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我有一个表，其中包含它指向的实体的条形码的字符串值。不幸的是，它不是外键，它只是一个字符串，所以不存在映射。这使得连接操作变得困难。我想知道如何将这个对象加入到另一个没有定义关系的表中。例如：

@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;
  ...
}

因此，我想使用 CriteriaQuery 可以返回所有样本并加入从中获取的实体。我已经开始写它，但是当我试图弄清楚如何去做时，我被卡住了。在 sql 中它会像这样：

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

我猜想关联的 jpa CriteriaQuery 看起来像这样：

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();
}

有什么想法可以执行左连接操作吗？我无法根据 CriteriaQuery api 弄清楚如何做到这一点。似乎应该存在的东西。

4

1 回答 1

1

我建议两个查询。对于第一个，获取与您的条形码字符串一致的主键值。然后在第二个查询中使用第一个查询中的数据。

于 2013-02-25T13:53:04.533 回答