2016-10-03 81 views
0

我试图做到这一点选择:JPA,春Webservice的选择与NOT IN和IN

SELECT c FROM Incident c 
WHERE c.incidentID IN 
    ( 
    SELECT DISTINCT d.incidentID FROM TagIncident d WHERE tagName IN (d.tagName=?1) 
    AND d.incidentID NOT IN 
    (SELECT a.incidentID FROM TagIncident a WHERE tagName IN (a.tagName=?2))  
) 

在我与JPA /弹簧系统,我发现了错误:

"HTTP Status 500 - Request processing failed; nested exception is org.springframework.dao.InvalidDataAccessApiUsageException: An exception occurred while creating a query in EntityManager:" 

是我在语法上做错了吗? 我在我的数据库(HANA)上测试过它,它工作正常。

感谢您的帮助!

编辑更多的错误日志

我最新的尝试是:

SELECT c FROM Incident c WHERE c.incidentID IN 
(SELECT DISTINCT d.incidentID FROM TagIncident d WHERE d.tagName IN 
(d.tagName=?1) AND d.incidentID NOT IN 
(SELECT a.incidentID FROM TagIncident a WHERE a.tagName IN (a.tagName=?2))) 

编辑

Exception Description: Syntax error parsing [SELECT c FROM Incident c WHERE c.incidentID IN (SELECT DISTINCT d.incidentID FROM TagIncident d WHERE d.tagName IN (d.tagName=?1) AND d.incidentID NOT IN (SELECT a.incidentID FROM TagIncident a WHERE a.tagName IN (a.tagName=?2)))]. [117, 131] 
The expression at index {0} is not a valid expression. [215, 229] 
The expression at index {0} is not a valid expression.; nested exception is java.lang.IllegalArgumentException: An exception occurred while creating a query in EntityManager: Exception Description: Syntax error parsing [SELECT c FROM Incident c WHERE c.incidentID IN (SELECT DISTINCT d.incidentID FROM TagIncident d WHERE d.tagName IN (d.tagName=?1) AND d.incidentID NOT IN (SELECT a.incidentID FROM TagIncident a WHERE a.tagName IN (a.tagName=?2)))]. [117, 131] 
The expression at index {0} is not a valid expression. [215, 229] 
The expression at index {0} is not a valid expression.] with root cause Local Exception Stack: Exception [EclipseLink-0] (Eclipse Persistence Services - 2.6.0.v20150309-bf26070): org.eclipse.persistence.exceptions.JPQLException Exception Description: Syntax error parsing [SELECT c FROM Incident c WHERE c.incidentID IN (SELECT DISTINCT d.incidentID FROM TagIncident d WHERE d.tagName IN (d.tagName=?1) AND d.incidentID NOT IN (SELECT a.incidentID FROM TagIncident a WHERE a.tagName IN (a.tagName=?2)))]. [117, 131] 
The expression at index {0} is not a valid expression. [215, 229] 
The expression at index {0} is not a valid expression. 

最新尝试:

List<String> list_add_tags = new ArrayList<String>(); 
List<String> list_remove_tags = new ArrayList<String>(); 

// creating custom sql_query 
String sql_query = "SELECT c FROM Incident c WHERE c.incidentID IN (SELECT DISTINCT(d.incidentID) FROM TagIncident d WHERE d.tagName IN (:add_tags) AND d.incidentID NOT IN (SELECT a.incidentID FROM TagIncident a WHERE a.tagName IN (:remove_tags)))"; 

TypedQuery<Incident> query = em.createQuery(sql_query, Incident.class); 

query.setParameter("add_tags", list_add_tags); 
query.setParameter("remove_tags", list_remove_tags); 

return query.getResultList(); 

仍然不起作用。 =(

错误:

You have attempted to set a value of type class java.util.ArrayList for parameter add_tags with expected type of class java.lang.String 
+0

哪些JPA提供商您使用的? Hibernate,EclipseLink,OpenJPA还是其中一个鲜为人知的DataNucleaus,Toplink,ObjectDB?在Hibernate中,您可以指定列表,如章节4.6.17.5 JPA规范中的示例所示,但用他们自己的话来说,他们认为他们支持它作为奖励功能。也许你应该尝试去除参数周围的括号?官方的例子确实意味着你不应该需要它们。 – coladict

+0

EclipseLink 2.6.0,从列表中删除了IN的括号,它的工作原理,通过这样做我能理解什么?我如何检查我的JPA版本? –

回答

1

通常我在LY使用本机查询,因为我能更容易测试,但试试这个:

SELECT c FROM Incident c 
WHERE c.incidentID IN 
    ( 
    SELECT DISTINCT d.incidentID FROM TagIncident d WHERE tagName IN :at 
    AND d.incidentID NOT IN 
    (SELECT a.incidentID FROM TagIncident a WHERE tagName IN :rt)  
) 

这应该与query.setParameter("tag", theListOfTags)工作。请注意,5.0.7之前的Hibernate版本在括号中带有参数的语法问题。

空列表也会产生语法错误。

JPA规范显示出他们的这个例子作为有效的语法,所以应该支持任何JPA提供商:

SELECT e 
FROM Employee e 
WHERE TYPE(e) IN :empTypes 
+0

标签是不同的,标签是我不想在最后选择的东西,这是我想在第二次选择一些标签。我会尝试没有括号。 谢谢! –

+0

它不必被称为'标签'。这只是使用命名参数,因此您不必使用位置并调用'setParameter'两次,或在具有大量参数的复杂查询中跟踪位置。我做的最大的修正是从那里删除'[ad] .tagName =?1',因为你正在与'IN(值列表)'部分进行比较。 – coladict

+0

你是对的!这是一个列表!我会尽力的,谢谢! –

0

尝试使用此JPA查询

SELECT c FROM Incident c, TagIncident t 
WHERE c.incidentID = t.incidentID 
AND t.tagName = ?1 
AND t.tagName != ?2 

此外,如果你启用休眠日志记录,那么你就可以看到生成的查询,看他们是否在工作外部SQL程序

logging.level.org.hibernate=DEBUG 
+0

该SQL将导致不同的结果。 =( 它不等价,我在什么地方在Bean中设置这个Logging? –

+0

如果您使用的是spring引导,您可以将此属性放入'application.properties'文件中。 – 11thdimension