Jersey, @QueryParam List

Question

I want to take a list as a parameter, but the value of the list does not correspond to java version list. For example @QueryParam("orderBy") List orderBy) This list has the value of [age, name] as an element.

Is it possible to have the value separately. Like 1st element age and 2nd element name.

`

@GET
@Path("/range")
@Produces({MediaType.APPLICATION_JSON,MediaType.APPLICATION_XML})
public Response searchForActivitiesWithRange(@QueryParam(value = "description")List<String> descriptions,
                                             @QueryParam(value = "durationFrom") int durationFrom,
                                             @QueryParam(value = "durationTo") int durationTo) {


    List<Activity> activities = activityRepository.findByDescriptionandDuration(descriptions,durationFrom,durationTo);

    if (activities == null || activities.size() == 0) {
        return Response.status(Status.NOT_FOUND).build();
    }

    return Response.ok().entity(new GenericEntity<List<Activity>>(activities){}).build();
}`

zuno · Accepted Answer · 2015-06-21T13:40:16.223

3

Not automatically. Every query param is read by jersey as a String and, after, it tries to convert in the data type of the parameter you specified. You have to split the param String or,client side, send multiple param with the same name and read them like a MultivaluedMap, in which the key is the param name and the value a list with all values.

edited Jun 21 '15 at 13:40

answered Jun 21 '15 at 13:26

zuno

563
4
19

QueryParam documentation says we can use List but I still get the param as a String instead of the List. https://docs.oracle.com/javaee/7/api/javax/ws/rs/QueryParam.html – ASR4 May 24 '22 at 23:35

score 0 · Answer 2 · answered Jan 15 '16 at 20:20

0

Use objects. You can create an ActivitiesRequest class and then define your variables in the class. You can also do that for your response.

answered Jan 15 '16 at 20:20

kyle

568
2
10
26

Jersey, @QueryParam List

2 Answers2

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