repository for Maven
  http://download.java.net/maven/2/
  default
 

  
    
   com.sun.jersey
   jersey-server
   1.8
    
  

This will essentially indicate the maven, the url to look for repositories while building. Also the dependency here is as required by the Jersey specification.
Step8: Now write the java service in a class file. Create a new class and add code similar to this. The annotations @Path along with @GET and @PathParam are very important, you can look into JSR311 API for details on this.

package com.rest.test;

import javax.ws.rs.GET;
import javax.ws.rs.Path;
import javax.ws.rs.PathParam;
import javax.ws.rs.core.Response;
 
@Path("/test")
public class TestRest {
 
 @GET
 @Path("/{name}")
 public Response getMsg(@PathParam("name") String name) {
 
  String output = "Hello, " + name;
 
  return Response.status(200).entity(output).build();
 
 }
 
}

 Step9: Now we are required to update web.xml according to Jersey reference implementation. web.xml should look like this.

 Restful Web Application
 
 
  TestServlet
  
                     com.sun.jersey.spi.container.servlet.ServletContainer
                
  
       com.sun.jersey.config.property.packages
       com.rest.test
  
  1
 
 
 
  TestServlet
  /resources/*
 
 

These steps should essentially deploy a RESTful java web service successfully and you can test it by hitting the following URL.

http://localhost:8080/resources/test/kausal

And you should see the output "Hello, kausal". Here "kausal" is the parameter passed to the method response().

I hope this gives basic idea about how to develop a RESTful java web service with Jersey reference implementation in eclipse using maven. Please feel free to post your queries/comments.

Resolving issues with Eclipse on Ubuntu upgrade

Hi folks,
Today I did a partial upgrade of my Ubuntu and after that, to my surprise, Eclipse stopped working properly.
It gave an error saying...
"An internal error occurred during: "Compute launch button tooltip".org/eclipse/cdt/core/model/CoreModel"
... while I was trying to run my Java App.
Re-install of Eclipse didn't work from Ubuntu Software Center, neither did it work from Terminal.
Even apt-get install eclipse-platform --fix-missing didn't fix the issue.
After going through several posts, I found that a package called tzdata-java has to be re-installed.
So, apt-get --reinstall install tzdata-java finally worked.
This was just the condensed information from different sources and not my discovery! Posted here so that it might be useful to someone.
Thanks!

Memory allocation upon malloc() in C

Hi All,
Many of you would have already known how memory allocation happens when malloc() function is used in C language. This article is just the gist of my study on malloc() function.

What happens when you invoke malloc()?
When the C compiler sees malloc(4) function in the program, it requests the underlying OS (Operating System) to allocate 4 bytes and the OS, upon allocation, returns the block containing the address to the next block in the memory + the starting address of the memory allocated + the size of the block (here 4 bytes), rounded off to the multiples of size of the most restricted datatype on the machine.

Example:

int *p;

p=(int*)malloc(sizeof(int));

printf("%u\n",p);

p=(int*)malloc(sizeof(int));

printf("%u\n",p);

p=(int*)malloc(sizeof(int));

printf("%u\n",p);

p=(int*)malloc(sizeof(int));

printf("%u\n",p);

returns...

152002568
152002584
152002600
152002616

These are separated by 16 bytes, on my 32-bit machine! Please find the math for this below.
I will try to illustrate it with another bigger example there and put math of both examples together.


What is the most restrictive datatype?
For each machine, there is a most restrictive type. It is not unique and varies on different machines. It means that if you can store the most restricted datatype in an address, you can store any primitive type. Generally it will be double (as it is on my machine) but I read somewhere that it can be int or float too.

Yet another example, with calculations?
Consider creating the following structure and allocate memory using malloc() function.

struct test

{

char c;

char f;

int a;

int b;

float d;

double e;

double g;

double h;

double i;

};

int main()

{

struct test *t;

t=(struct test*)malloc(sizeof(struct test));

printf("%d\n",sizeof(struct test));

printf("%u\n",t);

t=(struct test*)malloc(sizeof(struct test));

printf("%u\n",t);

t=(struct test*)malloc(sizeof(struct test));

printf("%u\n",t);

t=(struct test*)malloc(sizeof(struct test));

printf("%u\n",t);

t=(struct test*)malloc(sizeof(struct test));

printf("%u\n",t);

t=(struct test*)malloc(sizeof(struct test));

printf("%u\n",t);

return 0;

}

returns...

48

140615688

140615744

140615800

140615856

140615912

140615968

Reason goes here...
On a 32-bit machine, char occupies 1 byte, int and float, 4 bytes while double occupies 8 bytes. So the structure essentially needs 46 bytes but 48 bytes are allocated because it should be in the multiples of (8 bytes - most restricted datatype, double on a 32-bit machine).
Now, on a 32-bit machine, each word is addressed by 32 bits (==4 bytes). So, as per the explanation given above w.r.t. block returned by malloc(), it has to return the address to next block (4 bytes) + address of allocated block (4 bytes) + size of structure (48 bytes), i.e. 56 bytes. This is the reason why we got addresses separated by 56 bytes for each allocation!

Similar math follows for the first example I quoted...
On a 32-bit machine, int occupies 4 bytes, but address to the next block (4 bytes) and address of the currently allocated block (4 bytes) add up to it making it 12 bytes. But the actual memory allocated must be in the multiples of most restricted datatype on the underlying machine which is double ( 8 bytes) and hence 16 bytes are allocated.

Hope this information will be useful to you all. I would like to have your comments/feedback on this. I am open for discussion on this topic! :)

Presenting my blog on Java programming

Hi All,
Apart from the technical ideas and articles of mine in this blog, I am also presenting you all another blog of mine, which is dedicated to Java programming.
http://mkausal.blogspot.in/
Please visit the blog for any queries w.r.t. Java programming.

Thanks!

Solving problem with Update Manager in Ubuntu 12.04

Hi,
Yesterday I got a problem in my Ubuntu 12.04 machine after downloading few third party softwares. The Update Manager was giving the error...
"E:Encountered a section with no Package: header,
E:Problem with MergeList /var/lib/apt/lists/in.archive.ubuntu.com_ubuntu_dists_precise_main_binary-i386_Packages,
E:The package lists or status file could not be parsed or opened."
I tried different things like removing the newly added entries in /etc/apt/sources.list but it didn't work.
Then, I removed all the available listings in the directory.. /var/lib/apt/lists and updated the list using apt-get update and to my surprise it worked!
Remember, you need to do all the above operations as sudo!

Just posted if you would find it helpful and not waste time in trying different crazy things to fix this, like I did! :P :)

Thanks. Comments are welcome!