[英]jpa using a collection as a parameter for a IsContainingIgnoreCase query method
[英]JPA query to mix member of and collection as parameter
我从JPA查询开始。 我试图找出如何将集合设置为where子句“ member”的参数。 采用Ultimate JPA查询和技巧列表–第1部分中描述的模型
private static boolean isThisDogBelongingToAperson(EntityManager em, Dog dog, String name) {
Query query = em.createQuery('select count(p) from Person p where :dog member of p.dogs and p.name = :name');
query.setParameter('dog', dog);
query.setParameter('name', name);
try {
return query.getSingleResult() != null;
} catch (Exception e) {
return false;
}
}
这仅将Dog的一个实例作为参数。 如果我有一条狗和猫的清单来创建一个名为“ isAnyOfTheseDogsBelongingToAperson”之类的方法,该怎么办? 我是否需要为Dog列表的每个元素多次调用上述方法,或者是否可以将集合传递给查询? 例如:
private static boolean isAnyOfTheseDogsBelongingToAperson(EntityManager em, List<Dog> dogs, String name) {
Query query = em.createQuery('select count(p) from Person p where :dogs member of p.dogs and p.name = :name');
query.setParameter('dogs', dogs);
query.setParameter('name', name);
try {
return query.getSingleResult() != null;
} catch (Exception e) {
return false;
}
}
如果仅增加“ isAnyOfTheseDogsBelongingToAperson”方法的循环该怎么办? 尝试这样的事情:
private static boolean isAnyOfTheseDogsBelongingToAperson(EntityManager em, List<Dog> dogs, String name) {
for (Dog dog : dogs){
Query query = em.createQuery('select count(p) from Person p where :dog member of p.dog and p.name = :name');
query.setParameter('dog', dog);
query.setParameter('name', name);
try {
return query.getSingleResult() != null;
} catch (Exception e) {
return false;
}
}
}
否则,您可以尝试与列表的.contains方法进行比较。 为此,请参见以下代码:
private static boolean isAnyOfTheseDogsBelongingToAperson(EntityManager em, List<Dog> dogs){
boolean result = false;
List<Dog> dogsWithOwners = new ArrayList<Dog>();
Query query = em.createQuery('select p.dog from Person');
dogsWthOwners = query.getResultList();
for (Dog dog : dogs){
if (dogsWithOwners.contains(dog){
result = true;
}
}
return result;
}
希望这可以帮助您,祝您好运!
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