RestTemplate中使用ParameterizedTypeReference参数化类型支持泛型

使用普通的RestTemplate.postForObject是无法知道泛型的具体类型的，需要借助ParameterizedTypeReference参数化类型

public <T> T post(String url, @NotNull HttpEntity requestEntity, ParameterizedTypeReference<T> parameterizedTypeReference) {
    ResponseEntity<T> responseEntity = restTemplate.exchange(
            url,
            HttpMethod.POST,
            requestEntity,
            parameterizedTypeReference
    );
    return responseEntity.getBody();
}

ParameterizedTypeReference<TestModelType<List<TestModelBody>>> parameterizedTypeReference = 
    new ParameterizedTypeReference<TestModelType<List<TestModelBody>>>() {};
TestModelType<List<TestModelBody>> listTestModelType = 
    restTemplateUtils.get(testUrl, parameterizedTypeReference);

FastJson反序列化也支持泛型：

import com.alibaba.fastjson.JSONObject;
import com.alibaba.fastjson.TypeReference;

String jsonString = jsonStringEntity.getBody();
T body = JSONObject.parseObject(jsonString, clazz); // 直接传Class不能确定泛型
JSONObject.parseObject(
    jsonString, 
    new TypeReference<TestModelType<List<TestModelBody>>>(){}
);

更多示例：

HttpEntity<ChannelsPayRequest> requestEntity = new HttpEntity<>(request, getHttpHeaders());
ParameterizedTypeReference<Result<ChannelsPayResponse>> parameterizedTypeReference =
        new ParameterizedTypeReference<Result<ChannelsPayResponse>>() {};
ResponseEntity<Result<ChannelsPayResponse>> responseEntity = restTemplate.exchange(String.format("http://%s/channels/pay", channelService.getServiceAppName()),
        HttpMethod.POST, requestEntity, parameterizedTypeReference);
Result<ChannelsPayResponse> payResult = responseEntity.getBody();
log.info(JSON.toJSONString(payResult));