json - Spring MVC 3 : return a Spring-Data Page as JSON - issue with PagedResourcesAssembler

Question

所以，我对 spring 和 java 总体来说还很陌生

我尝试做的是在同一个渲染 View 上让表单发布数据以过滤表单下显示的结果列表。

我有一个简单的域类，如下所示:

@Entity
@Table(name = "SEC_PERSON")
public class Person {

@Id
@GeneratedValue(strategy = GenerationType.SEQUENCE, generator = "SEQ_SEC_PERSON")
@SequenceGenerator(name = "SEQ_SEC_PERSON", sequenceName = "SEQ_SEC_PERSON")
@Column(name = "ID")
private Long Id;

@Column(name = "CODE", nullable = false)
private String code;

@Column(name = "FIRSTNAME", nullable = false)
private String firstname;

@Column(name = "SURNAME", nullable = false)
private String surname;

@Column(name = "CREATIONDATE")
private DateTime creationDate;

//getters and setters

DTO，因为我希望我的域与演示文稿分离

public class PersonDTO {
private Long id;

@NotEmpty
private String code;

@NotEmpty
private String firstname;

@NotEmpty
private String surname;

private DateTime creationDate;

public PersonDTO() {
}
//getters and setters

扩展 Jpa 和 QueryDsl 的存储库

public interface PersonRepository extends JpaRepository<Person, Long>, QueryDslPredicateExecutor<Person> {}

用于我的搜索的数据访问类是 null 安全的(感谢 guava)及其等效的非 null 安全

人物标准:

public class PersonCriteria {

private String code;
private String surname;
private String firstname;
private LocalDate creationDateFrom;
private LocalDate creationDateTo;

//getters and setters
}

空安全版本

public class NullSafePersonCriteria {

private final PersonCriteria personCriteria;

public NullSafePersonCriteria(final PersonCriteria personCriteria) {
    checkArgument(personCriteria != null);
    this.personCriteria = personCriteria;
}

public Optional<String> getCode() {
    return Optional.fromNullable(this.personCriteria.getCode());
}

public Optional<String> getSurname() {
    return Optional.fromNullable(this.personCriteria.getSurname());
}

public Optional<String> getFirstname() {
    return Optional.fromNullable(this.personCriteria.getFirstname());
}

public Optional<LocalDate> getCreationDateFrom() {
    return Optional.fromNullable(this.personCriteria.getCreationDateFrom());
}

public Optional<LocalDate> getCreationDateTo() {
    return Optional.fromNullable(this.personCriteria.getCreationDateTo());
}

我的搜索谓词

public class PersonPredicates {
public static Predicate PersonLitstQuery(final PersonCriteria personCriteria) {
    final QPerson person = QPerson.person;
    final NullSafePersonCriteria nsPersonCriteria = new NullSafePersonCriteria(personCriteria);

    BooleanExpression criteria = QPerson.person.isNotNull();
    if (nsPersonCriteria.getCode().isPresent()) {
        criteria = criteria.and(person.code.matches(nsPersonCriteria.getCode().get()));
    }

    if (nsPersonCriteria.getSurname().isPresent()) {
        criteria.and(person.surname.startsWithIgnoreCase(nsPersonCriteria.getSurname().get()));
    }

    if (nsPersonCriteria.getFirstname().isPresent()) {
        criteria.and(person.firstname.startsWithIgnoreCase(nsPersonCriteria.getFirstname().get()));
    }

    if ((nsPersonCriteria.getCreationDateFrom().isPresent()) && (nsPersonCriteria.getCreationDateTo().isPresent())) {
        criteria.and(person.creationDate.between(nsPersonCriteria.getCreationDateFrom().get().toDateTimeAtStartOfDay(),
                nsPersonCriteria.getCreationDateTo().get().toDateTimeAtStartOfDay()));
    }

    return criteria;
}

我的服务实现如下:

@Service
public class PersonServiceImpl implements PersonService{
@Transactional(readOnly = true)
@Override
public Page<PersonDTO> search(final PersonCriteria criteria, final int pageIndex) {
    LOGGER.debug("Searching person with set of criterias");

    return new PersonPage(this.mapper.map(Lists.newArrayList(this.personRepository.findAll(PersonLitstQuery(criteria))),
            PersonDTO.class), constructPageSpecification(pageIndex), count(criteria));
}

我使用的映射器只是扩展了一点DozerMapper:

public class DozerMapper{
private final org.dozer.Mapper mapper;

public DozerMapper(final org.dozer.Mapper mapper) {
    this.mapper = mapper;
}

@Override
public <T> T map(final Object source, final Class<T> destinationClass) {
    return this.mapper.map(source, destinationClass);
}

@Override
public <T> List<T> map(final List<?> sources, final Class<T> destinationClass) {
    final List<T> result = Lists.newArrayList();
    for (final Object source : sources) {
        result.add(map(source, destinationClass));
    }
    return result;
}

现在，上述所有内容都有效，经过单元测试并返回我想要的结果。我的问题是 Controller 和 View ......

我仔细阅读了Oliver对这个问题的回答:Spring MVC 3: return a Spring-Data Page as JSON

尽管由于某种原因我无法让它工作。我已将以下依赖项添加到我的项目中以使用 HATEOAS 和 Spring-data-commons:

<dependency>
    <groupId>org.springframework.hateoas</groupId>
    <artifactId>spring-hateoas</artifactId>
    <version>0.7.0.RELEASE</version>
</dependency>
<dependency>
    <groupId>org.springframework.data</groupId>
<artifactId>spring-data-commons</artifactId>
<version>1.6.0.RELEASE</version>
</dependency>

我的 Controller 看起来像这样:

@Controller
@SessionAttributes("person")
public class PersonController

@RequestMapping(value = REQUEST_MAPPING_LIST, method = RequestMethod.GET)
public HttpEntity<PagedResources> persons(final Model model, @ModelAttribute final PersonCriteria searchCriteria,
        final Pageable pageable, final PagedResourcesAssembler assembler) {
    model.addAttribute(MODEL_ATTRIBUTE_SEARCHCRITERIA, searchCriteria);
    final Page<PersonDTO> persons = this.personService.search(searchCriteria, searchCriteria.getPageIndex());
    return new ResponseEntity<>(assembler.toResource(persons), HttpStatus.OK);
}

和我的jsp:

<html>
<head>
    <title>testing</title>
    <script src="jslinks for jqGrid and jquery" type="text/javascript"></script>
</head>
<body>
<form:form action="person" commandName="searchCriteria" method="POST">
    <div>
        <form:label path="code">Code: </form:label>
        <form:input path="code" type="text"/>
        <form:label path="surname">Surname: </form:label>
        <form:input path="surname" type="text"/>
        <form:label path="firstname">Firstname: </form:label>
        <form:input path="firstname" type="text"/>
        <form:label path="creationDateFrom">Creation Date From: </form:label>
        <smj:datepicker id="creationDateFrom" name="CreationDateFrom" />
        <form:label path="creationDateTo">Creation Date To: </form:label>
        <smj:datepicker id="creationDateTo" name="CreationDateTo" />
    </div>
    <div>
        <input type="submit" value="search"/>
    </div>
</form:form>

    <smjg:grid 
        gridModel="gridModel" 
        id="persons" 
        datatype="\"json\""
        url="\'person\'"
        jsonReader="{root:\"content\", repeatitems: false, records: \"numberOfElements\", total: \"totalPages\"}">
        <smjg:gridColumn name="code" />
        <smjg:gridColumn name="surname" align="left"/>
        <smjg:gridColumn name="firstname" align="left"/>
    </smjg:grid>
</body>
</html>

说明:smj 和 smjg 标签是我目前正在研究的标签库，它们将 jquery 链接到 spring mvc。例如: smjg:grid 将创建标签和调用 jqgrid 函数的 javascript。

与 Olivier 在这篇文章中的回答的第一个区别 Spring MVC 3: return a Spring-Data Page as JSON是如果我在 HttpEntity 中推断 PersonDTO 那么我会收到以下编译错误:

Type mismatch: cannot convert from ResponseEntity<PagedResources> to HttpEntity<PagedResources<PersonDTO>>

第二个区别是我似乎应该将我的 PersonDTO 推断到 PagedResourcesAssembler 中，对吗？

当我直接调用 url localhost:8081/app/person 时的结果我收到 http 500 错误:

org.springframework.http.converter.HttpMessageNotWritableException: Could not marshal [PagedResource { content: [Resource { content: com.app.admin.service.PersonDTO@60a349d0[id=2050,code=TEST2,firstname=ChadsdaTest,surname=Francois,creationDate=<null>], links: [] }, Resource { content: com.app.admin.service.PersonDTO@48462da5[id=5050,code=TESTNEW,firstname=Francois,surname=asdasdx,creationDate=<null>], links: [] }, Resource { content: com.app.admin.crediadmin.service.PersonDTO@5458c9fc[id=51,code=TEST,firstname=Francois,surname=asdawdsx,creationDate=<null>], links: [] }, Resource { content: com.app.admin.service.PersonDTO@de47c70[id=2051,code=TEST3,firstname=Chaqweqasdamsh,surname=Frasda,creationDate=<null>], links: [] }, Resource { content: com.app.admin.service.PersonDTO@7bd2085d[id=3053,code=TEST7,firstname=Francois,surname=Cadsdsx,creationDate=<null>], links: [] }, Resource { content: com.app.admin.service.PersonDTO@14676697[id=3050,code=TESTER,firstname=Francois,surname=CasdadsixChaix,creationDate=<null>], links: [] }, Resource { content: com.app.admin.service.PersonDTO@109de504[id=3051,code=TESTER3,firstname=FrancoisF,surname=Chtest,creationDate=<null>], links: [] }], metadata: Metadata { number: 0, total pages: 2, total elements: 7, size: 5 }, links: [<http://localhost:8081/app/person?page=1&size=5&sort=surname,asc>;rel="next"] }]: null; nested exception is javax.xml.bind.MarshalException
- with linked exception:
[com.sun.istack.SAXException2: unable to marshal type  "org.springframework.hateoas.Resource" as an element because it is not known to this context.]
org.springframework.http.converter.xml.Jaxb2RootElementHttpMessageConverter.writeToResult(Jaxb2RootElementHttpMessageConverter.java:99)
org.springframework.http.converter.xml.AbstractXmlHttpMessageConverter.writeInternal(AbstractXmlHttpMessageConverter.java:66)
org.springframework.http.converter.AbstractHttpMessageConverter.write(AbstractHttpMessageConverter.java:179)
org.springframework.web.servlet.mvc.method.annotation.AbstractMessageConverterMethodProcessor.writeWithMessageConverters(AbstractMessageConverterMethodProcessor.java:148)
org.springframework.web.servlet.mvc.method.annotation.HttpEntityMethodProcessor.handleReturnValue(HttpEntityMethodProcessor.java:124)
org.springframework.web.method.support.HandlerMethodReturnValueHandlerComposite.handleReturnValue(HandlerMethodReturnValueHandlerComposite.java:69)
org.springframework.web.servlet.mvc.method.annotation.ServletInvocableHandlerMethod.invokeAndHandle(ServletInvocableHandlerMethod.java:122)

根本原因:

javax.xml.bind.MarshalException
- with linked exception: [com.sun.istack.SAXException2: unable to marshal type "org.springframework.hateoas.Resource" as an element because it is not known to this context.]
com.sun.xml.bind.v2.runtime.MarshallerImpl.write(MarshallerImpl.java:318)
com.sun.xml.bind.v2.runtime.MarshallerImpl.marshal(MarshallerImpl.java:244)

我不确定我在这里做错了什么。

尽管如果我使用 .json 调用相同的 url，那么我会得到 json 输出，这看起来很奇怪，因为我仍然不生成 json。

Answer 1

您现在可能已经解决了这个问题，但既然我已经完成了这项工作，我想我应该为类似船上的其他人至少添加一个解决方案来解决您的问题。

Type mismatch: cannot convert from ResponseEntity<PagedResources> to HttpEntity<PagedResources<PersonDTO>> :

要解决此问题，请向返回类型添加一个附加类型参数:

@RequestMapping(value = REQUEST_MAPPING_LIST, method = RequestMethod.GET)
public HttpEntity<PagedResources<PersonDTO>> persons(final Model model, @ModelAttribute final PersonCriteria searchCriteria,
        final Pageable pageable, final PagedResourcesAssembler assembler) {
    ...
}

<小时/>

com.sun.istack.SAXException2: unable to marshal type "org.springframework.hateoas.Resource" as an element because it is not known to this context.]

看起来 Spring 正在尝试生成 XML，我认为如果它在类路径上找到 JAXB 实现，它会默认执行此操作。如果您不需要从此方法生成 XML，则可以添加 produces = {MediaType.APPLICATION_JSON_VALUE}到它的@RequestMapping。