当然，JAX-RS 应该无需任何额外的框架即可工作，就像 JAX-WS 一样。
但在 JAX-WS 中，我只是在 POJO 上添加一些注释，用一行代码创建一个自托管服务，仅此而已。
我找不到任何教程或资源来展示如何使用 JAX-RS 执行相同的操作。几乎每个教程都使用 Jersey（或 Easyrest 等）并且至少使用 Maven。是否没有一种简单的方法来设置基于 Rest 的服务，就像使用 JAX-WS 一样？
谢谢
编辑：嗯，我认为如果没有这些框架，@Path 等注释就无法使用？我的日食无法引用/找到它们:(
Of course JAX-RS should work without any additional framework like JAX-WS also does.
But in JAX-WS, I just put some annotations on a POJO, create a self hosted service with one single line of code and that's it.
I can't find any tutorials or resources that show how to do the same with JAX-RS. Nearly every tutorial uses Jersey (or Easyrest etc.) and at least Maven. Isn't there an easy way to set up a Rest based service like it can be done with JAX-WS?
Thank you
edit: Hm, I think annotations like @Path etc. aren't available without these frameworks? My eclipse cannot reference/find them :(
发布评论
评论（2）
JAX-RS 是一个 API。 Jersey 和 RESTEasy 是它的实现。 Jersey 是 JAX-RS 的“参考”实现，因此经常出现在教程中（毕竟，需要一些 API 实现才能使用它）。
http://en.wikipedia.org/wiki/Java_API_for_RESTful_Web_Services
JAX-RS is an API. Jersey and RESTEasy are its implementations. Jersey, is the 'reference' implementation of JAX-RS, and therefore shows up frequently in tutorials (after all, one needs some implementation of the API to use it).
http://en.wikipedia.org/wiki/Java_API_for_RESTful_Web_Services
可以轻松访问宁静的网络服务。您甚至可以使用：
URL url = new URL(docUrl);HttpsURLConnection connection = (HttpsURLConnection) url.openConnection();
您可以使用该连接，就像在其他地方使用它一样。
Restful webservices can be easily accessed. You can even use:
URL url = new URL(docUrl);HttpsURLConnection connection = (HttpsURLConnection) url.openConnection();
you can use the connection just like using it anywhere else.