如何在 spring queryDSL 中为 List 构建谓词？

Question

    Service.java

    private String sid;
    private String serviceName;
    private List<Feature> feature;

    Feature.java

    private Long id;
    private String featureName;

我构建如下谓词

     import com.querydsl.core.BooleanBuilder;
     import com.querydsl.core.types.Predicate;

     BooleanBuilder booleanBuilder = new BooleanBuilder();
     booleanBuilder.and(QService.service.id.eq(id));
     booleanBuilder.and(QFeature.feature.usoc.eq(featureUsoc));

     Predicate predicate = booleanBuilder;

     serviceRepository.findAll(predicate);

当我查询 repo 时，我得到以下错误

nested exception is java.lang.IllegalArgumentException: org.hibernate.QueryException: illegal attempt to dereference collection [{synthetic-alias}{non-qualified-property-ref}feature] with element property reference [featureName] [select service
from com.ws.app.model.Service service
where service.id = ?2 and feature.featureName = ?3]]

如果实体中的变量是List类型，如何建立谓词查询？

Answer 1

您需要一个 JOIN 或关联路径的子查询：

query().from(QService.service)
    .where(QService.service.id.eq(id))
    .where(query().from(QService.service.feature, QFeature.feature)
          .where(QFeature.feature.usoc.eq(featureUsoc))
          .exists())
    .fetch()

加入：

query().from(QService.service)
    .innerJoin(QService.service.feature, QFeature.feature)
    .on(QFeature.feature.usoc.eq(featureUsoc))
    .where(QService.service.id.eq(id))
    .fetch()