HATEOAS对Angular2的支持 [英] Angular2 support with HATEOAS
问题描述
在HATEOAS链接的支持下,我有一个宁静的Web服务.当我打电话 " http://localhost:8080/v1/bookings/1225380?lock = true 链接我得到了以下资源URL.我想将这些Hypermedia与我的Angular2应用程序集成(最近升级到最终版本).我发现在Angular HAL的支持下使用Angular1实现的资源很少(链接- https://github.com/LuvDaSun/angular-hal/tree/master/src ).但是我找不到Angular2的资源.
I have a restful web service with the support of HATEOAS links. When I call "http://localhost:8080/v1/bookings/1225380?lock=true" link I got following resource URLs. I want to integrate these Hypermedia with my Angular2 application (recently upgraded to final version). I found few resources which were implemented with Angular1 with the support of Angular HAL (links - https://paulcwarren.wordpress.com/2015/04/03/role-based-spas-with-angularjs-and-spring-hateoas/, https://github.com/LuvDaSun/angular-hal/tree/master/src). But I am unable to found a resource for Angular2.
"links": [
{
"rel": "client",
"href": "http://localhost:8080/v1/clients/10000"
},
{
"rel": "passengers",
"href": "http://localhost:8080/v1/bookings/1225380/passengers"
},
{
"rel": "itinerary",
"href": "http://localhost:8080/v1/bookings/1225380/itinerary"
},
{
"rel": "self",
"href": "http://localhost:8080/v1/bookings/1225380?lock=true"
},
{
"rel": "clientBookingHistory",
"href": "http://localhost:8080/v1/bookings/1225380/clientBookingHistory/10000"
}
]
推荐答案
您可以为此创建一个Injectable并使用该类而不是有角的http类.在这里,您可以过滤链接,然后使用正确的链接调用http.
You can create an Injectable for this and use this class instead of the angular http class. Here you filter the links and than call http with the right link.
@Injectable()
export class Hypermedia {
constructor(private http: Http) { }
get(links: any[], rel: String, body?: any, options?: RequestOptionsArgs): Observable<Response> {
var link = null;
var request = null;
// Find the right link
links.forEach(function (_link) {
if (_link.rel === rel) {
link = _link
return;
}
});
return this.http.get(link.href);
}
}
提供此注入器并将其添加到您需要的构造器中
Provide this injector and add it to the constructor where you need it
constructor(private hypermedia:Hypermedia)
然后,您可以像通常调用http类一样简单地调用它
Then you can simply call it like you normally would call the http class
this.hypermedia.get(myHypermediaLinksArray,myRel)
希望这会有所帮助:)
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