xml 매핑 오류를 사용하여 명명 된 네이티브 SQL 쿼리 사용

Question

xml 매핑을 사용하여 네이티브 원시 SQL 쿼리를 사용하고 있는데 mysql 함수 AES_ENCRYPT 및 AES_DECRYPT를 사용해야하지만 구문 오류가 표시되었지만 이유를 이해하지 못했습니다. 이제 모든 것을 검토하고 잘

내가

<?xml version="1.0"?> 
<!DOCTYPE hibernate-mapping PUBLIC "-//Hibernate/Hibernate Mapping DTD 3.0//EN" 
"http://www.hibernate.org/dtd/hibernate-mapping-3.0.dtd"> 
<!-- Generated 14/12/2013 03:41:41 AM by Hibernate Tools 3.6.0 --> 
<hibernate-mapping> 
    <class name="Model.TO.Usuario" table="usuario" catalog="sintaxhighlighter"> 
     <id name="correo" type="string"> 
      <column name="Correo" length="50" /> 
      <generator class="assigned" /> 
     </id> 
     <many-to-one name="roles" class="Model.TO.Roles" fetch="select"> 
      <column name="Roles_idRoles" length="2" not-null="true" /> 
     </many-to-one> 
     <property name="password" type="java.lang.String"> 
      <column name="Password" not-null="true" /> 
     </property> 
     <property name="nombres" type="string"> 
      <column name="Nombres" length="50" not-null="true" /> 
     </property> 
     <property name="apellidoP" type="string"> 
      <column name="ApellidoP" length="25" not-null="true" /> 
     </property> 
     <property name="apellidoM" type="string"> 
      <column name="ApellidoM" length="25" not-null="true" /> 
     </property> 
     <set name="codigoses" table="usuarios_has_codigos" inverse="false" lazy="true" fetch="select"> 
      <key> 
       <column name="Usuarios_Correo" length="50" not-null="true" /> 
      </key> 
      <many-to-many entity-name="Model.TO.Codigos"> 
       <column name="Codigos_CodName" length="4" not-null="true" /> 
      </many-to-many> 
     </set> 
    </class> 

     <sql-query name="finByCorreo" > 
     <return alias="Usuario" class="Model.TO.Usuario" /> 
     SELECT use.Correo AS {Usuario.correo}, 
     AES_DECRYPT (use.Password,'escom') AS {Usuario.password}, 
     use.Nombres AS {Usuario.nombres}, 
     use.ApellidoP AS {Usuario.apellidoP}, 
     use.ApellidoM AS {Usuario.apellidoM}  
     FROM usuario use 
     WHERE use.Correo LIKE :tags 
    </sql-query> 
</hibernate-mapping> 

FindByCorreo

Usuario.hbm.xml AES_ENCRYPT

의 암호화 된 데이터를 저장하는 BLOB 필드를 사용하고

public Usuario finByCorreo(Usuario usuario) { Usuario model = null; Session session = HibernateUtil.getSessionFactory().openSession(); System.out.println(usuario.getPassword() + " " + usuario.getCorreo()); try { session.beginTransaction(); Query query=session.getNamedQuery("finByCorreo"); query.setString("tags", usuario.getCorreo()); model = (Usuario) query.uniqueResult(); session.getTransaction().commit(); } catch (HibernateException e) { System.out.println(e.getMessage()); session.getTransaction().rollback(); } return model; } 

Usuario.java

public class Usuario implements java.io.Serializable { 
    private String correo; 
    private Roles roles; 
    private String password; 
    private String nombres; 
    private String apellidoP; 
    private String apellidoM; 
    private Set codigoses = new HashSet(0); 
    . 
} 

결과 오류 내가 거기에 볼

WARN: SQL Error: 1064, SQLState: 42000 
dic 14, 2013 11:13:14 PM org.hibernate.engine.jdbc.spi.SqlExceptionHelper logExceptions 
ERROR: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'WHERE use.Correo LIKE 'amezcua92@hotmail.com'' at line 7 

Answer 1

한 가지 문제는, 당신의 usuario 테이블의 별명입니다. use은 (는) DB 엔진에 대한 키워드입니다. MySQL을 사용하고 있다면 이것을 확인하십시오. 9.3. Reserved Words

<sql-query name="finByCorreo" > 
    <return alias="Usuario" class="Model.TO.Usuario" /> 
    SELECT ut.Correo AS {Usuario.correo}, 
    AES_DECRYPT (ut.Password,'escom') AS {Usuario.password}, 
    ut.Nombres AS {Usuario.nombres}, 
    ut.ApellidoP AS {Usuario.apellidoP}, 
    ut.ApellidoM AS {Usuario.apellidoM}  

    FROM usuario ut ------------------- here ut instead of use 

    WHERE ut.Correo LIKE :tags 
</sql-query>