JPA Select Count Distinct

Question

我需要一個相對簡單的查詢，但JPA使它很難創建它。

的SQL變種是這樣的：

SELECT COUNT(DISTINCT OrderID) AS DistinctOrders FROM Orders WHERE CustomerID=7; 

[編輯：訂單是不是主鍵。可以有更多的OrderId在表中是相等的]

我需要設置CustomerID與傳遞給我的方法的變量。

我發現有關CriteriaQuerydistinct()的文檔，但我似乎無法將它們全部包裝在一起。

這是我到目前爲止有：

CriteriaBuilder cb = this.em.getCriteriaBuilder(); 
CriteriaQuery<Order> c = cb.createQuery(Order.class); 
Root<Order> order = c.from(Order.class); 
Path<String> customerID = order.get("customerID"); 
c.where(cb.equal(customerID, theId)); 

Answer 1

CriteriaBuilder cb = this.em.getCriteriaBuilder(); 
CriteriaQuery<Long> c = builder.createQuery(Long.class);//the query returns a long, and not Orders 
Root<Order> order = c.from(Order.class); 
//c.select(cb.countDistinct(order));//and this is the code you were looking for 
c.select(cb.countDistinct(order.get("orderID")));//for counting distinct fields other than the primary key 
Path<String> customerID = order.get("customerID"); 
c.where(cb.equal(customerID, theId));