JPA 动态条件-api 查询

Question

我想知道是否有一种通用方法可以将标准 api 与更复杂的模型结合使用？

我有一个与其他实体具有一对一关系的实体 class。我的服务包装器通过条件 api 进行数据库查询，从前端获取参数以计算分页、排序和过滤。

实体

@Entity
public class Person implements Serializable {
    @Id
    private Long id;
    private String name;
    private String givenName;

    @Temporal(TemporalType.DATE)
    private Date birthdate;

    @OneToOne(cascade = CascadeType.ALL)
    @JoinColumn(name = "INFORMATION_ID")
    private Information information;

    @OneToOne(cascade = CascadeType.ALL)
    @JoinColumn(name = "ADDRESS_ID")
    private Address address;
    ...
}

@Entity
public class Information implements Serializable {
    @Id
    private Long id;
    private String detail;
    ...
}

@Entity
public class Address implements Serializable {
    @Id
    private Long id;
    private String street;
    private String city;
    ...
}

服务

@Stateless
public class PersonService {
    @PersistenceContext(unitName = "ProblemGenericDatatableFilterPU")
    private EntityManager em;
    public List<Person> findAllPersons222(int first, int pageSize, String sortField, SortOrder sortOrder, Map<String, Object> filters) {
        CriteriaBuilder builder = em.getCriteriaBuilder();
        CriteriaQuery<Person> criteriaQuery = builder.createQuery(Person.class);
        Root<Person> rootPerson = criteriaQuery.from(Person.class);
        Join<Person, Information> joinPersonInformation = rootPerson.join(Person_.information);
        Join<Person, Address> joinPersonAddress = rootPerson.join(Person_.address);

        // select
        criteriaQuery.select(rootPerson);

        // filter
        List<Predicate> allPredicates = new ArrayList<>();
        for(Entry<String, Object> currentEntry : filters.entrySet()) {
            Predicate currentPredicate;

            if(currentEntry.getKey().startsWith("information_")) {
                currentPredicate = builder.like(
                        builder.lower(joinPersonInformation.<String>get(currentEntry.getKey())),
                    builder.lower(builder.literal(String.valueOf(currentEntry.getValue())))
                );
            }
            else if(currentEntry.getKey().startsWith("address_")) {
                currentPredicate = builder.like(
                        builder.lower(joinPersonAddress.<String>get(currentEntry.getKey())),
                    builder.lower(builder.literal(String.valueOf(currentEntry.getValue())))
                );
            }
            else {
                currentPredicate = builder.like(
                        builder.lower(rootPerson.<String>get(currentEntry.getKey())),
                    builder.lower(builder.literal(String.valueOf(currentEntry.getValue())))
                );
            }
            allPredicates.add(currentPredicate);
        }
        criteriaQuery.where(builder.and(allPredicates.toArray(new Predicate[0])));

        // order
        if(sortField != null && !sortField.isEmpty()) {
            Order orderBy;
            if(sortField.startsWith("information_")) {
                orderBy = (sortOrder == SortOrder.DESCENDING
                        ? builder.desc(joinPersonInformation.get(sortField))
                        : builder.asc(joinPersonInformation.get(sortField)));
            }
            else if(sortField.startsWith("address_")) {
                orderBy = (sortOrder == SortOrder.DESCENDING
                        ? builder.desc(joinPersonAddress.get(sortField))
                        : builder.asc(joinPersonAddress.get(sortField)));
            }
            else {
                orderBy = (sortOrder == SortOrder.DESCENDING
                        ? builder.desc(rootPerson.get(sortField))
                        : builder.asc(rootPerson.get(sortField)));
            }
            criteriaQuery.orderBy(orderBy);
        }

        Query query = em.createQuery(criteriaQuery);
        // pagination
        query.setFirstResult(first);
        query.setMaxResults(pageSize);
        return query.getResultList();
    }
}

我需要根据访问属性的 root/join 来区分过滤和排序的情况。另外，我需要在 facelet 中使用命名约定。 count-query 除了排序之外也是如此。

现在我问自己是否有一些 "dot-notation" 或任何东西可以让这个案子变得可有可无。在 e。 G。本机 SQL 我会做一些事情，比如创建一个子查询和 select 来自内部投影的所有别名值 (select * from (select person.name as name, address.street as street, ...) where name = ... and street like ...).

如有任何建议，我将不胜感激。

Answer 1

我终于有时间来处理我的问题了。我找到了一个不完美的解决方案，但对我有用。

当我搜索另一个问题时，我找到了 this article by Leonardo Shikida 并找到了一个非常方便的 Path<?> getPath(...) 方法（我还深入研究了 CriteriaAPI 中出色的继承关系：路径、根、加入、来自等）。考虑到这一点，我想起了我以前的问题，并想出了一种更通用的方法。所以这就是我所做的：

首先我创建了我需要的所有连接（即 Root<?> 和 Join<? ?>）并将它们放在 Map<String, From<?, ?>> 中，其中 String 是一个元素以点分符号查询属性（命名约定和完整解决方案的缺点）并且 From 是相应的来源。

使用地图，我可以以或多或少通用的方式进行过滤和排序。

为了使其工作，前端需要使用完全相同的命名约定并相应地传递 filters-Map（即 JSF 在 p:column 中使用 primefaces field 属性）。

public List<Person> newFindAllPersons(int first, int pageSize, String sortField, SortOrder sortOrder, Map<String, Object> filters)
{
    CriteriaBuilder builder = em.getCriteriaBuilder();
    CriteriaQuery<Person> criteriaQuery = builder.createQuery(Person.class);

    // setting up the required joins
    Root<Person> rootPerson = criteriaQuery.from(Person.class);
    Join<Person, Information> joinPersonInformation = rootPerson.join(Person_.information);
    Join<Person, Address> joinPersonAddress = rootPerson.join(Person_.address);
    Join<Address, Information> joinAddressInformation = joinPersonAddress.join(Address_.information);

    // putting all joins into a map with a dot`ted name
    Map<String, From<?, ?>> mapFieldToFrom = new HashMap<>();
    mapFieldToFrom.put("person", rootPerson);
    mapFieldToFrom.put("person.address", joinPersonAddress);
    mapFieldToFrom.put("person.information", joinPersonInformation);
    mapFieldToFrom.put("person.address.information", joinAddressInformation);

    // select
    criteriaQuery.select(rootPerson);

    // filter
    List<Predicate> allPredicates = new ArrayList<>();
    for(Entry<String, Object> currentEntry : filters.entrySet())
    {
        Predicate currentPredicate = builder.like(
                builder.lower(getStringPath(currentEntry.getKey(), mapFieldToFrom)),
                builder.lower(builder.literal("%" + String.valueOf(currentEntry.getValue()) + "%"))
        );
        allPredicates.add(currentPredicate);
    }
    criteriaQuery.where(builder.and(allPredicates.toArray(new Predicate[0])));

    // order
    if(sortField != null && !sortField.isEmpty())
    {
        Path<?> actualPath = getStringPath(sortField, mapFieldToFrom);
        Order orderBy = (sortOrder == SortOrder.DESCENDING
                ? builder.desc(actualPath)
                : builder.asc(actualPath));

        criteriaQuery.orderBy(orderBy);
    }

    Query query = em.createQuery(criteriaQuery);
    // pagination
    query.setFirstResult(first);
    query.setMaxResults(pageSize);
    return query.getResultList();
}

/**
 * divides the given field at the last dot and takes <br>
 * -   the first part as the key in the map to retrieve the From<?, ?> <br>
 * -   the last part as the name of the column in the entity
 */
private Path<String> getStringPath(String field, Map<String, From<?, ?>> mapFieldToFrom)
{
    if(!field.matches(".+\..+"))
    {
        throw new IllegalArgumentException("field '" + field + "' needs to be a dotted path (i. e. customer.address.city.zipcode)");
    }
    String fromPart = field.substring(0, field.lastIndexOf('.'));
    String fieldPart = field.substring(field.lastIndexOf('.') + 1);

    From<?, ?> actualFrom = mapFieldToFrom.get(fromPart);
    if(actualFrom == null)
    {
        throw new IllegalStateException("the given map does not contain a from or for the value '" + fromPart + "' or is null");
    }
    return actualFrom.get(fieldPart);
}

前端示例

<p:dataTable>
    <!-- mapFieldToFrom.put("person", rootPerson); -->
    <p:column field="person.name"> 
    </p:column>

    <!-- mapFieldToFrom.put("person.address", joinPersonAddress); -->
    <p:column field="person.address.street">
    </p:column>
</p:dataTable>

JPA 动态条件-api 查询

JPA dynamic criteria-api query

jpa

dynamic

criteria-api