java - Spring Boot Data JPA Paging 以正确的方式实现

Question

我想在 Spring Boot + Spring Data JPA 应用程序中添加分页功能，我有 Repository :

@Repository
public interface PostRepository extends JpaRepository<Post, Long> {
    List<Post> findAllByOrderByPostIdDesc(Pageable pageable);
}

在 Service 类中我有:

@Service
public class PostService {

    @Autowired
    private PostRepository postRepository;

    public List<Post> findAll(Pageable pageable) {
        return postRepository.findAllByOrderByPostIdDesc(pageable);
    }
}

最后Controller是

@Controller
public class HomeController {

    @Autowired
    private PostService postService;

    @GetMapping(value = "/")
    public String getHomePage(Model model, Pageable pageable){
        model.addAttribute("postList", postService.findAll(pageable));
        return "home";
    }
}

上面在http://localhost:8080映射上的代码我得到所有记录作为响应，但是如果我传递page和size params as http://localhost:8080/?page=0&size=1 我可以进行分页工作。

现在我想将 http://localhost:8080/?page=0&size=1 设置为默认登陆页面地址，并且在 HTML 代码中如何动态更新参数 的值页面和大小是:

 <ul class="pagination justify-content-center mb-4">
            <li class="page-item">
              <a class="page-link" th:href="@{/(page=0,size=1)}">&larr; Older</a>
            </li>
            <li class="page-item disabled">
              <a class="page-link" th:href="@{/(page=0,size=1)}">Newer &rarr;</a>
            </li>
          </ul>

我尝试了很多技巧，但没有一个对我有用!

更新

我尝试按照建议添加此类，但是@Nyle Hassan，

@Configuration
public class DefaultView extends WebMvcConfigurerAdapter {

    @Override
    public void addViewControllers( ViewControllerRegistry registry ) {
        registry.addViewController( "/" ).setViewName( "forward:/?page=0&size=10" );
        registry.setOrder( Ordered.HIGHEST_PRECEDENCE );
        super.addViewControllers( registry );
    }
}

结果是:

2019-01-01 17:26:43.484 ERROR 3824 --- [nio-8080-exec-1] o.a.c.c.C.[.[.[/].[dispatcherServlet]    : Servlet.service() for servlet [dispatcherServlet] threw exception

java.lang.StackOverflowError: null
    at javax.servlet.http.HttpServletRequestWrapper.getSession(HttpServletRequestWrapper.java:231) ~[tomcat-embed-core-8.5.29.jar:8.5.29]
    at org.apache.catalina.core.ApplicationHttpRequest.getSession(ApplicationHttpRequest.java:597) ~[tomcat-embed-core-8.5.29.jar:8.5.29]
    at javax.servlet.http.HttpServletRequestWrapper.getSession(HttpServletRequestWrapper.java:231) ~[tomcat-embed-core-8.5.29.jar:8.5.29]
    at org.apache.catalina.core.ApplicationHttpRequest.getSession(ApplicationHttpRequest.java:597) ~[tomcat-embed-core-8.5.29.jar:8.5.29]
    at javax.servlet.http.HttpServletRequestWrapper.getSession(HttpServletRequestWrapper.java:231) ~[tomcat-embed-core-8.5.29.jar:8.5.29]
    at org.apache.catalina.core.ApplicationHttpRequest.getSession(ApplicationHttpRequest.java:597) ~[tomcat-embed-core-8.5.29.jar:8.5.29]
    at javax.servlet.http.HttpServletRequestWrapper.getSession(HttpServletRequestWrapper.java:231) ~[tomcat-embed-core-8.5.29.jar:8.5.29]
    at org.apache.catalina.core.ApplicationHttpRequest.getSession(ApplicationHttpRequest.java:597) ~[tomcat-embed-core-8.5.29.jar:8.5.29]
    at javax.servlet.http.HttpServletRequestWrapper.getSession(HttpServletRequestWrapper.java:231) ~[tomcat-embed-core-8.5.29.jar:8.5.29]
    at org.apache.catalina.core.ApplicationHttpRequest.getSession(ApplicationHttpRequest.java:597) ~[tomcat-embed-core-8.5.29.jar:8.5.29]

Answer 1

the code above on http://localhost:8080 mapping I get all the records as response

你确定吗？您使用的是哪个版本的 Spring Boot？我检查了最新版本(2.1.1)并调用 http://localhost:8080没有任何查询参数将实例化一个 Pageable 对象，其中 page = 0，size = 20，不排序。只需尝试将调试点设置为 org.springframework.data.web.PageableHandlerMethodArgumentResolver#resolveArgument了解 Spring 如何执行此绑定(bind)。

此参数解析器还附带一个注释，允许您更改在查询参数不存在时使用的默认值。您可以像这样使用它:

@GetMapping(value = "/")
public String getHomePage(Model model, @PageableDefault(page=0, size = 1) Pageable pageable) {
    ...
}

第 0 页是默认注释，因此具有相同的效果:

@GetMapping(value = "/")
public String getHomePage(Model model, @PageableDefault(size = 1) Pageable pageable) {
    ...
}

in HTML code how can I dynamically update values of params page and size

下次，请发布两个单独的问题，但这里是:

首先，更改List<Post>至Page<Post>在您的存储库和服务中。在您的 Controller 中，将结果绑定(bind)到模型:

Page<Post> page = postService.findAll(pageable);
model.addAttribute("page", page); // page.getContent() contains the List<Page> now

并在 View 中使用page呈现分页链接，例如像这样:

<li class="page-item">
    <a class="page-link" th:if="${page.hasPrevious()}" th:href="@{/(page=${page.previousPageable().getPageNumber()},size=${page.getSize()})}">&larr; Older</a>
</li>
<li class="page-item disabled">
    <a class="page-link" th:if="${page.hasNext()}" th:href="@{/(page=${page.nextPageable().getPageNumber()},size=${page.getSize()})}">Newer &rarr;</a>
</li>