'Spring RestTemplate and generic types ParameterizedTypeReference collections like List<T>

An Abstract controller class requires List of objects from REST. While using Spring RestTemplate its not mapping it to required class instead it returns Linked HashMAp

 public List<T> restFindAll() {

    RestTemplate restTemplate = RestClient.build().restTemplate();
    ParameterizedTypeReference<List<T>>  parameterizedTypeReference = new ParameterizedTypeReference<List<T>>(){};
    String uri= BASE_URI +"/"+ getPath();

    ResponseEntity<List<T>> exchange = restTemplate.exchange(uri, HttpMethod.GET, null,parameterizedTypeReference);
    List<T> entities = exchange.getBody();
    // here entities are List<LinkedHashMap>
    return entities;

}

If I use,

ParameterizedTypeReference<List<AttributeInfo>>  parameterizedTypeReference = 
    new ParameterizedTypeReference<List<AttributeInfo>>(){};
    ResponseEntity<List<AttributeInfo>> exchange =
  restTemplate.exchange(uri, HttpMethod.GET, null,parameterizedTypeReference);

It works fine. But can not put in all subclasses, any other solution.

javaresttemplatespring-restspring-web


Solution 1:[1]

I worked around this using the following generic method:

public <T> List<T> exchangeAsList(String uri, ParameterizedTypeReference<List<T>> responseType) {
    return restTemplate.exchange(uri, HttpMethod.GET, null, responseType).getBody();
}

Then I could call:

List<MyDto> dtoList = this.exchangeAsList("http://my/url", new ParameterizedTypeReference<List<MyDto>>() {});

This did burden my callers with having to specify the ParameterizedTypeReference when calling, but meant that I did not have to keep a static mapping of types like in vels4j's answer 

Solution 2:[2]

Using ParameterizedTypeReference for a List<Domain>, when Domain is an explicit class, that ParameterizedTypeReference works well, like this:

@Override
public List<Person> listAll() throws Exception {
    ResponseEntity<List<E>> response = restTemplate.exchange("http://example.com/person/", HttpMethod.GET, null,
            new ParameterizedTypeReference<List<Person>>() {});
    return response.getBody();
}

However, if a method listAll is used in generic flavor, that list should be parameterized itself. The best way I found for this is:

public abstract class WebServiceImpl<E> implements BaseService<E> {

    private Class<E> entityClass;

    @SuppressWarnings("unchecked")
    public WebServiceImpl() {
        this.entityClass = (Class<E>) ((ParameterizedType) getClass().getGenericSuperclass())
            .getActualTypeArguments()[0];
    }


    @Override
    public List<E> listAll() throws Exception {
        ResponseEntity<List<E>> response =  restTemplate.exchange("http://example.com/person/", HttpMethod.GET, null,
                new ParameterizedTypeReference<List<E>>() {
                    @Override
                    public Type getType() {
                        Type type = super.getType();
                        if (type instanceof ParameterizedType) {
                            Type[] responseWrapperActualTypes = { entityClass };
                            ParameterizedType responseWrapperType = new ParameterizedTypeImpl(List.class,
                                    responseWrapperActualTypes, null);
                            return responseWrapperType;
                        }
                        return type;
                    }
                });
        return response.getBody();
    }
}

Solution 3:[3]

Couldnt find a solution from Spring, hence I have done it with ParameterizedTypeReference in HashMap like 

 public final static HashMap<Class,ParameterizedTypeReference> paramTypeRefMap = new HashMap() ;
 static {
    paramTypeRefMap.put(AttributeDefinition.class, new ParameterizedTypeReference<List<AttributeDefinition>>(){} );
    paramTypeRefMap.put(AttributeInfo.class, new ParameterizedTypeReference<List<AttributeInfo>>(){} );
 }

and used it 

ParameterizedTypeReference parameterizedTypeReference = paramTypeRefMap.get(requiredClass);
ResponseEntity<List> exchange = restTemplate.exchange(uri, HttpMethod.POST, entity, parameterizedTypeReference);

Solution 4:[4]

I did this a bit different. In my situation, I had a base class where I was implementing a set of CRUD operations and then using derived classes to implement specific resource types.

In the base class, I was trying to define a ParameterizedTypeReference as follows:

ParameterizedTypeReference<ServicePagedResult<R>> typeRef = 
  new ParameterizedTypeReference<ServicePagedResult<R>>() {};

This didn't work so I ended up creating an abstract method in the base class:

protected abstract ParameterizedTypeReference<ServicePagedResult<R>> 
getServicePagedResultTypeRef();

and then in the derived classes:

@Override
protected ParameterizedTypeReference<ServicePagedResult<AccountResource>>
getServicePagedResultTypeRef() {
  return new ParameterizedTypeReference<ServicePagedResult<AccountResource>>() {};
}

I could then use that in the base class like:

ResponseEntity<ServicePagedResult<R>> response = lbRestTemplate.exchange(
  uri, HttpMethod.GET, null, getServicePagedResultTypeRef(), uriVariables);

Solution 5:[5]

The easiest solution for me is to define an object MyOperationResult containing the list you expect as field and use restTemplate.getForObject to get this result.

Solution 6:[6]

In case someone need a Kotlin solution, you can do:

val responseType = object : ParameterizedTypeReference<Map<String, Any?>>() {}         
val request = HttpEntity<Any?>(data)
val response = restTemplate.exchange(url, HttpMethod.POST, request, responseType)
val responseMap = response?.body as Map<String, Any>

Solution 7:[7]

You can use TypeUtil from apache's commons-lang3 to build the generic Type to pass to exchange:

 public List<T> restFindAll() {

    RestTemplate restTemplate = RestClient.build().restTemplate();
    String uri= BASE_URI +"/"+ getPath();

    ResponseEntity<List<T>> exchange = restTemplate.exchange(uri, HttpMethod.GET, null, ParameterizedTypeReference.forType(TypeUtils.parameterize(List.class, clazz)));
    List<T> entities = exchange.getBody();
    return entities;

}

Sources

This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.

Source: Stack Overflow

Solution Source
Solution 1 Marcin
Solution 2 Moesio
Solution 3 vels4j
Solution 4 Ken Joyner
Solution 5 JRA_TLL
Solution 6 ELavicount
Solution 7

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