Question

您好，我是JPA的新手。我试图使用一列作为标准选择一组特定的记录。我的实体代码是根据表结构自动生成的，如下所示：

create table TEST.COMPUTERS(
       "COLUMN1" VARCHAR2(6) not null,
       "COLUMN2" VARCHAR2(10) not null,
       "COLUMN3" VARCHAR2(5) not null,
       "COLUMN4" VARCHAR2(8) not null,
       "COLUMN5" VARCHAR2(48),
        constraint "PK" primary key ("COLUMN1","COLUMN2")
    );

Entiry Class的构造函数代码：

@Table(name = "COMPUTERS", schema = "TEST") 
public class Computers implements java.io.Serializable {


 /** full constructor */
        public Computers(ComputersId id, String column3, String column4,
                String column5) {
            this.id = id;
            this.column3 = column3;
            this.column4 = column4;
            this.column5 = column5;
        }

          @EmbeddedId
          public ComputersId getId() {
            return this.id;
       }

      public void setId(ComputersId id) {
        this.id = id;
      }
        // and then ....getter and setter methods for Column 3-5

然后我执行以下操作：

Query query  =   EntityManagerHelper.getEntityManager().createQuery("SELECT s from Computers s where s.column1 = :column1").setParameter("column1", "SONY LAPTOPS");

运行上述内容后，我收到以下错误：

An error occurred while parsing the query filter "SELECT s from Computers where s.column1 = :column1". Error message: No field named "column1" in class "class Computers".

对此有何指示？非常感谢..

Answer 1

由于Column1是embededId的一部分，因此您必须通过id字段：

Query query  =   EntityManagerHelper.getEntityManager().createQuery("SELECT s from Computers s where s.id.column1 = :column1").setParameter("column1", "SONY LAPTOPS");

PS：您的专栏似乎选择了相当糟糕的名字..？也许你应该给他们一些更具描述性的名字？

JPA 1.2 createQuery Where Clause Criteria

1 个答案: