Saturday, January 28, 2012

Spring MVC 3.1, jqGrid, and Spring Data JPA Integration Guide (Part 2)

Review

In the previous section, we have laid down the functional specs of the application. In this section, we will discuss the project's structure, write the Java classes, and organize them in layers.


Project Structure

Our application is a Maven project and therefore follows Maven structure. As we create the classes, we've organized them in logical layers: domain, repository, service, and controller.

Here's a preview of our project's structure:

The Layers

Domain

This layer contains two domain classes, User and Role. They represent our database tables, user and role respectively. Because we're developing a JPA-based repository, both classes must be annotated with JPA annotations.




Data Transfer Object (DTO)

This layer contains three DTO classes:
  • UserDto is a POJO for mapping User objects to and from the presentation layer
  • StatusResponse is a POJO for sending boolean responses to the presentation layer
  • JqgridResponse is a container for UserDto objects for sending records to the jqGrid table. It contains information regarding the number of rows, current page, and total pages.




What is the principle behind JqgridResponse's structure?

By default, the jqGrid plugin can process data in various formats: XML, JSON, Arrays, and etc. For this tutorial, we will use JSON because it's lightweight and simple. If we're going to use JSON, our JSON format must match our jqGrid's jsonReader property. Consequently, our DTO object must match as well.

Below is a sample jqGrid jsonReader property declaration:

Below is a sample JSON string that's acceptable to jqGrid (Notice the fields match the jsonReader fields and JqgridResponse's fields):
{"page":"1","total":"2","records":"2","rows":[{"id":1,"firstName":"John","lastName":"Smith","username":"john","role":1},{"id":2,"firstName":"Jane","lastName":"Adams","username":"jane","role":2}]}

Controller

This layer contains two controllers, MediatorController and UserController.
  • MediatorController is responsible for redirecting requests from the root path to the Users page
  • UserController is responsible for handling user related requests




The UserController in all cases, except for the getUsersPage() method, returns a JSON string as indicated in the @RequestMapping annotation:
produces="application/json"


The methods create(), update(), delete(), and get() are pretty much straightforward. However, the records() and getFilteredRecords() methods are somewhat more involved.

The records() method basically returns a list of UserDto objects as JSON strings. If the search prooperty is true, it will call the getFilteredRecords() method. Otherwise, it will retrieve all records.

The getFilteredRecords() is quite interesting. The basic algorithm is as follows:
  1. Convert a JSON String filter to a JqgridFilter object
  2. Use JqgridObjectMapper.map() method to do the conversion
    (After the conversion, a list of Rule objects are produced)
  3. Loop these list.
  4. If any of the fields match "username", "firstName", "lastName", and "role", store its value
    (This means we can only search within these fields.)
  5. Do a repository search based on non-empty field parameters
  6. Return the results to the presentation layer

Introducing Jsonquery

Question: What if I want to have a dynamic search on all fields? For example, instead of doing a "like" comparison, I want to do a "greater than" or "less than" and combine various operators, i.e. "and", "or".

Answer: This is not possible with the way I coded that here. And even if I could, it would be a big mess of if-else conditions. Luckily, there's a library that would simplify that for us: jsonquery

Jsonquery is a framework that translates SQL-like JSON queries to type-safe JPQL queries through Querydsl which means each query is type-safe, fluent, and SQL-like. Currently, the framework is designed for JPA-based backends.

To see the actual project, please visit https://github.com/markdevcode/jsonquery
To see the samples, please visit https://github.com/markdevcode/jsonquery-samples

However there are caveats:
  • It's not yet available from the Maven repository (it's on its way)
  • You must build and deploy the project from Github
  • You must use JPA and QueryDSL

Repository

This layer contains a single interface, UserRepository. This is our data access object (DAO). With the help of Spring Data JPA, Spring will automatically provide the actual implementation.

What is Spring Data JPA?

Spring JPA is part of the umbrella Spring Data project that makes it easy to easily implement JPA based repositories.

Implementing a data access layer of an application has been cumbersome for quite a while. Too much boilerplate code has to be written to execute simple queries as well as perform pagination, and auditing. Spring JPA aims to significantly improve the implementation of data access layers by reducing the effort to the amount that's actually needed. As a developer you write your repository interfaces, including custom finder methods, and Spring will provide the implementation automatically.

Source: http://www.springsource.org/spring-data/jpa

Service

This layer contains a single service, UserService. Its main purpose is to handle the CRUD operations for the User object. Notice all operations are eventually delegated to the UserRepository.

Utilities

  • JqgridFilter is a Java representation of a jqGrid filter
  • JqgridObjectMapper is used to convert a jqGrid filter to a JqgridFilter object
  • UserMapper is used to map User objects to UserDto objects
  • TraceInterceptor is an AOP-based utility class to help us debug our application. This is a subclass of CustomizableTraceInterceptor (see Spring Data JPA FAQ)

Next

In the next section, we will discuss the presentation layer and write the HTML and JavaScript files. Click here to proceed.
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21 comments:

  1. nice tutorial.

    i have a column with user_id , how to write a method for this column in repository class.

    in my code

    Page findByUserid(Integer user_id, Pageable pageable);

    Error
    Caused by: java.lang.IllegalArgumentException: No property user found for type class org.krams.domain.User

    ReplyDelete
  2. @Anonymous, I think it should be:

    Page findByUser_Id(Integer user_id, Pageable pageable);

    ReplyDelete
    Replies
    1. Thanks it worked out for me.

      Delete
  3. Thanks You

    One comment : Sort is missing ;)

    ReplyDelete
  4. Thanks for sharing it works for Me...:)

    ReplyDelete
  5. Awesome tutorial, thanks!

    ReplyDelete
  6. This comment has been removed by the author.

    ReplyDelete
  7. @krams :hi i would like to ask that send me entire project to my email id:sag38.keerthi@gmail.com
    if u wont mind...
    eagerly waiting for this code..

    ReplyDelete
    Replies
    1. I don't understand why you need it to be sent by email? You can download the whole project from Github.

      Delete
  8. @krams :hi i kram i need pagination in spring 3.x
    if u know then farward me that project also
    but i just kept project in hold because of that issue..so please send me that code to mail id
    :sag38.keerthi@gmail.com

    ReplyDelete
    Replies
    1. Why don't you use Spring Data? It has built-in pagination support. Are you asking something in the frontend layer? I suggest using one of the existing jQuery table plugins.

      Delete
  9. This comment has been removed by the author.

    ReplyDelete
  10. Hi @Krams: First i want to thank you as your tutorial and example are the best, the best as i have tried many as new to Spring and hibernate all other post by many famous people gave many error and it never execute properly, yours is the best, thanks again for sharing such great knowledge

    Question:for some reason @RequestMapping(value="/update", produces="application/json", method=RequestMethod.POST) is giving me error under Produces , can you help me with this, how can i resolve it?

    ReplyDelete
    Replies
    1. What specific error are you getting? Can you copy and paste it here using pastebin.com?

      Delete
  11. Hi @Krams: Thank you for the tutorial. I am wondering why you use UserDto DTO in the tutorial. Can you just use domain object User alone instead?

    ReplyDelete
  12. can you pls provide the link from where we can download the project

    ReplyDelete
  13. why you dont just extend your Entity as Serializable and use it in your Jqgrid response?

    ReplyDelete
  14. I have read your blog its very attractive and impressive. I like it your blog.

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    ReplyDelete