播放框架：将数据显示为html表

Question

I am developing a small project which has a couple of tables. One of them is Person table which has id and name as attributes. I am successfully getting the resultset from database (using MySql) and can convert them into json. However, i am unable to find a way to display the json as a html table. Is there a way to pass the list of Persons to the html and just display it in a table ??

Application.java

public class Application extends Controller {

@Inject
FormFactory formFactory;

public Result index() {
    return ok(index.render());
}

@Transactional(readOnly = true)
public Result getPersons() {
    List<Person> persons = (List<Person>) JPA.em().createQuery("select p from Person p").getResultList();
    return ok(toJson(persons));
}

index.scala.html

 @()
@main("Welcome to Play") {
<script type='text/javascript' src='@routes.Assets.at("javascripts/index.js")'></script>

<ul id="persons"></ul>

//logic to display table
}

index.coffee

$ ->
  $.get "/persons", (persons) ->
    $.each persons, (index, person) ->
      $("#persons").append $("<li>").text person.id + person.name 

This displays a list of person objects. But i need to display a html table

Answer 1

However, i am unable to find a way to display the json as a html table. Is there a way to pass the list of Persons to the html and just display it in a table ??

It doesn't appear you have read the Play manual very well. However, I would not send JSON to the template. Instead, I would pass the

List<Persons> 

(Java Object) to the scala template and use the Scala Templating language to display the table. I think you would benefit from reading the template help page here.

Pay special attention to the Iterating section for doing a for loop. This is how I would do it (untested):

index.scala.html

@(persons: List[Person])
@main("Welcome to Play") {
<script type='text/javascript' src='@routes.Assets.at("javascripts/index.js")'></script>

<ul id="persons"></ul>

//logic to display table
<table>
   @for(person <- persons){
      <tr>
          <td>@person.getName()</td>
          <td>@person.getId()</td>
      </tr>
    }
</table>
}

Note that in a Scala Template, instead of using the syntax of:

List<Person>

(like in Java), you use

List[Person]  

And for Application Controller, change the relevant part to:

public Result getPersons() {
    List<Person> persons = (List<Person>) JPA.em().createQuery("select p    from Person p").getResultList();
return ok(index.render(persons);

This assumes you have two fields in class Person: name and id. It also assumes that you have "get" methods for them. If they are public fields, you can use @person.name and @person.id. Change those variables to match what you have in your Person class.

You shouldn't need index.coffee then. You also won't need the tag in your index.scala.html (unless you need it for other reasons).

On the help page mentioned above, you might also check out the "Declaring reusable blocks" section to create a resuable block to display each Person object. However, I would get the simplest version working first and then attempt this.