2016-07-28 66 views
0

我使用Spring 4在Tomcat 8 JDK 8如何传入URI请求定制RequestObject绑定在Spring MVC 4

控制器:

@RestController 
public class MyController { 
    @RequestMapping(value = "/abc/{id}", method = RequestMethod.GET) 
    public ResponseEntity<MyResponse> myMethod(@PathVariable("id") String id, 
      @RequestParam(value = "add", required = false, defaultValue = "1") String add, 
      @RequestParam(value = "sub", required = false, defaultValue = "1") String sub, 
      @RequestParam(value = "mul", required = false, defaultValue = "1") String mul, 
      @RequestParam(value = "div", required = false, defaultValue = "1") String div) { 

     MyRequest request = new MyRequest(); 
     request.setId(id); 
     request.setAdd(add); 
     request.setSub(sub); 
     request.setMul(mul); 
     request.setDiv(div); 

     // call service implementation.. 
     service.myMethodImpl(request); 
    } 

所以基本上,我的URI看起来像这样: /ABC/100 =添加10 &分= 10 & MUL = 10 & DIV = 10

是否有任何绑定/ API /用SpringMVC中可获得的注释,所以我不必MyRequest手动绑定?

+0

可能重复的自动绑定这些值发送的请求参数[Spring MVC:复杂对象作为GET @RequestParam](http://stackoverflow.com/questions/16942193/spring-mvc-complex-object-as-get-requestparam) –

+0

@ dustin.schultz谢谢。我应该删除我的帖子,如果它是重复的? –

回答

0

没有,

处理请求,我们需要唯一可以做的事情是将它们都作为一个POJO这些参数,然后使用POST方法那么春天将