Question

我正在使用Java8和JPA2 / Hibernate5，Spring4和mySQL。

如果有人可以提供帮助，我将不胜感激。

我有以下表格。

+--------+          +------------+        +---------+
| job    |          | person_job |        | person  |
+--------+          +------------+        +---------+
|  ID    |          |  JOB_ID    |        |   ID    |
|        |          |  PER_ID    |        |         |
+--------+          +------------+        +---------+

person可以包含多个jobs。

Job.java

@ManyToOne(cascade = CascadeType.ALL, fetch = FetchType.EAGER)
@JoinTable(name = "person_job", joinColumns = {
        @JoinColumn(name = "PER_ID", referencedColumnName = "ID") }, inverseJoinColumns = {
                @JoinColumn(name = "JOB_ID", referencedColumnName = "ID", unique = true) })
private Person person;

当我直接对数据库运行时，以下SQL工作正常：

select
         e.*, p.*
     from
         www.job as e  
     inner join
         www.person_job as pj 
             on e.id=pj.JOB_ID  
     inner join
         www.person as p 
             on pj.PER_ID=p.ID  
...

当我运行本机查询（使用上面的sql）时，我收到以下错误：

java.sql.SQLException：找不到列'JOB_ID'。

在 Job.java 中，如果我在JOB_ID中交换PER_ID和@JoinTable，那么我得到：

java.sql.SQLException：找不到列'PER_ID'。

因为上面的SQL对数据库运行正常，我认为问题在于 Job.java 中的连接配置。

更新

我也有以下正常工作：

+--------+          +--------------+        +----------+
| job    |          | job_location |        | location |
+--------+          +--------------+        +----------+
|  ID    |          |  JOB_ID      |        |   ID     |
|        |          |  LOC_ID      |        |          |
+--------+          +--------------+        +----------+

job可以包含多个locations。

jobs.java

@OneToMany(cascade = CascadeType.ALL, fetch = FetchType.EAGER)
@JoinTable(name = "job_location", joinColumns = {
        @JoinColumn(name = "JOB_ID", referencedColumnName = "ID") }, inverseJoinColumns = {
                @JoinColumn(name = "LOC_ID", referencedColumnName = "ID", unique = true) })
private Set<Location> locations;

更新

执行本机查询时，我只收到此错误。当我做下面的例子时，它完美地运作。

    return (T) entityManager.find(entityClass, id);

我的原生查询有问题吗？（如果person上没有Jobs.java列，则此查询有效。

    StringBuilder sb = new StringBuilder();
    sb.append(" select e.* ");
    sb.append(" from ");
    sb.append("    www.job as e ");
    //sb.append(" inner join www.person_job as pj on e.id = pj.JOB_ID ");
    //sb.append(" inner join www.person as p on pj.PER_ID = p.id ");
    sb.append("  where e.id = :id ");
    Query q = entityManager.createNativeQuery(sb.toString(), JobWithDistance.class);
    q.setParameter("id", id);
    List<Job> jobs = (List<Job>) q.getResultList();

添加2个注释掉的行没有区别。

Answer 1

SOLUTION:

I needed to change:

sb.append(" select e.* ");

to

sb.append(" select * ");

Answer 2

试试这个

// in Job entity
@ManyToOne(cascade = CascadeType.ALL, fetch = FetchType.EAGER)
@JoinTable(name = "person_job", joinColumns = { @JoinColumn(name = "JOB_ID") }, 
                                inverseJoinColumns = { @JoinColumn(name = "PER_ID") }
)
private Person person;

...

// in Person entity
@OneToMany(mappedBy = "person")
private List<Job> jobs;

java.sql.SQLException：未找到列'JOB_ID'

2 个答案: