在jpa标准api中使用NOT EXISTS构建查询

Question

我有两个名为table1,table2的表.这两个表都有相同的字段.这两个表之间没有关系.我的要求是我想要table1中的所有记录,而不是table2中的记录.所以我使用Criteria API编写了一个查询.但它没有给出正确的结果.由于我是这个JPA和标准API的新手,任何人都可以指出我在做错的地方.以下代码我用来做这个.

CriteriaBuilder cb = mediationEntityManager.getCriteriaBuilder();
CriteriaQuery<Table1>  cq = cb.createQuery(Table1.class);
Root<Table1> table1 = cq.from(Table1.class);
cq.select(table1)

Subquery<Table2> subquery =  cq.subquery(Table2.class)
Root table2 = subquery.from(Table2.class)
subquery.select(table2)
cq.where(cb.not(cb.exists(subquery)))
TypedQuery<Table1> typedQuery = mediationEntityManager.createQuery(cq); 
List<Table1> resultList = typedQuery.getResultList();

MySQL查询:

SELECT table1 
FROM   table1 table1 
WHERE  NOT EXISTS (SELECT table2 
                   FROM   table2 table2 
                   WHERE  table2.name = table1.name 
                          AND table2.education = table1.education 
                          AND table2.age = table1.age) 
       AND table1.name = 'san' 
       AND table1.age = '10'; 

我需要针对上述MySQL查询的JPA条件API查询.

Answer 1

您可以使用Criteria API尝试以下代码.我没试过,但您可以尝试相应地修改代码.

CriteriaBuilder cb = mediationEntityManager.getCriteriaBuilder();  
CriteriaQuery<Table1> query = cb.createQuery(Table1.class); 
Root<Table1> table1 =  query.from(Table1.class); 
query.select(table1);
//--  
Subquery<Table2> subquery = query.subquery(Table2.class); 
Root<Table2> table2 = subquery.from(Table2.class);  
subquery.select(table2);  
//--
List<Predicate> subQueryPredicates = new ArrayList<Predicate>(); 
subQueryPredicates.add(cb.equal(table1.get(Table1_.name), table2.get(Table2_.name)));
subQueryPredicates.add(cb.equal(table1.get(Table1_.age), table2.get(Table2_.age)));
subQueryPredicates.add(cb.equal(table1.get(Table1_.education), table2.get(Table2_.education)));
subquery.where(subQueryPredicates.toArray(new Predicate[]{})); 
//--
List<Predicate> mainQueryPredicates = new ArrayList<Predicate>(); 
mainQueryPredicates.add(cb.equal(table1.get(Table1_.name), "san");
mainQueryPredicates.add(cb.equal(table1.get(Table1_.age), "10");
mainQueryPredicates.add(cb.not(cb.exists(subquery))); 
//--
query.where(mainQueryPredicates.toArray(new Predicate[]{})); 
TypedQuery<Table1> typedQuery =  mediationEntityManager.createQuery(query); 
List<Table1> resultList = typedQuery.getResultList();

此外,您可以尝试以下JPQL查询,这更容易理解,更改和调试.

SELECT t1 
FROM   table1 t1, 
       table2 t2 
WHERE  t1.name = 'san' 
       AND t1.age = '10' 
       AND (t2.name <> t1.name 
             AND t2.education <> t1.education 
             AND t2.age <> t1.age);