Rendering Views as String with Spring MVC and Apache Tiles - json

I am trying to reuse some of my tiles in a controller which is returning a json response to the client. I would like to return a json response similar to the following format:
{
'success': <true or false>,
'response': <the contents of an apache tile>
}
In my controller I would like to perform logic similar to this pseudocode:
boolean valid = validator.validate(modelObj)
String response = ""
if(valid){
response = successView.render() // im looking for a way to actually accomplish
// this, where the successView is the apache tiles view.
// I would also need to pass a model map to the view somehow.
}else{
response = errorView.render()
}
writeJsonResponse(httpResponse, /* a Map whose json representation looks like the one I described above. */)

I belive that you want to implement a view class that will wrap the output of a jsp in json. The class in question may be org.springframework.web.servlet.view.tiles2.TilesView.
Another option may be to extend the JSON converter. org.springframework.http.converter.json.MappingJacksonHttpMessageConverter

If you need to render the view using Apache Tiles 2, you must use
org.springframework.web.servlet.view.tiles2.TilesViewResolver
See the example tutorial here: http://krams915.blogspot.com/2010/12/spring-mvc-3-tiles-2-integration.html
If you need to render the response as JSON, you can use the #ResponseBody which requires Jackson in your classpath. See the example here http://krams915.blogspot.com/2011/01/spring-mvc-3-and-jquery-integration.html (The controller returns JSON). You can also see a similar example of the #ResponseBody at http://krams915.blogspot.com/2010/12/jqgrid-and-spring-3-mvc-integration.html

Related

Returning XML and Json formatted data from MVC controller

I need to create an MVC 4 controller that can return the same data either in a JSON format for an XML format depending on the request from the view. Would it be better to create two controllers or can this be done with one controller? If it can be done with one controller, can someone show me how this is done?
THanks.
You can certainly have add an Action to your existing controller that will return a JSON object as the result; luckily MVC has a nice, built-in way of returning JSON:
public JsonResult MyJsonAction()
{
var result = "my data";
return Json(result);
}
As for returning XML, there seems to be some useful answers on this question

Return JSON Result from an Api Controller

I'm building an Api Controller and I need to serialize my List to JSON as my action result.but It seems that such statements doesn't work
return Json(data, JsonRequestBehavior.AllowGet);
How can I achieve this ?
As you mentioned that you are using WEB API, I'm assuming it has the JsonFormatter configured. With that said, the responsibility to convert you action result into a JSON is not of your action but from the Media Type Formatter chosen as part of the Content Negotiation process.
That said, it's enough for your Action to return the actual List type and the Web API Media Type formatter will take care of formatting it to JSON.
For example, let's say that data is a List<Foo> where Foo is some type that you created. It is enough for your controller action to be:
public List<Foo> GetFoo()
{
var data = GetListOfFoo();
return data;
}
Have you tried using a JSON serializtion class?
I have had success using the ideas put forward in this article:
Serializing a list to JSON
Or, if you don't want to use serialization, the example for an action result using JSON in MSDN just uses a generic list object.

Pass Json Object from Play! framework to HighCharts JS

http://www.playframework.com/documentation/2.1.x/JavaTodoList
Using the above tutorial as a reference, I have created an application which sends data from the model to view via the Application controller.
I have managed to display the model(Tasks) as a high chart. The code is here.
public static Result format(){
return ok(views.html.frmt.render("Visualize it",Task.all()));
}
This goes to this view page.
http://ideone.com/ycz9ko
Currently, I use scala templating inside the javascript code itself. Refer to lines 9-14 and lines 20-24.This unelegant style of doing things is not really optimal.
I want to be able to accomplish the above using Json instead.
public static Result jsonIt(){
List<Task> tasks = Task.all();
return ok(Json.toJson(tasks));
}
My Qns are how to send the JSON objects to a view template.
And how to parse it into a Highcharts format. Is there some standard procedure to do this ? Or else I have to write my own method to do this ?
It'll great if someone can show me a code snippet. Also I would prefer a post not using Ajax. I would just want to know how to do this first.
I also found this stackoverflow post useful.how to parse json into highcharts. However, it didnt answer the part about converting from Play format to Highcharts format.
Thanks in advance
You don't need to pass a json object to your template, instead you might do an ajax call from your client side javascript (your template) and get json response that you could use futher in javascript code to build a chart. For example :
You have some path that is bind to your controller jsonIt() like so /chartsdata/json
then using jquery shorthand for ajax request:
var chart_data = $.get('/chartsdata/json', function(data) {
return data;
});
now you can use a chart_data that is an array of objects where each object represents a Task, in your further javascript code to build a chart.

Laravel return json or view

I'm developing an API where if the user specifies the action with .json as a suffix (e.g. admin/users.json), they get the response in the return of json, otherwise they get a regular html View.
Some actions may not have a json response, in which case they would just return a html View.
Does anyone have advice on how this can be implemented cleanly? I was hoping it could be achieved via the routing.
I suggest you to create your application as an api.
Foreach page, you need two controllers. Each controller use a different route (in your case, one route ending by .json, and one without).
The json controller return data in json form. The "normal" controller call the corresponding json route, deserialize the json, then pass the resulting array to the view.
This way, you've got a standardized api (and maintained, because your own app use it) available, as well as a "normal" website.
More information:
Consuming my own Laravel API
Edit: Maybe it's doable with a filter, but I'm not sure about that and I don't have time to try it myself right now.
In Laravel 5.x, to implement both capabilities like sending data for AJAX or JSON request and otherwise returning view template for others, all you have to do is check $request->ajax() or $request->isJson().
public function controllerMethod(Request $request)
{
if ($request->ajax() || $request->isJson()) {
//Get data from your Model or whatever
return $data;
} else {
return view('myView.index');
}
}

How to enrich a spring controller json's response

I've got a simple Spring MVC app where controllers are generating Json objects and returning them as Strings.
I'd like to return those json trees as-is out of the controllers and have a kind of servlet filter which will enrich them. Basically it will take the json node and move it as a child of a brand new json root. Think of a 'body' encapsulated in a complete response including also a 'head' child node that would be generated by this filter.
It is possible to do so within Spring ?
Thanks for your advices !
I would advise instead of returning plain Strings, return POJO and convert this to JSON using HTTP Message converters (e.g. MappingJackson2HttpMessageConverter) and #ResponseBody annotation. Then you can decorate the return object the way you want.