2017-10-19 124 views
0

一名工作人员向一名报告人报告,一名报告人员可能拥有多名工作人员。Spring Hibernate查询无法解析属性

虽这么说,我有自我映射他们一起Staff.java

//reporting officer 
@ManyToOne (cascade = {CascadeType.MERGE, CascadeType.DETACH, CascadeType.REFRESH}) 
@JoinColumn(name = "reporting_officer_id") 
private Staff reportingOfficer; 

而且在我的数据库

CREATE TABLE `tbl_staff` (
    `id` INT NOT NULL AUTO_INCREMENT, 
    `staff_no` INT(5) NOT NULL, 
    `staff_name` VARCHAR(25) NOT NULL, 
    `login_id` VARCHAR(16) NOT NULL, 
    `date_of_birth` DATE NULL DEFAULT NULL, 
    `joined_date` DATE NULL DEFAULT NULL, 
    `department_id` INT NULL DEFAULT NULL, 
    `designation` VARCHAR(255) NULL DEFAULT NULL, 
    `reporting_officer_id` INT NULL DEFAULT NULL, 
    `photograph` BLOB NULL DEFAULT NULL, 
    FOREIGN KEY (`department_id`) REFERENCES tbl_department(`id`), 
    PRIMARY KEY (`id`) 
); 

我试图执行此查询...

@Query("FROM Staff t WHERE " + "t.reporting_officer_id=:reportingOfficerId") 
    public List<Staff> findStaffsWithSameReportingOfficer(@Param("reportingOfficerId") int reportingOfficerId); 

而我得到这个错误...

Caused by: org.hibernate.QueryException: could not resolve property: reporting_officer_id of: com.staff.domain.Staff [FROM com.staff.domain.Staff t WHERE t.reporting_officer_id=1] 
    at org.hibernate.QueryException.generateQueryException(QueryException.java:120) 
    at org.hibernate.QueryException.wrapWithQueryString(QueryException.java:103) 
    at org.hibernate.hql.internal.ast.QueryTranslatorImpl.doCompile(QueryTranslatorImpl.java:217) 
    at org.hibernate.hql.internal.ast.QueryTranslatorImpl.compile(QueryTranslatorImpl.java:141) 
    at org.hibernate.engine.query.spi.HQLQueryPlan.<init>(HQLQueryPlan.java:115) 
    at org.hibernate.engine.query.spi.HQLQueryPlan.<init>(HQLQueryPlan.java:77) 
    at org.hibernate.engine.query.spi.QueryPlanCache.getHQLQueryPlan(QueryPlanCache.java:153) 
    at org.hibernate.internal.AbstractSharedSessionContract.getQueryPlan(AbstractSharedSessionContract.java:545) 
    at org.hibernate.internal.AbstractSharedSessionContract.createQuery(AbstractSharedSessionContract.java:654) 
    ... 124 common frames omitted 
Caused by: org.hibernate.QueryException: could not resolve property: reporting_officer_id of: com.staff.domain.Staff 
    at org.hibernate.persister.entity.AbstractPropertyMapping.propertyException(AbstractPropertyMapping.java:62) 
    at org.hibernate.persister.entity.AbstractPropertyMapping.toType(AbstractPropertyMapping.java:56) 
    at org.hibernate.persister.entity.AbstractEntityPersister.toType(AbstractEntityPersister.java:1859) 
    at org.hibernate.hql.internal.ast.tree.FromElementType.getPropertyType(FromElementType.java:393) 
    at org.hibernate.hql.internal.ast.tree.FromElement.getPropertyType(FromElement.java:512) 
    at org.hibernate.hql.internal.ast.tree.DotNode.getDataType(DotNode.java:660) 
    at org.hibernate.hql.internal.ast.tree.DotNode.prepareLhs(DotNode.java:264) 
    at org.hibernate.hql.internal.ast.tree.DotNode.resolve(DotNode.java:204) 
    at org.hibernate.hql.internal.ast.tree.FromReferenceNode.resolve(FromReferenceNode.java:109) 
    at org.hibernate.hql.internal.ast.tree.FromReferenceNode.resolve(FromReferenceNode.java:104) 
    at org.hibernate.hql.internal.ast.HqlSqlWalker.resolve(HqlSqlWalker.java:1033) 
    at org.hibernate.hql.internal.antlr.HqlSqlBaseWalker.expr(HqlSqlBaseWalker.java:1286) 
    at org.hibernate.hql.internal.antlr.HqlSqlBaseWalker.exprOrSubquery(HqlSqlBaseWalker.java:4699) 
    at org.hibernate.hql.internal.antlr.HqlSqlBaseWalker.comparisonExpr(HqlSqlBaseWalker.java:4169) 
    at org.hibernate.hql.internal.antlr.HqlSqlBaseWalker.logicalExpr(HqlSqlBaseWalker.java:2134) 
    at org.hibernate.hql.internal.antlr.HqlSqlBaseWalker.whereClause(HqlSqlBaseWalker.java:813) 
    at org.hibernate.hql.internal.antlr.HqlSqlBaseWalker.query(HqlSqlBaseWalker.java:607) 
    at org.hibernate.hql.internal.antlr.HqlSqlBaseWalker.selectStatement(HqlSqlBaseWalker.java:311) 
    at org.hibernate.hql.internal.antlr.HqlSqlBaseWalker.statement(HqlSqlBaseWalker.java:259) 
    at org.hibernate.hql.internal.ast.QueryTranslatorImpl.analyze(QueryTranslatorImpl.java:261) 
    at org.hibernate.hql.internal.ast.QueryTranslatorImpl.doCom 
+1

Jpql以'select {alias}'开头。它也使用字段而不是列。所有基本的jpa文档都会提到这 – DN1

+0

您好,我使用“private staff reportingOfficer”作为字段而不是指定的reportingOfficerId。那么我怎么能够仍然能够为reporting_officer_id选择表格呢? @DN1 – DoubleClickOnThis

+0

以下是HQL/JPQL的文档。它涵盖了所有你问的问题:https://docs.jboss.org/hibernate/orm/current/userguide/html_single/Hibernate_User_Guide.html#hql。有些事情需要学习。 JPQL就是其中之一。 '从s中选择s.reportingOfficer.id =:id'。 –

回答

1
@Query("SELECT t FROM Staff t WHERE t.reportingOfficer.id = :reportingOfficerId") 
public List<Staff> findStaffsWithSameReportingOfficer(@Param("reportingOfficerId") Integer reportingOfficerId); 
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