Jakarta ee 意外标记“&引用；在JPA查询期间？

我正在尝试查询db2数据库中的Api键列表

问题是，JPA提供的实际查询有一个额外的比较字段，导致语法错误

错误：

[ERROR   ] CNTR0020E: EJB threw an unexpected (non-declared) exception during invocation of method "getApikeys" on bean "BeanId(WebApiConsole#WebApiConsole.war#ConsoleREST, null)". Exception data: javax.persistence.PersistenceException: Exception [EclipseLink-4002] (Eclipse Persistence Services - 2.5.2.v20140222-22988a5): org.eclipse.persistence.exceptions.DatabaseException
Internal Exception: com.ibm.db2.jcc.am.SqlSyntaxErrorException: An unexpected token "?" was found following "KEYS t1 WHERE ( LIKE".  Expected tokens may include:  "IN".. SQLCODE=-104, SQLSTATE=42601, DRIVER=4.16.53
Error Code: -104
Call: SELECT t1.APIKEYID, t1.APIKEY, t1.CREATEDAT, t1.PROJECTID FROM DB2ADMIN.PROJECTS t0, DB2ADMIN.APIKEYS t1 WHERE ( LIKE ? AND (t0.PROJECTID = t1.PROJECTID))
    bind => [1 parameter bound]
Query: ReadAllQuery(name="Apikey.queryAllByProject" referenceClass=Apikey sql="SELECT t1.APIKEYID, t1.APIKEY, t1.CREATEDAT, t1.PROJECTID FROM DB2ADMIN.PROJECTS t0, DB2ADMIN.APIKEYS t1 WHERE ( LIKE ? AND (t0.PROJECTID = t1.PROJECTID))")

摘录。我不知道为什么会出现问号

SELECT t1.APIKEYID, t1.APIKEY, t1.CREATEDAT, t1.PROJECTID FROM DB2ADMIN.PROJECTS t0, DB2ADMIN.APIKEYS t1 WHERE ( LIKE ? AND (t0.PROJECTID = t1.PROJECTID))

命名查询：

@NamedQuery(name    = "Apikey.queryAllByProject",
            query   = "SELECT k FROM Apikey k WHERE k.project LIKE :project")

最后，这里是jax rs应用程序代码：

Project p = (Project) em.createNamedQuery("Project.queryProjectById")
        .setParameter("projectid", Integer.parseInt(projectid)).getResultList().get(0);
List<JsonObject> apikeys = em.createNamedQuery("Apikey.queryAllByProject")
                        .setParameter("project", p)
                        .getResultList();

Project p=（Project）em.createNamedQuery（“Project.queryProjectById”）
.setParameter（“projectid”，Integer.parseInt（projectid））.getResultList（）.get（0）；
List apikees=em.createNamedQuery（“Apikey.queryalByProject”）
.setParameter（“项目”，p）
.getResultList（）；

奇怪的是，如果我将其更改为以下内容，一切都会解决：

@NamedQuery(name    = "Apikey.queryAllByProject",
            query   = "SELECT k FROM Apikey k WHERE k.project.projectid LIKE :projectid")


List<JsonObject> apikeys = em.createNamedQuery("Apikey.queryAllByProject")
                        .setParameter("projectid", 10000)
                        .getResultList();

@NamedQuery（name=“Apikey.queryalByProject”，
query=“从Apikey k中选择k，其中k.project.projectid类似于：projectid”）
List apikees=em.createNamedQuery（“Apikey.queryalByProject”）
.setParameter（“projectid”，10000）
.getResultList（）；

LIKE

运算符不能用于比较两个实体，无法用SQL语言表达这一点。您需要根据匹配的id或其他条件获取它们。另请参见列出JPA运算符及其功能的表

不能在实体比较中使用“like”。由于行“（t0.PROJECTID=t1.PROJECTID）”，我以为JPA会自动使用id/fk列。非常感谢。