我试图搜索,但没有找到准确的解决方案。我有Address
个实体。对于每个新的Address请求,首先我要检查数据库中是否存在相同的地址请求。我的申请是仓库和有可能会有多次同一地址请求。
地址实体
@Entity
@NamedQuery(name="Address.findAll", query="SELECT a FROM Address a")
public class Address implements Serializable {
private static final long serialVersionUID = 1L;
@Id
@GeneratedValue(strategy=GenerationType.IDENTITY)
private Integer id;
private String firstname;
private String lastname;
private String address1;
private String address2;
private String address3;
private String city;
private String postcode;
@JsonProperty(value="county")
private String state;
private String country;
private String telephoneno;
private String mobileno;
private String email;
//bi-directional many-to-one association to Collection
@OneToMany(mappedBy="address")
@JsonIgnore
private List<Collection> collections;
//bi-directional many-to-one association to Delivery
@OneToMany(mappedBy="address")
@JsonIgnore
private List<Delivery> deliveries;
public Address() {
}
public Integer getId() {
return id;
}
public void setId(Integer id) {
this.id = id;
}
public String getAddress1() {
return this.address1;
}
public void setAddress1(String address1) {
this.address1 = address1;
}
public String getAddress2() {
return this.address2;
}
public void setAddress2(String address2) {
this.address2 = address2;
}
public String getAddress3() {
return this.address3;
}
public void setAddress3(String address3) {
this.address3 = address3;
}
public String getCity() {
return this.city;
}
public void setCity(String city) {
this.city = city;
}
public String getCountry() {
return this.country;
}
public void setCountry(String country) {
this.country = country;
}
public String getEmail() {
return this.email;
}
public void setEmail(String email) {
this.email = email;
}
public String getPostcode() {
return this.postcode;
}
public void setPostcode(String postcode) {
this.postcode = postcode;
}
public String getState() {
return this.state;
}
public void setState(String state) {
this.state = state;
}
public String getFirstname() {
return firstname;
}
public void setFirstname(String firstname) {
this.firstname = firstname;
}
public String getLastname() {
return lastname;
}
public void setLastname(String lastname) {
this.lastname = lastname;
}
public String getTelephoneno() {
return telephoneno;
}
public void setTelephoneno(String telephoneno) {
this.telephoneno = telephoneno;
}
public String getMobileno() {
return mobileno;
}
public void setMobileno(String mobileno) {
this.mobileno = mobileno;
}
public List<Collection> getCollections() {
return this.collections;
}
public void setCollections(List<Collection> collections) {
this.collections = collections;
}
public Collection addCollection(Collection collection) {
getCollections().add(collection);
collection.setAddress(this);
return collection;
}
public Collection removeCollection(Collection collection) {
getCollections().remove(collection);
collection.setAddress(null);
return collection;
}
public List<Delivery> getDeliveries() {
return this.deliveries;
}
public void setDeliveries(List<Delivery> deliveries) {
this.deliveries = deliveries;
}
public Delivery addDelivery(Delivery delivery) {
getDeliveries().add(delivery);
delivery.setAddress(this);
return delivery;
}
public Delivery removeDelivery(Delivery delivery) {
getDeliveries().remove(delivery);
delivery.setAddress(null);
return delivery;
}
}
我知道一个解决方案可能是在And
包含所有字段的存储库中声明一个方法。例如。
public Address findByFirstnameAndLastnameAndAddress1AndAddress2AndAddress3AndCityAndPostcode....();
但我想知道是否有更好的方法来做到这一点。有没有什么用于我只是传递新的Address
对象来检查数据库中是否存在相同的Address
。
修改
根据Manish的回答,以下是我的理解:
1&GT;如答案中所述创建界面ExtendedJpaRepository
。
2 - ;为此接口创建实现类,如下所示(参考:Spring Data Jpa Doc)
public class MyRepositoryImpl<T, ID extends Serializable>
extends SimpleJpaRepository<T, ID> implements MyRepository<T, ID> {
List<T> findByExample(T example){
//EclipseLink implementation for QueryByExample
}
}
3&GT;然后,对于每个存储库接口,扩展ExtendedJpaRepository
。这应该使findByExample
在每个存储库中随时可用。
4&GT;创建自定义存储库工厂以替换Spring data JPA doc的步骤4中所述的默认RepositoryFactoryBean。
5个声明自定义工厂的bean。(Spring Data JPA Doc的步骤5)
答案 0 :(得分:3)
您要找的是Query-by-Example
。如this post中所述,此功能已考虑用于JPA 2.0,但未包含在最终版本中。该帖还解释了大多数JPA提供商都具备实现此功能的必要功能。
您可以创建自定义JPA存储库实现,该实现提供开箱即用的此功能。详细信息在Spring Data JPA documentation。
中提供一个起点是创建一个新界面,例如:
public interface ExtendedJpaRepository<T, ID extends Serializable>
extends JpaRepository<T, ID> {
List<T> findByExample(T example);
}
然后,插入使用底层JPA提供程序的此接口的实现。最后,配置您的自定义实现以用于所有存储库接口。
之后,您应该可以致电addressRepository.findByExample(address)
,前提是AddressRepository
延长ExtendedJpaRepository
。
答案 1 :(得分:2)
您可以使用Spring-data为您提供开箱即用的规格。并且能够使用条件API以编程方式构建查询。要支持规范,您可以使用JpaSpecificationExecutor接口扩展存储库接口
public interface CustomerRepository extends SimpleJpaRepository<T, ID>, JpaSpecificationExecutor {
}
附加接口(JpaSpecificationExecutor)带有允许您以各种方式执行规范的方法。
例如,findAll方法将返回与规范匹配的所有实体:
List<T> findAll(Specification<T> spec);
规范界面如下:
public interface Specification<T> {
Predicate toPredicate(Root<T> root, CriteriaQuery<?> query,
CriteriaBuilder builder);
}
好的,那么典型的用例是什么?可以轻松地使用规范在实体之上构建可扩展的谓词集,然后可以将其与JpaRepository结合使用,而无需为每个所需组合声明查询(方法)。这是一个例子: 例2.15。客户规格
public class CustomerSpecs {
public static Specification<Customer> isLongTermCustomer() {
return new Specification<Customer>() {
public Predicate toPredicate(Root<Customer> root, CriteriaQuery<?> query,
CriteriaBuilder builder) {
LocalDate date = new LocalDate().minusYears(2);
return builder.lessThan(root.get('dateField'), date);
}
};
}
public static Specification<Customer> hasSalesOfMoreThan(MontaryAmount value) {
return new Specification<Customer>() {
public Predicate toPredicate(Root<T> root, CriteriaQuery<?> query,
CriteriaBuilder builder) {
// build query here
}
};
}
}
您在业务需求抽象级别上表达了一些标准,并创建了可执行规范。所以客户端可能会使用如下规范: 列出customers = customerRepository.findAll(isLongTermCustomer());
您还可以组合规范 例2.17。组合规格
MonetaryAmount amount = new MonetaryAmount(200.0, Currencies.DOLLAR);
List<Customer> customers = customerRepository.findAll(
where(isLongTermCustomer()).or(hasSalesOfMoreThan(amount)));
正如您所看到的,规格提供了一些链式胶合代码方法 并结合规格。因此,扩展您的数据访问层是 只需创建新的规范实现和 将它们与已经存在的那些相结合。
您可以创建复杂规范,这是一个示例
public class WorkInProgressSpecification {
public static Specification<WorkInProgress> findByCriteria(final SearchCriteria searchCriteria){
return new Specification<WorkInProgress>() {
@Override
public Predicate toPredicate(Root<WorkInProgress> root,
CriteriaQuery<?> query, CriteriaBuilder cb) {
List<Predicate> predicates = new ArrayList<Predicate>();
if(searchCriteria.getView()!=null && !searchCriteria.getView().isEmpty()){
predicates.add(cb.equal(root.get("viewType"), searchCriteria.getView()));
}if(searchCriteria.getFeature()!=null && !searchCriteria.getFeature().isEmpty()){
predicates.add(cb.equal(root.get("title"), searchCriteria.getFeature()));
}if(searchCriteria.getEpic()!=null && !searchCriteria.getEpic().isEmpty()){
predicates.add(cb.equal(root.get("epic"), searchCriteria.getEpic()));
}if( searchCriteria.getPerformingGroup()!=null && !searchCriteria.getPerformingGroup().isEmpty()){
predicates.add(cb.equal(root.get("performingGroup"), searchCriteria.getPerformingGroup()));
}if(searchCriteria.getPlannedStartDate()!=null){
System.out.println("searchCriteria.getPlannedStartDate():" + searchCriteria.getPlannedStartDate());
predicates.add(cb.greaterThanOrEqualTo(root.<Date>get("plndStartDate"), searchCriteria.getPlannedStartDate()));
}if(searchCriteria.getPlannedCompletionDate()!=null){
predicates.add(cb.lessThanOrEqualTo(root.<Date>get("plndComplDate"), searchCriteria.getPlannedCompletionDate()));
}if(searchCriteria.getTeam()!=null && !searchCriteria.getTeam().isEmpty()){
predicates.add(cb.equal(root.get("agileTeam"), searchCriteria.getTeam()));
}
return cb.and(predicates.toArray(new Predicate[]{}));
}
};
}
}