我有一个表的字符串值为它所指向的实体的条形码。不幸的是,它不是一个外键,它只是一个字符串,所以不存在映射。这使得联合操作变得困难。我想知道如何将这个对象加入另一个没有定义关系的表中。例如:CriteriaQuery加入字符串值
@Entity
@Table(name = "TblSample", schema = SCHEMA, catalog = CATALOG)
public class Sample {
@Id
@Column(name = "id", nullable = false)
private int id;
@Column(name = "barcodeEntity", nullable = false)
private String barcodeEntity;
@OneToOne
@JoinColumn(name = "barcodeContainer", nullable = false)
private Container container;
...
}
@Entity
@Table(name = "TblSoil", schema = SCHEMA, catalog = CATALOG)
public class Soil {
@Column(name = "barcode", nullable = false)
private String barcode;
@Column(name = "name", nullable = false)
private String name;
...
}
@Entity
@Table(name = "TblLeaf", schema = SCHEMA, catalog = CATALOG)
public class Leaf {
@Column(name = "barcode", nullable = false)
private String barcode;
@Column(name = "name", nullable = false)
private String name;
...
}
@Entity
@Table(name = "TblContainer", schema = SCHEMA, catalog = CATALOG)
public class Container {
@Column(name = "barcode", nullable = false)
private String barcode;
@Column(name = "name", nullable = false)
private String name;
@Column(name = "location", nullable = false)
private String location;
...
}
所以,我想用一个CriteriaQuery中可以返回所有的样品和加入的,它是取自实体。我已经开始编写它,但是当我试图找出如何去做时,我陷入了困境。在SQL它想是这样的:
SELECT TOP 100
sample.Id
, sample.barcodeEntity
, leaf.name
, soil.name
, sample.barcodeContainer
, container.name
, container.location
FROM TblSample sample
LEFT JOIN TblSoil leaf on
soil.barcode = sample.barcodeEntity
LEFT JOIN TblLeaf leaf on
leaf.barcode = sample.barcodeEntity
JOIN TblContainer container on
container.barcode = sample.barcodeContainer
我想,相关的JPA CriteriaQuery中会是这个样子:
public void findSamples(Map<String, String> filterCriteria) {
final CriteriaBuilder builder = getEntityManager().getCriteriaBuilder();
final CriteriaQuery<SampleLocation> query = builder.createQuery(SampleLocation.class);
final Root<Sample> derivation = query.from(Sample.class);
// Note that the next two lines don't work
final Join<Leaf> joinOnLeaf = derivation.join(Sample_.barcodeEntity, JoinType.LEFT);
final Join<Soil> joinOnSoil = derivation.join(Sample_.barcodeEntity, JoinType.LEFT);
final Join<Container> joinOnContainer = derivation.join(Sample_.barcodeContainer);
CompoundSelection<SampleLocation> cSelect =
builder.construct(SampleLocation.class, sample.Id, sample.entitybarcode, joinOnLeaf.get(Leaf_.name), joinOnLeaf.get(Soil_.name), sample.barcodeContainer, joinOnContainer.get(Container_.name), joinOnContainer.get(Container_.location));
query.select(cSelect);
TypedQuery<SampleLocation> typedQuery = entityManager.createQuery(query);
typedQuery.setMaxResults(100);
return typedQuery.getResults();
}
任何想法如何,我可以执行左连接操作?我无法根据CriteriaQuery API做出决定。看起来像是应该存在的东西。
老实说,一旦查询得到这个复杂度,我会诉诸于HQL(假设你使用的是Hibernate ...) – 2013-02-25 13:56:59
作为最后的手段,我可能会试试这个。不过,我宁愿不要因为它不安全。我们大多数是Java开发人员,所以我们从这个角度来看更适合执行数据库工作。说了这也许是这些特殊情况之一。如果我无法使用JPA,那么我认为我们将开发一个独立的项目来执行复杂的查询(比如这个),以便将其与代码库的其余部分隔离开来。一旦将SQL引入代码库中,维护起来可能会变得非常困难(正如我发现的那样)。 – Craig 2013-02-26 07:06:14