使用JPA解析普通查询时出现语法错误

dan*_*max 7 jpa eclipselink jpql java-ee

我使用netbeans向导创建了实体bean,并尝试从数据库中获取数据.无论我使用什么SQL查询,它都不起作用.我尝试使用由向导创建的命名查询:

@NamedQuery(name = "Usr.findAll", query = "SELECT u FROM Usr u")
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它返回:

Caused by: Exception [EclipseLink-8025] (Eclipse Persistence Services - 2.0.1.v20100213-r6600): org.eclipse.persistence.exceptions.JPQLException
Exception Description: Syntax error parsing the query [Usr.findAll], line 1, column 0: unexpected token [Usr].
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如果我试试;

SELECT uid FROM usr;
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我明白了:

Caused by: java.lang.IllegalArgumentException: An exception occurred while creating a query in EntityManager: 
Exception Description: Syntax error parsing the query [SELECT uid FROM usr;], line 0, column -1: unexpected end of query.
Internal Exception: MismatchedTokenException(-1!=78)
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即使我尝试:

SELECT * FROM usr
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我明白了:

Caused by: Exception [EclipseLink-8025] (Eclipse Persistence Services - 2.0.1.v20100213-r6600): org.eclipse.persistence.exceptions.JPQLException
Exception Description: Syntax error parsing the query [SELECT * FROM usr], line 1, column 7: unexpected token [*].
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我获取数据的路线是:

@PersistenceContext
EntityManager em;
....

em=Persistence.createEntityManagerFactory("SchoolPU").createEntityManager();
List users = em.createQuery("SELECT * FROM usr").getResultList();
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任何人都可以帮助我解决这个微不足道的问题?

Usr实体类:

import java.io.Serializable;
import javax.persistence.Basic;
import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.Id;
import javax.persistence.Lob;
import javax.persistence.NamedQueries;
import javax.persistence.NamedQuery;
import javax.persistence.Table;

/**
 *
 * @author danizmax
 */
@Entity
@Table(name = "USR")
@NamedQueries({
    @NamedQuery(name = "Usr.findAll", query = "SELECT u FROM Usr u"),
    @NamedQuery(name = "Usr.findByUid", query = "SELECT u FROM Usr u WHERE u.uid = :uid"),
    @NamedQuery(name = "Usr.findByPassword", query = "SELECT u FROM Usr u WHERE u.password = :password"),
    @NamedQuery(name = "Usr.findByFistname", query = "SELECT u FROM Usr u WHERE u.fistname = :fistname"),
    @NamedQuery(name = "Usr.findByLastname", query = "SELECT u FROM Usr u WHERE u.lastname = :lastname"),
    @NamedQuery(name = "Usr.findByAddress1", query = "SELECT u FROM Usr u WHERE u.address1 = :address1"),
    @NamedQuery(name = "Usr.findByAddress2", query = "SELECT u FROM Usr u WHERE u.address2 = :address2"),
    @NamedQuery(name = "Usr.findByPostcode", query = "SELECT u FROM Usr u WHERE u.postcode = :postcode"),
    @NamedQuery(name = "Usr.findByEmail", query = "SELECT u FROM Usr u WHERE u.email = :email"),
    @NamedQuery(name = "Usr.findByPhone", query = "SELECT u FROM Usr u WHERE u.phone = :phone")})
public class Usr implements Serializable {
    private static final long serialVersionUID = 1L;
    @Id
    @Basic(optional = false)
    @Column(name = "UID", nullable = false, length = 8)
    private String uid;
    @Basic(optional = false)
    @Column(name = "PASSWORD", nullable = false, length = 20)
    private String password;
    @Basic(optional = false)
    @Column(name = "FISTNAME", nullable = false, length = 30)
    private String fistname;
    @Basic(optional = false)
    @Column(name = "LASTNAME", nullable = false, length = 60)
    private String lastname;
    @Basic(optional = false)
    @Column(name = "ADDRESS1", nullable = false, length = 100)
    private String address1;
    @Column(name = "ADDRESS2", length = 100)
    private String address2;
    @Basic(optional = false)
    @Lob
    @Column(name = "CITY", nullable = false)
    private byte[] city;
    @Basic(optional = false)
    @Column(name = "POSTCODE", nullable = false, length = 10)
    private String postcode;
    @Column(name = "EMAIL", length = 50)
    private String email;
    @Column(name = "PHONE")
    private Integer phone;

    public Usr() {
    }

    public Usr(String uid) {
        this.uid = uid;
    }

    public Usr(String uid, String password, String fistname, String lastname, String address1, byte[] city, String postcode) {
        this.uid = uid;
        this.password = password;
        this.fistname = fistname;
        this.lastname = lastname;
        this.address1 = address1;
        this.city = city;
        this.postcode = postcode;
    }

    public String getUid() {
        return uid;
    }

    public void setUid(String uid) {
        this.uid = uid;
    }

    public String getPassword() {
        return password;
    }

    public void setPassword(String password) {
        this.password = password;
    }

    public String getFistname() {
        return fistname;
    }

    public void setFistname(String fistname) {
        this.fistname = fistname;
    }

    public String getLastname() {
        return lastname;
    }

    public void setLastname(String lastname) {
        this.lastname = lastname;
    }

    public String getAddress1() {
        return address1;
    }

    public void setAddress1(String address1) {
        this.address1 = address1;
    }

    public String getAddress2() {
        return address2;
    }

    public void setAddress2(String address2) {
        this.address2 = address2;
    }

    public byte[] getCity() {
        return city;
    }

    public void setCity(byte[] city) {
        this.city = city;
    }

    public String getPostcode() {
        return postcode;
    }

    public void setPostcode(String postcode) {
        this.postcode = postcode;
    }

    public String getEmail() {
        return email;
    }

    public void setEmail(String email) {
        this.email = email;
    }

    public Integer getPhone() {
        return phone;
    }

    public void setPhone(Integer phone) {
        this.phone = phone;
    }

    @Override
    public int hashCode() {
        int hash = 0;
        hash += (uid != null ? uid.hashCode() : 0);
        return hash;
    }

    @Override
    public boolean equals(Object object) {
        // TODO: Warning - this method won't work in the case the id fields are not set
        if (!(object instanceof Usr)) {
            return false;
        }
        Usr other = (Usr) object;
        if ((this.uid == null && other.uid != null) || (this.uid != null && !this.uid.equals(other.uid))) {
            return false;
        }
        return true;
    }



    @Override
    public String toString() {
        return "org.danizmax.Usr[uid=" + uid + "]";
    }

}
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persistance.xml

<?xml version="1.0" encoding="UTF-8"?>
<persistence version="2.0" xmlns="http://java.sun.com/xml/ns/persistence" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://java.sun.com/xml/ns/persistence http://java.sun.com/xml/ns/persistence/persistence_2_0.xsd">
  <persistence-unit name="SchoolPU" transaction-type="JTA">
    <jta-data-source>jdbc/school</jta-data-source>
    <properties>
    </properties>
  </persistence-unit>
</persistence>
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我使用实体的类:

import java.util.Iterator;
import java.util.List;
import javax.ejb.Stateless;
import javax.persistence.EntityManager;
import javax.persistence.Persistence;
import javax.persistence.PersistenceContext;

/**
 *
 * @author danizmax
 */
@Stateless
public class ValidatorBean {

    @PersistenceContext
    EntityManager em;

    public ValidatorBean() {


    }

    public boolean validate(String user, String pass) {

        List users = em.createQuery("SELECT * FROM usr").getResultList();

        Iterator it = users.iterator();

        //ignore the stupid validation it's only to try out JPA
        while(it.hasNext()){
            Usr u = (Usr) it.next();

            if(u.getUid().equals(user) && u.getPassword().equals(pass)){
                return true;
            }
        }


        return false;
    }
}
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更新:为了对那些回答并奖励你的努力的人们公平,既然我实际上已经学会了这项技术并在现实世界中使用它,我决定通过给出最好的答案来结束这个答案,这是我找到自己的最可能的解决方案.很久以前.

小智 5

我有同样的问题.在我的例子中,我使用了
em.createQuery 方法("Usr.findAll");
而不是
em.createNamedQuery("Usr.findAll");


Jam*_*mes 4

只有您的第一个查询是正确的,其他查询不是 JPQL,因此应该会导致错误。对 SQL 使用 @NamedNativeQuery。

对于第一个查询,这似乎并不是您真正用来获取错误的内容,

异常描述:解析查询 [Usr.findAll],第 1 行,第 0 列时出现语法错误:意外标记 [Usr]。

请注意,错误显示“column 0”是“Usr”,这似乎是您将名称而不是名称放入查询中。我猜你正在做的,

em.createQuery("Usr.findAll").getResultList();

但应该做的,

em.createNamedQuery("Usr.findAll").getResultList();

或者,

em.createQuery("从用户中选择 u").getResultList();

请参阅 http://wiki.eclipse.org/EclipseLink/UserGuide/JPA/Basic_JPA_Development/Querying/JPQL