Our Grails 2.4.4 application is using RESTful URLs throughout. Given the following URL:
/stores/123/products/456
I'd like to validate that there is a store with an ID of 123, and if not, redirect to a 404 on every request to the Product controller. I don't want to have to put that store lookup code in each action method, nor do I want to create a controller base class, because then I have to put a method call in every action method.
Can this be done with a interceptor somehow?
Interceptors are introduced in Grails 3.0. You would need filters in Grails 2.4.4.
before = { } is what will be needed here.
Also look at the docs which variable are available to filters by default (eg: params, request, response etc). If this is still unclear, I can add an answer as an example. But I hope docs will be self explanatory. As an example i would do it as
class EntityCheckFilters {
def filters = {
storeExistCheck( controller:'product' ) {
before = {
if ( !params.storeId || !Store.exists( params.sotreId as Long ) ) {
response.sendError(404)
// or for example if you have a separate action to handle 404
// redirect(action: 'handle404')
// this is important,
// because we do not want to pass through with the original call
return false
}
}
}
}
}
I think you can do this using:
URL Mapping
Filters
But I dont think putting a logic (checking if there is a valid store with the given id) in URL Mapping is good idea, so better to use Filters.
So you url mapping will look like this:
"/stores/$storeId/products/$productId" (controller = "product")
And your filter:
class YourFilterNameFilters {
def filters = {
secureReports(controller:'*', action:'*') {
before = {
if(parmas.controller == "product" && params.storeId){
Store store = Store.get(params.sotreId)
if(!store){
response.sendError(404)
}
}
}
}
}
Related
I have developed Small Spring boot Rest api app. I can able to get the data or create new record and search with paging and sorting.
Now i'm looking for provide input data in body to get the data instead of providing in URL with GET method. Is this method also default function ? Please advise.
public interface CodeTextRepository extends PagingAndSortingRepository<CodeText, Long> {
}
How to write POST method to just get the data ?
http://localhost:8080/api/code
method : POST
{
"code":1
}
If I understand you correctly, you want to create a controller that will get the a model as body parameter ({ "code": 1 }) in a POST method and then do something with it.
To do that, you can create a controller that looks like the following (I inserted pseudo-code as an example):
#RestController
#RequestMapping(value = "/api/code")
public class CodeTextController {
private CodeTextRepository codeTextRepository;
// constructor injection
public CodeTextController(CodeTextRepository codeTextRepository) {
this.codeTextRepository = codeTextRepository;
}
#PostMapping
public CodeText postCodeText(#RequestBody CodeTextRequest codeTextRequest) {
// some code to get from the DB
return codeText;
}
}
public class CodeTextRequest {
private int code;
// getters and setters
}
Simply add Accept header to the request, like
accept: application/json
Spring Data-Rest will return the body after a POST request if either the returnBodyOnCreate flag was explicitly set to true in the RepositoryRestConfiguration OR if the flag was NOT set AND the request has an Accept header.
You can set the flag directly during configuration, or you can set it via the application.properties:
spring.data.rest.returnBodyOnCreate = true
you can also set it separately for update:
spring.data.rest.returnBodyOnUpdate = true
---- edit
Maybe I misunderstood your question. If you simply want to GET an existing data using POST method, then DO NOT DO IT AT ALL! That's not a REST API any more. There must be some reason you want to do it, but you should try do resolve that original problem instead in another way!
I have an issue with my app, that is "hybrid", what I mean by "hybrid" controllers have to manage both views ans APIs.
So, basically, for each controller, I must check:
if $request->wantsJson(){
... // Client rendering using Angular, return json
}else{
// Server rendering Using blade, return view
}
I don't like the fact to have a conditional in every controller method.
I also wouldn't like to have a API folder with a copy of all my controller, there would be a lot of duplicated code.
How should I do it?
I would suggest to create a separate class to handle output ex: class ResultOutput with the method, output.
So, in your controller, when you are ready to output your data, just create a new instance of ResultOutput class, and call method output with relevant data.
In ResultOutput class, inject Request object so you can determine the method of the output based on above logic.
Ex: In your controller:
return (new ResultOutput())->output($data);
In ResultOutput class:
class ResultOutput()
{
private $type;
public __construct(Request $request) {
$this->output = 'view';
if ($request->wantsJson()) {
$this->output = 'json';
}
}
public method output($data) {
if ($this->type =='view') {
// return the view with data
} else {
// return the json output
}
}
}
In this way, if you need to introduce new output method (ex: xml), you can do it without changing all your controllers.
I am new to grails. I have recently used session in my controller. But for only one page. Now I want to use session for a number of pages. But I have no idea how to do it. Here is my code below which works for one page. Can anyone please help me on this ?
def index() {
def user = springSecurityService.currentUser
if (user){
redirect(controller: 'admistratorAction', action: 'createUser')
}else{
redirect(controller: 'login', action: 'index')
}
}
You may be new to Grails, I hope you are not new to HttpSession. :)
Session information is scoped only to the current web application
(ServletContext), so information stored in one context will not be
directly visible in another.
As long as you are in the same ServletContext you should be able to access session variable directly. Also look at Servlet API in grails.
#Alidad- Scroll back to last question from OP.
In that case you can take advantage of grails filter to do the check before execution of each action. As dmahapatro mentioned you can use session across your app to store user object and with this filter you can do a check before any action.
Something like this can help you achieve it.:
class SecurityFilters {
def filters = {
loginCheck(controller: '*', action: '*') {
before = {
if (!session.user)) {
redirect(action: 'login')
return false
}
}
}
}
}
I have a standard CI web app, but I've decided to get the chaotic javascript in order using backbone. I had a whole pile of serialized forms/jQuery AJAX requests to various controller methods: authenticate, change_password, register_member, request_new_password, etc.., and don't quite understand how REST works instead. I'm using Phil Sturgeon's REST library for CI https://github.com/philsturgeon/codeigniter-restserver
Should every backbone model have a different api url? And what am I supposed to actually call the controller methods?
<?php
require(APPPATH.'/libraries/REST_Controller.php');
class RestApi extends REST_Controller
{
function get()
{
But it just 404s.
I just don't get how to replace the routing to fifty of my old methods based on a handful of HTTP methods. Does the name of the backbone model need to match something on the server side?
You have to name your functions index_HTTPMETHOD. In your example it would be:
class RestApi extends REST_Controller {
// this will handle GET http://.../RestApi
function index_get() {
}
// additionally this will handle POST http://.../RestApi
function index_post() {
}
// and so forth
// if you want to POST to http://.../RestApi/somefunc
function somefunc_post() {
}
}
the url-attribute of the model should match the server-side 'url' which returns the JSON that will make up the model's attributes. Backbone.js has default functionality to this, which is to match the model's collection url with it's id attribute. The collection url requirement can be foregone by overriding the urlRoot -function, in order to operate model's outside of collections.
If you want to be independent of the id -attribute as well, you sould override the url -attribute/function to provide your own url that matches to the model on the server, like this:
url: 'path/to/my/model'
or
url: function() { // Define the url as a function of some model properties
var path = this.model_root + '/' + 'some_other_url_fragment/' + this.chosen_model_identifier;
return path;
}
Looking for a way to construct or generate a url for a specific resource in asp.net web api. It can be done in the controller since it inherits from ApiController hence you get the UrlHelper.
I am looking to construct resource url out of the context of the ApiController.
Here is what I did:
Requires HttpContext/Request, so might not work in Application_Start.
Only tested in WebApi 1
Only works for routes registered in GlobalConfiguration (but if you have some other one, just pass it in instead)
// given HttpContext context, e.g. HttpContext.Current
var request = new HttpRequestMessage(HttpMethod.Get, context.Request.Url) {
Properties = {
{ HttpPropertyKeys.HttpConfigurationKey, GlobalConfiguration.Configuration },
{ HttpPropertyKeys.HttpRouteDataKey, new HttpRouteData(new HttpRoute()) },
{ "MS_HttpContext", new HttpContextWrapper(context) }
}
};
var urlHelper = new UrlHelper(request);
What about the UrlHelper classes:
System.Web.Http.Routing.UrlHelper;
System.Web.Mvc.UrlHelper
The MVC one has some useful static methods accepting routing information or it can be used as an instance created by passing in a RequestContext (which is available in most MVC filters and various other places). The instance methods should be exactly what you need to generate urls.
The HTTP one accepts a ControllerContext (which is also available in most HTTP filters and various other places).
I'm not sure about the ApiController, as I haven't used it before. This may then be redundant for you, but then again, it may not be. Check out your Global.asax.cs file, specifically the RegisterRoutes function. Initially, you should see the following mapping:
routes.MapRoute ("Default", "{controller}/{action}/{id}", new { controller = "MyController", action = "Index", id = "" });
So by default your application is set up to handle routes in the following format:
{ControllerName}/{ActionName}/{ResourceId}
A controller class set up like the following should enable you to receive requests in that format.
class {ControllerName}Controller : ApiController
{
public ActionResult {ActionName} (string id)
{
// fetch your resource by its unique identifier
}
}