JPA 查询语言标准生成器

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【中文标题】JPA 查询语言标准生成器【英文标题】:JPA query language criteriaBuilder 【发布时间】:2011-06-12 00:25:01 【问题描述】:

我在 EJB 容器中使用 JPA 构建了一个应用程序。 这是我的代码

@PersistenceContext(unitName = "damate-pu")
private EntityManager   em;

@Override
public Workspace find(String username, String path) 
    CriteriaBuilder criteriaBuilder = em.getCriteriaBuilder();
    CriteriaQuery<Workspace> criteriaQuery = criteriaBuilder.createQuery(Workspace.class);
    Root<Workspace> from = criteriaQuery.from(Workspace.class);
    Predicate condition = criteriaBuilder.equal(from.get("Username"), username);
    Predicate condition2 = criteriaBuilder.equal(from.get("Path"), path);
    Predicate condition3 = criteriaBuilder.and(condition, condition2);
    criteriaQuery.where(condition3);
    Query query = em.createQuery(criteriaQuery);

    return (Workspace) query.getSingleResult();

当我尝试从 Web 服务运行此方法时,我收到以下错误: java.lang.IllegalArgumentException: The attribute [Username] from the managed type....

可能是什么问题?我觉得from.get("Username") 有问题... 你怎么看?以及如何解决?


编辑:Workspace.java

package com.ubb.damate.model;

import java.io.Serializable;
import javax.persistence.*;
import java.util.Date;
import java.util.Set;


/**
 * The persistent class for the workspace database table.
 * 
 */
@Entity
@Table(name="workspace")
public class Workspace implements Serializable 
    private static final long serialVersionUID = 1L;

    @Id
    @GeneratedValue(strategy=GenerationType.AUTO)
    @Column(name="WorkspaceId", unique=true, nullable=false)
    private int workspaceId;

    @Temporal( TemporalType.DATE)
    @Column(name="CreationDate", nullable=false)
    private Date creationDate;

    @Lob()
    @Column(name="Path", nullable=false)
    private String path;

    @Column(name="Username", nullable=false, length=20)
    private String username;

    //bi-directional many-to-one association to Project
    @OneToMany(mappedBy="workspace")
    private Set<Project> projects;

    public Workspace() 
    

    public int getWorkspaceId() 
        return this.workspaceId;
    

    public void setWorkspaceId(int workspaceId) 
        this.workspaceId = workspaceId;
    

    public Date getCreationDate() 
        return this.creationDate;
    

    public void setCreationDate(Date creationDate) 
        this.creationDate = creationDate;
    

    public String getPath() 
        return this.path;
    

    public void setPath(String path) 
        this.path = path;
    

    public String getUsername() 
        return this.username;
    

    public void setUsername(String username) 
        this.username = username;
    

    public Set<Project> getProjects() 
        return this.projects;
    

    public void setProjects(Set<Project> projects) 
        this.projects = projects;
    

【问题讨论】:

【参考方案1】:

您是否尝试过使用元模型?

CriteriaBuilder criteriaBuilder = em.getCriteriaBuilder();
Metamodel m = em.getMetamodel();
EntityType<Workspace> WS = m.entity(Workspace.class);
CriteriaQuery<Workspace> criteriaQuery = criteriaBuilder.createQuery(Workspace.class);
Root<Workspace> from = criteriaQuery.from(Workspace.class);
Predicate condition = criteriaBuilder.equal(from.get(WS.username), username);

http://download.oracle.com/javaee/6/tutorial/doc/gjivm.html

【讨论】:

我无法从 WS 获取用户名。为什么会这样? 抱歉,错过了正确的 MetaModel 文档页面 download.oracle.com/javaee/6/tutorial/doc/gjiup.html【参考方案2】:

在构建条件查询(或在字符串中构建 jpql)时,您希望使用实体属性名称,而不是列名称。您的数据库列名为“用户名”,但 Workspace 对象的属性是“用户名”,没有大写 U。

【讨论】:

【参考方案3】:
CriteriaBuilder criteriaBuilder = entityManager.getCriteriaBuilder();
        CriteriaQuery<> criteriaQuery = criteriaBuilder
                .createQuery(Date.class);
        Root<test> root = criteriaQuery.from(test.class);

        criteriaQuery.select(criteriaBuilder.greatest(root
                .<Date> get("Starttime")));
        criteriaQuery.where(
                criteriaBuilder.equal(root.get("columnName 1"), filtervalue),
                criteriaBuilder.equal(root.get("columnName 2"), Filtervalue));

        TypedQuery<Date> query = entityManager.createQuery(criteriaQuery);
        Date date = query.getSingleResult();

【讨论】:

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