POST complex parameters to REST service what would request URL and Body look like

问题

So I am standing up a rest service. It is going to request a post to a resource that is pretty complex. The actual backend requires multiple (19!) parameters. One of which is a byte array. It sounds like this is going to need to be serialized by the client and sent to my service.

I am trying to understand how I can right a method that will handle this. I am thinking something like this

@POST
@Path("apath")
@Consumes(MediaType.APPLICATION_JSON, MediaType.TEXT_HTML)
public Response randomAPI(@Parameter apiID, WhatParamWouldIPutHere confused){
}

What parameter types would I do to capture that inbound (Serialized) Post data. And what would the client URI look like?

回答1:

In order to get all the parameters of the request you can use @Context UriInfoas parameter of your randomAPI method.

Then use UriInfo#getQueryParameters() to get the full MultivaluedMap of parameters.

if you wish to convert the MultivaluedMap to a simple HashMap I added the code for that too.

so your method will basically look like something like this:

@POST
@Path("apath")
@Consumes(MediaType.APPLICATION_JSON, MediaType.TEXT_HTML)
public Response randomAPI(@Context UriInfo uriInfo){
    Map params= (HashMap) convertMultiToHashMap(uriInfo.getQueryParameters());
    return service.doWork(params);
}

public Map<String, String> convertMultiToHashMap(MultivaluedMap<String, String> m) {
        Map<String, String> map = new HashMap<String, String>();
        for (Map.Entry<String, List<String>> entry : m.entrySet()) {
            StringBuilder sb = new StringBuilder();
            for (String s : entry.getValue()) {
                sb.append(s);
            }
            map.put(entry.getKey(), sb.toString());
        }
        return map;
    }

Additional Info :

The @Context annotation allows you to inject instances of javax.ws.rs.core.HttpHeaders, javax.ws.rs.core.UriInfo, javax.ws.rs.core.Request, javax.servlet.HttpServletRequest, javax.servlet.HttpServletResponse, javax.servlet.ServletConfig, javax.servlet.ServletContext, and javax.ws.rs.core.SecurityContext objects.

回答2:

What I am thinking is you can simply use the httpservlet request and all the parameters can be retrieved as below

@RequestMapping(value = "/yourMapping", method = RequestMethod.POST)
public @ResponseBody String yourMethod(HttpServletRequest request) {
          Map<String, String[]> params = request.getParameterMap();
          //Loop through the parameter maps and get all the paramters one by one including byte array
          for(String key : params){
            if(key == "file"){  //This param is byte array/ file data, specially handle it
            byte[] content = params.get(key);
             //Do what ever you want with the byte array

            else if(key == "any of your special params") {
             //handle
            }

            else {
            }

          }
}

http://docs.oracle.com/cd/E17802_01/products/products/servlet/2.3/javadoc/javax/servlet/ServletRequest.html#getParameterMap%28%29

来源：https://stackoverflow.com/questions/38401820/post-complex-parameters-to-rest-service-what-would-request-url-and-body-look-lik