英文:
how to write this query using jpa criteria api?
问题
选择 c.id，(选择 c2.value 从 customer_table c2 其中 c2.id = c.id 和 c2.key = 'test') 作为 "test"，(选择 c2.value 从 customer_table c2 其中 c2.id = c.id 和 c2.key = 'service-category') 作为 "service-category"，(选择 c2.value 从 customer_table c2 其中 c2.id = c.id 和 c2.key = 'exam') 作为 "exam"从 customer_table c分组 按 c.id;
英文:
Select c.id,(Select c2.value from customer_table c2 Where c2.id = c.id And c2.key = 'test') as "test",(Select c2.value from customer_table c2 Where c2.id = c.id And c2.key = 'service-category') as "service-category",(Select c2.value from customer_table c2 Where c2.id = c.id And c2.key = 'exam') as "exam"From customer_table cGroup By c.id;
答案1
得分: 3
假设客户表实体 customerTable 的存在和正确建模，其关系以及 value 是 String 类型，实现如下所示：
CriteriaBuilder cb = entityManager.getCriteriaBuilder();CriteriaQuery<YourPojo> cq = cb.createQuery(YourPojo.class);Root<CustomerTable> root = cq.from(CustomerTable.class);// 子查询 1Subquery<String> sqVal1 = cq.subquery(String.class);Root<CustomerTable> sq1Root = sqVal1.from(CustomerTable.class);sqVal1.where(cb.and(cb.equal(root.get("id"), sq1Root.get("id")),cb.equal(sq1Root.get("key"), "test")));sqVal1.select(sq1Root.get("value"));// 子查询 2Subquery<String> sqVal2 = cq.subquery(String.class);Root<CustomerTable> sq2Root = sqVal2.from(CustomerTable.class);sqVal2.where(cb.and(cb.equal(root.get("id"), sq2Root.get("id")),cb.equal(sq2Root.get("key"), "service-category")));sqVal2.select(sq2Root.get("value"));// 子查询 3Subquery<String> sqVal3 = cq.subquery(String.class);Root<CustomerTable> sq3Root = sqVal3.from(CustomerTable.class);sqVal3.where(cb.and(cb.equal(root.get("id"), sq3Root.get("id")),cb.equal(sq3Root.get("key"), "exam")));sqVal3.select(sq3Root.get("value"));cq.groupBy(root.get("id"));cq.multiselect(root.get("id"),sqVal1.getSelection(),sqVal2.getSelection(),sqVal3.getSelection());
你需要一个带有与 multiselect 子句中参数相同（顺序和类型相同）的构造函数的 POJO（普通 Java 对象）。
public class YourPojo {public YourPojo(String id, String val1, String val2, String val3){// 构造函数实现部分}}
推荐使用元模型来访问实体的属性，这将会替换下面的代码：
root.get("id");
使用以下代码：
root.get(CustomerTable_.id);
使用元模型的众多优势之一是能够自动补全属性名称，并降低此处出错的机会。
英文:
Assuming the existence and correct modeling of the customerTable entity, its relationships and that value is of type String, the implementation would be like this:
CriteriaBuilder cb = entityManager.getCriteriaBuilder();CriteriaQuery<YourPojo> cq = cb.createQuery(YourPojo.class);Root<CustomerTable> root = cq.from(CustomerTable.class);//Subquery 1Subquery<String> sqVal1 = cq.subquery(String.class);Root<CustomerTable> sq1Root = sqVal1.from(CustomerTable.class);sqVal1.where(cb.and(cb.equal(root.get("id"),sq1Root.get("id")),cb.equal(sq1Root.get("key"),"test")));sqVal1.select(sq1Root.get("value"));//Subquery 2Subquery<String> sqVal2 = cq.subquery(String.class);Root<CustomerTable> sq2Root = sqVal2.from(CustomerTable.class);sqVal2.where(cb.and(cb.equal(root.get("id"),sq2Root.get("id")),cb.equal(sq2Root.get("key"),"service-category")));sqVal2.select(sq2Root.get("value"));//Subquery 3Subquery<String> sqVal3 = cq.subquery(String.class);Root<CustomerTable> sq3Root = sqVal3.from(CustomerTable.class);sqVal3.where(cb.and(cb.equal(root.get("id"),sq3Root.get("id")),cb.equal(sq3Root.get("key"),"exam")));sqVal3.select(sq3Root.get("value"));cq.groupBy(root.get("id"));cq.multiselect(root.get("id"),sqVal1.getSelection(),sqVal2.getSelection(),sqVal3.getSelection());
You need a pojo with a constructor with the same parameters (in order and type) as the multiselect clause
public class YourPojo {public YourPojo(String id, String val1, String val2, String val3){[...]}}
It is recommended to use metamodels to access the properties of the entities, which would lead to replace the following code
root.get("id");
with this other
root.get(CustomerTable_.id);
One of the many advantages of using metamodels without getting too into the subject is the ability to auto-complete the property name and reduce the chance of error at this point.
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