與JPA2的多個聯接

Question

我正試圖抓住JPA2並試圖做一些聯接來獲得單個結果。這是我目前試過的：

CriteriaBuilder cb = getEntityManager().getCriteriaBuilder(); 
CriteriaQuery<FileCollection> cq = cb.createQuery(getEntityClass()); // getEntityClass() will return FileCollection.class 

Root<FileCollection> collectionRoot = cq.from(getEntityClass()); 
Join<FileCollection, Repository> repositories = collectionRoot.join(FileCollection_.repository); 
Join<Repository, Customer> customers = repositories.join(Repository_.customer); 

cq.select(collectionRoot); 
cq.where(cb.equal(customers.get(Customer_.name), customerName), 
    cb.equal(repositories.get(Repository_.name), repositoryName) , 
    cb.equal(collectionRoot.get(FileCollection_.folderName), folderName) 
); 

return getEntityManager().createQuery(cq).getSingleResult(); 

雖然這不行。如果我將第二個參數的第二個參數註釋掉，它就可以工作（所以我只提供一個客戶名）。所以我出錯了。我只是不知道什麼！以下是我試圖用SQL表達的查詢：

SELECT f.* 
FROM filecollection f 
JOIN repository r ON f.REPOSITORY_ID = r.REPOSITORY_ID 
JOIN customer c ON r.CUSTOMER_ID = c.CUSTOMER_ID 
WHERE c.NAME = 'X' AND r.NAME = 'Y' AND f.FOLDER_NAME = 'Z'; 

任何人都可以幫助我並指出我的錯誤。與此同時，我會回到我的JPA2書籍，看看我能否解決這個問題！

Answer 1

我想應該是這樣的：

cq.where(cb.and(cb.equal(customers.get(Customer_.name), customerName), 
       cb.equal(repositories.get(Repository_.name), repositoryName) , 
       cb.equal(collectionRoot.get(FileCollection_.folderName), folderName) 
));