问题

我开始使用 MapStruct `1.4.0.CR1` 版本。我还在使用 Gradle：

```groovy
dependencies {
  annotationProcessor("org.mapstruct:mapstruct-processor:${project.property("mapstruct.version")}")

  implementation("org.mapstruct:mapstruct:${project.property("mapstruct.version")}")
}

我有一些 JPA 实体需要进行映射：

public class Exam implements Serializable {
  // 更多类成员在这里

  @OneToMany(cascade = CascadeType.ALL, fetch = FetchType.LAZY, mappedBy = "exam", orphanRemoval = true)
  private Set<Scan> scans;

  public Exam() { } // ...JPA 需要的无参构造函数

  public Exam(final Builder builder) {
    // ...设置其余内容
    scans = builder.scans;
  }

  // 这里有 getters（没有 setters）、hashCode、equals，以及 builder
}

public class Scan implements Serializable {
  // 更多类成员在这里

  @OneToMany(cascade = CascadeType.ALL, fetch = FetchType.LAZY, mappedBy = "scan", orphanRemoval = true)
  private Set<Alarm> alarms;

  @OneToMany(cascade = CascadeType.ALL, fetch = FetchType.LAZY, mappedBy = "scan", orphanRemoval = true)
  private Set<Isotope> isotopes;

  protected Scan() { } // ...JPA 需要的无参构造函数

  public Scan(final Builder builder) {
    // ...设置其余内容
    alarms = builder.alarms;
    isotopes = builder.isotopes;
  }

  // 这里有 getters（没有 setters）、hashCode、equals，以及 builder
}

我有类似的映射类，但它们的字段/成员没有 JPA 实体多，而且它们位于一个完全不同的子系统中（因此需要映射）。问题是 MapStruct 告诉我在 Scan 中没有 isotopes：java: 源参数中不存在名为 "scans.isotopes" 的属性。是否意味着 "scans.empty"。

基本上，在（新的）映射的 Exam 类中，并不是通过一个 scans 的 Set 包含了 isotopes 和 alarms。这是我的 ExamMapper：

@FunctionalInterface
@Mapper(componentModel = "spring", injectionStrategy = InjectionStrategy.CONSTRUCTOR)
public interface ExamMapper {
  // @Mapping(source = "scans.alarms", target = "alarms")
  @Mapping(source = "scans.isotopes", target = "isotopes")
  Exam valueFrom(tld.domain.Exam entity);
}

是否有办法实现这一点？我认为这可能是微不足道的，但我对 MapStruct 还相对新手；)

<details>
<summary>英文:</summary>

I started using MapStruct `1.4.0.CR1`. I&#39;m also using Gradle:

```groovy
dependencies {
  annotationProcessor(&quot;org.mapstruct:mapstruct-processor:${project.property(&quot;mapstruct.version&quot;)}&quot;)

  implementation(&quot;org.mapstruct:mapstruct:${project.property(&quot;mapstruct.version&quot;)}&quot;)
}

I have some JPA entities I'm trying to map:

public class Exam implements Serializable {
  // More class members here

  @OneToMany(cascade = CascadeType.ALL, fetch = FetchType.LAZY, mappedBy = "exam", orphanRemoval = true)
  private Set<Scan> scans;

  public Exam() { } // ...no-argument constructor required by JPA

  public Exam(final Builder builder) {
    // ...set the rest also
    scans = builder.scans;
  }

  // getters (no setters), hashCode, equals, and builder here
}

public class Scan implements Serializable {
  // More class members here

  @OneToMany(cascade = CascadeType.ALL, fetch = FetchType.LAZY, mappedBy = "scan", orphanRemoval = true)
  private Set<Alarm> alarms;

  @OneToMany(cascade = CascadeType.ALL, fetch = FetchType.LAZY, mappedBy = "scan", orphanRemoval = true)
  private Set<Isotope> isotopes;

  protected Scan() { } // ...no-argument constructor required by JPA

  public Scan(final Builder builder) {
    // ...set the rest also
    alarms = builder.alarms;
    isotopes = builder.isotopes;
  }

  // getters (no setters), hashCode, equals, and builder here
}

I have similar classes for mapping, but they don't have as many fields/members as the JPA entities, moreover, they are on a completely different sub-system (hence the mapping). The problem is that MapStruct is telling me there are no isotopes within Scans: java: No property named "scans.isotopes" exists in source parameter(s). Did you mean "scans.empty"?.

Basically, isotopes and alarms are not contained within a Set of scans in the (new) mapped Exam class. This is my ExamMapper:

@FunctionalInterface
@Mapper(componentModel = "spring", injectionStrategy = InjectionStrategy.CONSTRUCTOR)
public interface ExamMapper {
  // @Mapping(source = "scans.alarms", target = "alarms")
  @Mapping(source = "scans.isotopes", target = "isotopes")
  Exam valueFrom(tld.domain.Exam entity);
}

Is there a way to accomplish this? I think this may be trivial, but I'm fairly new to MapStruct

答案1

得分: 1

@Mapping(source = "scans", target = "isotopes")
Exam valueFrom(tld.domain.Exam entity);

Isotope valueFrom(tld.domain.Isotope isotope);

default Set<Isotope> flatMapIsotopes(Set<Scan> scans) {
  return scans.stream()
      .flatMap(scan -> scan.getIsotopes().stream())
      .map(this::valueFrom)
      .collect(Collectors.toSet());
}

@Mapper(componentModel = "spring", injectionStrategy = InjectionStrategy.CONSTRUCTOR)
public interface ExamMapper {

  @Mapping(source = "scans", target = "isotopes")
  Exam valueFrom(tld.domain.Exam entity);

  Isotope valueFrom(tld.domain.Isotope isotope);

  default Set<Isotope> flatMapIsotopes(Set<Scan> scans) {
    return scans.stream()
        .flatMap(scan -> scan.getIsotopes().stream())
        .map(this::valueFrom)
        .collect(Collectors.toSet());
  }
 
}