JAX-RS参数转换器-使用它的编译时错误

Question

This is my param converter

import org.springframework.data.domain.Sort;

public class MyParamConverter implements ParamConverter<Sort> {
    @Override
    public Sort fromString(String s){
        return new Sort(new Sort.Order(Sort.Direction.ASC, "ds"));
    }

    @Override
    public String toString(Sort mo){
        return mo.toString();
    }

}

this is my paramconverter provider

@Provider
public class MyParamConverterProvider implements ParamConverterProvider {

@Override
public <T> ParamConverter<T> getConverter(Class<T> rawType, Type genericType, Annotation[] annotations) {
    if(rawType.equals(Sort.class)){
        return (ParamConverter<T>) new MyParamConverter();
    }
    return null;
}

I am trying to use in my API as

@GET
@Path("/")
Response read(@QueryParam("sort") Sort order);

I am expecting the jax to map string that I pass in my url eg &sort="asc" to Sort object. But I am getting an compile time error that have a registered implementation of paramconverter provider . I need to find a way when I pass a query param as &sort="somethung" it gets convert to SORT automatically either by using custom annotation or by using Param Converter .

Answer 1

with reference to your comment, try registering your provider like:

@ApplicationPath("/")
public class MyApplication extends ResourceConfig {

    @Override
    public Set<Class<?>> getClasses() {
        final Set<Class<?>> classes = new HashSet<Class<?>>();
        classes.add(MyParamConverterProvider.class);

        return classes;
    }
}

or, if you are using Jersey

@ApplicationPath("/")
public class MyApplication extends ResourceConfig {

    public MyApplication() {
        packages("my.package");

        // or without package scanning
        register(MyParamConverterProvider.class);
    }
}