spring，hibernate，postgres數據庫查詢

Question

我的spring，hibernate app有問題。我正在嘗試使用spring security進行登錄，並且在將我的用戶帳戶查詢映射到數據庫時遇到了一些麻煩。

問題是我的代碼將達到「test1」，但它不會達到「test2」，因爲我沒有收到任何錯誤控制檯也應用程序將繼續運行我不知道問題可能是什麼。

當我按下「登錄」按鈕時，我將導向登錄失敗頁面。另外我會指出，我是春季和冬眠的新人。

任何人有任何想法我做錯了什麼？

UserAccountService.java

package main.java.services; 

import javax.persistence.EntityManager; 
import javax.persistence.PersistenceContext; 
import javax.persistence.Query; 

import main.java.model.UserAccount; 

import org.springframework.stereotype.Service; 
import org.springframework.transaction.annotation.Transactional; 

@Service("userAccountService") 
@Transactional 
public class UserAccountService { 

    private EntityManager entityManager; 

    @PersistenceContext 
    public void setEntityManager(EntityManager entityManager){ 
      this. entityManager = entityManager; 
     } 
    public UserAccount get(Integer id) 
    { 
     Query query = entityManager.createQuery("FROM user_account as ua WHERE ua.id="+id); 
     return (UserAccount) query.getSingleResult(); 
    } 

    public UserAccount get(String username) 
    { 
     System.out.println("test1"); 
     Query query = entityManager.createQuery("FROM user_account as ua WHERE ua.username='"+username+"'"); 
     System.out.println("test2"); 
     return (UserAccount) query.getSingleResult(); 
    } 

    public void add(UserAccount userAccount) 
    { 
     entityManager.persist(userAccount); 
    } 
} 

LoginService.java

package main.java.services; 

import javax.annotation.Resource; 

import main.java.model.UserAccount; 
import main.java.security.CustomUserDetails; 
import main.java.security.UserGrantedAuthority; 

import org.springframework.security.core.GrantedAuthority; 
import org.springframework.security.core.userdetails.UserDetails; 
import org.springframework.security.core.userdetails.UserDetailsService; 

public class LoginService implements UserDetailsService{ 

    @Resource(name="userAccountService") 
    private UserAccountService userAccountService; 

    public LoginService(){ } 

    public UserDetails loadUserByUsername(String username){ 
    if (username != null && !username.equals("")){ 
     UserAccount userAccount = userAccountService.get(username); 
     System.out.println(userAccount); 
     if (userAccount == null) { 
      return null; 
     } 
     GrantedAuthority grantedAuth = new UserGrantedAuthority(userAccount.getAuthority()); 
     System.out.println(userAccount.getId() + userAccount.getAuthority()+userAccount.getPassword()); 
     return new CustomUserDetails(userAccount.getId(), userAccount.getUsername(), userAccount.getPassword(), new GrantedAuthority[]{ grantedAuth }); 
    } else { 
     return null; 
    } 
} 
} 

hibernateContext.xml

<?xml version="1.0" encoding="UTF-8"?> 
<beans xmlns="http://www.springframework.org/schema/beans" 
     xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" 
     xmlns:context="http://www.springframework.org/schema/context" xmlns:tx="http://www.springframework.org/schema/tx" 
     xmlns:p="http://www.springframework.org/schema/p" 
     xsi:schemaLocation="http://www.springframework.org/schema/beans 
     http://www.springframework.org/schema/beans/spring-beans.xsd 
     http://www.springframework.org/schema/context 
     http://www.springframework.org/schema/context/spring-context-3.1.xsd 
     http://www.springframework.org/schema/tx 
     http://www.springframework.org/schema/tx/spring-tx-3.1.xsd"> 

    <context:property-placeholder location="/WEB-INF/jdbc.properties" /> 

    <tx:annotation-driven transaction-manager="transactionManager" /> 

    <!-- Declare a datasource that has pooling capabilities--> 
    <bean id="dataSource" class="com.mchange.v2.c3p0.ComboPooledDataSource" 
      destroy-method="close" 
      p:driverClass="${jdbc.driverClassName}" 
      p:jdbcUrl="${jdbc.url}" 
      p:user="${jdbc.username}" 
      p:password="${jdbc.password}" 
      p:acquireIncrement="5" 
      p:idleConnectionTestPeriod="60" 
      p:maxPoolSize="100" 
      p:maxStatements="50" 
      p:minPoolSize="10" /> 

    <!-- Declare a JPA entityManagerFactory--> 
    <bean id="entityManagerFactory" class="org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean" > 
     <property name="persistenceXmlLocation" value="classpath*:META-INF/persistence.xml"></property> 
     <property name="persistenceUnitName" value="hibernatePersistenceUnit" /> 
     <property name="dataSource" ref="dataSource"/> 
     <property name="jpaVendorAdapter"> 
      <bean class="org.springframework.orm.jpa.vendor.HibernateJpaVendorAdapter" > 
       <property name="databasePlatform"> 
        <value>${jdbc.dialect}</value> 
       </property> 
       <property name="showSql" value="true"/> 
      </bean> 
     </property> 
    </bean> 

    <!-- Declare a transaction manager--> 
    <bean id="transactionManager" class="org.springframework.orm.jpa.JpaTransactionManager"> 
     <property name="entityManagerFactory" ref="entityManagerFactory" /> 
    </bean> 

</beans> 

Answer 1

爲了閱讀你的文章，我第一次可以檢測到您的查詢是錯誤的... 您的查詢是一個SQL查詢。在這種情況下，您應該使用createNativeQuery（） 而不是createQuery（）。正確的查詢（假設我不知道你 班）：

Query query = entityManager.createQuery("SELECT us FROM UserAccount as ua WHERE ua.username='"+username+"'"); 

其中UserAccount是類，表中沒有名稱的名稱。此外 它更好地使用準備的語句（谷歌它）傳遞參數的查詢。