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Spring 项目（以其生产形式）将使用 MySQL 数据库。我希望在项目上运行的测试使用 HYPERSONIC_IN_MEMORY。

我在 test/resources 文件夹中添加了以下新文件。

  • 数据库属性
  • 应用程序上下文.xml
  • 测试持久性.xml

但是，该项目引发了 MYSQL 语法错误（它试图以某种方式将 MySQL 与 HYPERSONIC 一起使用）。

考虑到我是一个完全的初学者（Spring 中的第一个项目），我想知道切换数据库进行测试的步骤是什么（我可能做错了什么）。

LE：这是我在运行 maven 测试时得到的结果：

Tests run: 9, Failures: 0, Errors: 9, Skipped: 0, Time elapsed: 12.29 sec <<< FAILURE!
testCountAllAccountAllocations(com.mms.pone.portal.domain.AccountAllocationIntegrationTest)  Time elapsed: 11.813 sec  <<< ERROR!
org.springframework.transaction.CannotCreateTransactionException: Could not open JPA EntityManager for transaction; nested exception is javax.persistence.PersistenceException: org.hibernate.exception.GenericJDBCException: Cannot open connection
    at org.springframework.orm.jpa.JpaTransactionManager.doBegin(JpaTransactionManager.java:427)
    at org.springframework.transaction.support.AbstractPlatformTransactionManager.getTransaction(AbstractPlatformTransactionManager.java:371)
    at org.springframework.test.context.transaction.TransactionalTestExecutionListener$TransactionContext.startTransaction(TransactionalTestExecutionListener.java:513)
    at org.springframework.test.context.transaction.TransactionalTestExecutionListener.startNewTransaction(TransactionalTestExecutionListener.java:271)
    at org.springframework.test.context.transaction.TransactionalTestExecutionListener.beforeTestMethod(TransactionalTestExecutionListener.java:164)
    at org.springframework.test.context.TestContextManager.beforeTestMethod(TestContextManager.java:358)
    at org.springframework.test.context.junit4.statements.RunBeforeTestMethodCallbacks.evaluate(RunBeforeTestMethodCallbacks.java:73)
    at org.springframework.test.context.junit4.statements.RunAfterTestMethodCallbacks.evaluate(RunAfterTestMethodCallbacks.java:83)
    at org.springframework.test.context.junit4.statements.SpringRepeat.evaluate(SpringRepeat.java:72)
    at org.springframework.test.context.junit4.SpringJUnit4ClassRunner.runChild(SpringJUnit4ClassRunner.java:231)
    at org.junit.runners.BlockJUnit4ClassRunner.runChild(BlockJUnit4ClassRunner.java:47)
    at org.junit.runners.ParentRunner$3.run(ParentRunner.java:231)
    at org.junit.runners.ParentRunner$1.schedule(ParentRunner.java:60)
    at org.junit.runners.ParentRunner.runChildren(ParentRunner.java:229)
    at org.junit.runners.ParentRunner.access$000(ParentRunner.java:50)
    at org.junit.runners.ParentRunner$2.evaluate(ParentRunner.java:222)
    at org.springframework.test.context.junit4.statements.RunBeforeTestClassCallbacks.evaluate(RunBeforeTestClassCallbacks.java:61)
    at org.springframework.test.context.junit4.statements.RunAfterTestClassCallbacks.evaluate(RunAfterTestClassCallbacks.java:71)
    at org.junit.runners.ParentRunner.run(ParentRunner.java:300)
    at org.springframework.test.context.junit4.SpringJUnit4ClassRunner.run(SpringJUnit4ClassRunner.java:174)
    at org.apache.maven.surefire.junit4.JUnit4TestSet.execute(JUnit4TestSet.java:53)
    at org.apache.maven.surefire.junit4.JUnit4Provider.executeTestSet(JUnit4Provider.java:123)
    at org.apache.maven.surefire.junit4.JUnit4Provider.invoke(JUnit4Provider.java:104)
    at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
    at sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:57)
    at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:43)
    at java.lang.reflect.Method.invoke(Method.java:601)
    at org.apache.maven.surefire.util.ReflectionUtils.invokeMethodWithArray(ReflectionUtils.java:164)
    at org.apache.maven.surefire.booter.ProviderFactory$ProviderProxy.invoke(ProviderFactory.java:110)
    at org.apache.maven.surefire.booter.SurefireStarter.invokeProvider(SurefireStarter.java:172)
    at org.apache.maven.surefire.booter.SurefireStarter.runSuitesInProcessWhenForked(SurefireStarter.java:104)
    at org.apache.maven.surefire.booter.ForkedBooter.main(ForkedBooter.java:70)
Caused by: javax.persistence.PersistenceException: org.hibernate.exception.GenericJDBCException: Cannot open connection
    at org.hibernate.ejb.AbstractEntityManagerImpl.convert(AbstractEntityManagerImpl.java:1387)
    at org.hibernate.ejb.AbstractEntityManagerImpl.convert(AbstractEntityManagerImpl.java:1315)
    at org.hibernate.ejb.AbstractEntityManagerImpl.throwPersistenceException(AbstractEntityManagerImpl.java:1397)
    at org.hibernate.ejb.TransactionImpl.begin(TransactionImpl.java:63)
    at org.springframework.orm.jpa.DefaultJpaDialect.beginTransaction(DefaultJpaDialect.java:70)
    at org.springframework.orm.jpa.JpaTransactionManager.doBegin(JpaTransactionManager.java:377)
    ... 31 more
Caused by: org.hibernate.exception.GenericJDBCException: Cannot open connection

还有更多。我觉得 maven 没有使用新的 xml 文件并且仍然坚持旧的。

我有什么办法可以弄清楚吗？

4

2 回答 2

2

搜索org.hibernate.dialect.MySQLDialect并将其替换为org.hibernate.dialect.HSQLDialect( docs )

[编辑]错误显示“无法打开连接”。检查 JDBC URL、用户+密码以及输出中的更多错误。

于 2012-11-08T11:21:01.620 回答
1

我假设这是您原来的 applicationContext.xml 的外观（还假设 persistence.xml 文件定义了一个带有 jndi 名称“testEM”的持久性单元）

<jee:jndi-lookup id="entityManagerFactory" jndi-name="java:comp/env/testEM" />

<bean id="transactionManager" class="org.springframework.orm.jpa.JpaTransactionManager">
    <property name="entityManagerFactory" ref="entityManagerFactory" />
</bean>

出于测试目的，您基本上需要覆盖 entityManagerFactory 的创建方式。在您的 applicationContext-test.xml 文件中覆盖“entityManagerFactory”bean 定义，如下面的代码片段所示。在 src/test/resources 中创建一个“META-INF/test-persistence.xml”，以创建与内存中 hsql 数据库的连接。

测试持久性.xml 

    <persistence-unit name="testPu">
    <properties>
        <property name="hibernate.connection.driver_class" value="com.mysql.jdbc.Driver"/>
        <property name="hibernate.connection.password" value=""/>
        <property name="hibernate.connection.url" value="jdbc:hsqldb:mem:schemaname"/>
        <property name="hibernate.connection.username" value="sa"/>
        <property name="hibernate.dialect" value="org.hibernate.dialect.HSQLDialect"/>
    </properties>
</persistence-unit>

applicationContext-test.xml

    <!--  Import the original applicationContext and override properties for test purposes -->
<import resource="classpath:applicationContext" />

<!--  In our case, we need to override entityManagerFactory definition -->
<bean id="entityManagerFactory" class="org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean">
    <property name="persistenceXmlLocation" value="classpath*:META-INF/test-persistence.xml" />
    <!-- other properties (omitted) -->
</bean>
于 2012-11-08T14:01:00.253 回答