Java 如何使用 Spring RestTemplate 使用 Page<Entity> 响应
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How to consume Page<Entity> response using Spring RestTemplate
提问by bgalek
I'm using spring data (mongoDb) and I've got my repository:
我正在使用 spring 数据（mongoDb）并且我有我的存储库：
public interface StoriesRepository extends PagingAndSortingRepository<Story, String> {}
Then i have a controller:
然后我有一个控制器：
@RequestMapping(method = RequestMethod.GET)
public ResponseEntity<Page<StoryResponse>> getStories(Pageable pageable) {
Page<StoryResponse> stories = storiesRepository.findAll(pageable).map(StoryResponseMapper::toStoryResponse);
return ResponseEntity.ok(stories);
}
Everything works fine, but I can't consume my endpoint using RestTemplate getForEntity method:
一切正常，但我无法使用 RestTemplate getForEntity 方法使用我的端点：
def entity = restTemplate.getForEntity(getLocalhost("/story"), new TypeReference<Page<StoryResponse>>(){}.class)
What class should I provide to successfully deserialize my Page of entities?
我应该提供什么类来成功反序列化我的实体页面？
采纳答案by Ali Dehghani
new TypeReference<Page<StoryResponse>>() {}
The problem with this statement is that Hymanson cannot instantiate an abstract type. You should give Hymanson the information on how to instantiate Pagewith a concrete type. But its concrete type, PageImpl, has no default constructor or any @JsonCreators for that matter, so you can notuse the following code either:
此语句的问题在于 Hymanson 无法实例化抽象类型。您应该向 Hymanson 提供有关如何Page使用具体类型实例化的信息。但是它的具体类型 ,PageImpl没有默认构造函数或任何@JsonCreators ，因此您也不能使用以下代码：
new TypeReference<PageImpl<StoryResponse>>() {}
Since you can't add the required information to the Pageclass, It's better to create a custom implementation for Pageinterface which has a default no-arg constructor, as in this answer. Then use that custom implementation in type reference, like following:
由于您无法向Page类添加所需的信息，因此最好为Page具有默认无参数构造函数的接口创建自定义实现，如本答案所示。然后在类型引用中使用该自定义实现，如下所示：
new TypeReference<CustomPageImpl<StoryResponse>>() {}
Here are the custom implementation, copied from linked question:
以下是从链接问题复制的自定义实现：
public class CustomPageImpl<T> extends PageImpl<T> {
private static final long serialVersionUID = 1L;
private int number;
private int size;
private int totalPages;
private int numberOfElements;
private long totalElements;
private boolean previousPage;
private boolean firstPage;
private boolean nextPage;
private boolean lastPage;
private List<T> content;
private Sort sort;
public CustomPageImpl() {
super(new ArrayList<>());
}
@Override
public int getNumber() {
return number;
}
public void setNumber(int number) {
this.number = number;
}
@Override
public int getSize() {
return size;
}
public void setSize(int size) {
this.size = size;
}
@Override
public int getTotalPages() {
return totalPages;
}
public void setTotalPages(int totalPages) {
this.totalPages = totalPages;
}
@Override
public int getNumberOfElements() {
return numberOfElements;
}
public void setNumberOfElements(int numberOfElements) {
this.numberOfElements = numberOfElements;
}
@Override
public long getTotalElements() {
return totalElements;
}
public void setTotalElements(long totalElements) {
this.totalElements = totalElements;
}
public boolean isPreviousPage() {
return previousPage;
}
public void setPreviousPage(boolean previousPage) {
this.previousPage = previousPage;
}
public boolean isFirstPage() {
return firstPage;
}
public void setFirstPage(boolean firstPage) {
this.firstPage = firstPage;
}
public boolean isNextPage() {
return nextPage;
}
public void setNextPage(boolean nextPage) {
this.nextPage = nextPage;
}
public boolean isLastPage() {
return lastPage;
}
public void setLastPage(boolean lastPage) {
this.lastPage = lastPage;
}
@Override
public List<T> getContent() {
return content;
}
public void setContent(List<T> content) {
this.content = content;
}
@Override
public Sort getSort() {
return sort;
}
public void setSort(Sort sort) {
this.sort = sort;
}
public Page<T> pageImpl() {
return new PageImpl<>(getContent(), new PageRequest(getNumber(),
getSize(), getSort()), getTotalElements());
}
}
回答by pathfinder
You can probably use exchange method of restTemplate and get the body from it..
您可能可以使用 restTemplate 的交换方法并从中获取正文..
Check the following answer https://stackoverflow.com/a/31947188/3800576. This might help you
检查以下答案https://stackoverflow.com/a/31947188/3800576。这可能会帮助你
回答by jtcotton63
I know this thread is a little old, but hopefully someone will benefit from this.
我知道这个线程有点旧，但希望有人能从中受益。
@Ali Dehghani's answer is good, except that it re-implements what PageImpl<T>has already done. I considered this to be rather needless. I found a better solution by creating a class that extends PageImpl<T>and specifies a @JsonCreatorconstructor:
@Ali Dehghani 的回答很好，只是它重新实现了PageImpl<T>已经完成的工作。我认为这是相当没有必要的。我通过创建一个扩展PageImpl<T>并指定@JsonCreator构造函数的类找到了一个更好的解决方案：
import com.fasterxml.Hymanson.annotation.JsonCreator;
import com.fasterxml.Hymanson.annotation.JsonProperty;
import com.company.model.HelperModel;
import org.springframework.data.domain.PageImpl;
import org.springframework.data.domain.PageRequest;
import java.util.List;
public class HelperPage extends PageImpl<HelperModel> {
@JsonCreator
// Note: I don't need a sort, so I'm not including one here.
// It shouldn't be too hard to add it in tho.
public HelperPage(@JsonProperty("content") List<HelperModel> content,
@JsonProperty("number") int number,
@JsonProperty("size") int size,
@JsonProperty("totalElements") Long totalElements) {
super(content, new PageRequest(number, size), totalElements);
}
}
Then:
然后：
HelperPage page = restTemplate.getForObject(url, HelperPage.class);
This is the same as creating a CustomPageImpl<T>class but allows us to take advantage of all the code that's already in PageImpl<T>.
这与创建CustomPageImpl<T>类相同，但允许我们利用PageImpl<T>.
回答by Vladimir Mitev
As "pathfinder" mentioned you can use exchangemethod of RestTemplate. However instead of passing ParameterizedTypeReference<Page<StoryResponse>>()you should pass ParameterizedTypeReference<PagedResources<StoryResponse>>(). When you get the response you could retrieve the content - Collection<StoryResponse>.
作为“探路者”中提到，您可以使用exchange的方法RestTemplate。但是，ParameterizedTypeReference<Page<StoryResponse>>()您应该通过而不是通过ParameterizedTypeReference<PagedResources<StoryResponse>>()。当您收到响应时，您可以检索内容 - Collection<StoryResponse>。
The code should look like this:
代码应如下所示：
ResponseEntity<PagedResources<StoryResponse>> response = restTemplate.exchange(getLocalhost("/story"),
HttpMethod.GET, null, new ParameterizedTypeReference<PagedResources<StoryResponse>>() {});
PagedResources<StoryResponse> storiesResources = response.getBody();
Collection<StoryResponse> stories = storiesResources.getContent();
Apart from the content storiesResourcesholds page metadata and links too.
除了内容也storiesResources包含页面元数据和链接。
A more step-by-step explanation is available here: https://stackoverflow.com/a/46847429/8805916
此处提供了更多分步说明：https: //stackoverflow.com/a/46847429/8805916
回答by pama
I can only make it work downgrading Spring library to 1.* and not using 2.* I had to create my own code for the Page which does not extends PageImpl
我只能使它工作将 Spring 库降级到 1.* 而不是使用 2.* 我必须为不扩展 PageImpl 的页面创建自己的代码
回答by Terrence Wei
If you looking at this thread, and if you try this answer https://stackoverflow.com/a/44895867/8268335
如果您查看此线程，并且尝试此答案 https://stackoverflow.com/a/44895867/8268335
You will meet the 2nd problem:
你会遇到第二个问题：
Can not construct instance of org.springframework.data.domain.Pageable
Then I find the perfect solution from here: https://stackoverflow.com/a/42002709/8268335
然后我从这里找到了完美的解决方案：https: //stackoverflow.com/a/42002709/8268335
I create the class RestPageImplfrom the answer above and problem solved.
我根据RestPageImpl上面的答案创建了类并解决了问题。
