从 JpaRepository 方法返回一个布尔值

Question

I have a native query in an interface which extends JpaRepository . The method should ideally return a boolean value, but I can't figure out how to SELECT anything that gets automatically translated into boolean .

This works, although I have to call it as Boolean.valueOf(hasKids(id)) :

// yuck. I wanted a boolean
@Query(nativeQuery = true, value = "select 'true' from dual where exists("
          + "select * from child_table where parent_id = ?)")
String hasKids(long parentId);

How can I change this to the more natural return type?

boolean hasKids(long parentId);  // throws ClassCastException

Update:

the stacktrace is not very helpful IMHO because it's the usual nightmare of Hibernate proxies and AspectJ closures, but here's the relevant portion anyway.

Caused by: java.lang.ClassCastException: java.lang.String cannot be cast to java.lang.Boolean
    at com.sun.proxy.$Proxy1025.hasKids(Unknown Source)
    at com.bela.foo.bar.Service.ThingyServiceImpl.recordHasKids_aroundBody4(ThingyServiceImpl.java:85)
    at com.bela.foo.bar.Service.ThingyServiceImpl$AjcClosure5.run(ThingyServiceImpl.java:1)
...

Answer 1

I ran into a similar problem. My solution was to use a projection of java.lang.Boolean.

@Query("select new java.lang.Boolean(count(*) > 0) from child_table where parent_id = ?")
Boolean hasKids(long parentId);

Hope this helps someone.

Answer 2

I think you want to check the row exist or not for the parent id,and return true and false on the basis of that, then go for the case .

Changes to made in query

    "select case when (count(*) >0) then true else false end from dual where exists("
      + "select * from child_table where parent_id = ?)

Answer 3

I tested this by removing the single quotes around true and it works.

@Query(nativeQuery = true, value = "select true from dual where exists("
      + "select * from child_table where parent_id = ?)")
String hasKids(long parentId);

Answer 4

With my Oracle constraint and a combination of all the suggestions here, I found a solution that worked for my situation without having to call Boolean.valueOf(hasKids(id)):

@Query(nativeQuery = true, value = "select case when exists(select * from child_table " 
    + "where parent_id = :parentId) then 'true' else 'false' end from dual")
Boolean hasKids(@Param("parentId") long parentId);

Answer 5

Actually the query should be something like this:

@Query("SELECT CASE WHEN COUNT(u) > 0 THEN true ELSE false END FROM User u WHERE rs = :user")
    boolean userExsist(@Param("user") User user);

Of course for an example like the above is only valid when you want to run a custom query.

Answer 6

似乎有一个问题，至少当查询是非本地查询时，mysql (count( ) >0) 转换为布尔值很好 (count( ) >0) 返回“BigInteger cannot be cast to java.lang.Boolean”当查询是原生的