打电话来的getNextException看到原因：如何使休眠/ JPA显示DB服务器消息异常(Ca

我使用PostgreSQL，Hibernate和JPA。每当有数据库中的一个例外，我得到这样的事情是不是非常有帮助，因为它并没有显示真正出了错DB服务器上。

Caused by: java.sql.BatchUpdateException: Batch entry 0 update foo set ALERT_FLAG='3' was aborted.  Call getNextException to see the cause.
    at org.postgresql.jdbc2.AbstractJdbc2Statement$BatchResultHandler.handleError(AbstractJdbc2Statement.java:2621)
    at org.postgresql.core.v3.QueryExecutorImpl.processResults(QueryExecutorImpl.java:1837)
    at org.postgresql.core.v3.QueryExecutorImpl.execute(QueryExecutorImpl.java:407)
    at org.postgresql.jdbc2.AbstractJdbc2Statement.executeBatch(AbstractJdbc2Statement.java:2754)
    at com.mchange.v2.c3p0.impl.NewProxyPreparedStatement.executeBatch(NewProxyPreparedStatement.java:1723)
    at org.hibernate.jdbc.BatchingBatcher.doExecuteBatch(BatchingBatcher.java:70)
    at org.hibernate.jdbc.AbstractBatcher.executeBatch(AbstractBatcher.java:268)
    ... 82 more

我想从数据库异常消息出现在应用程序的日志。

我遇到了这篇文章，它使用的看点来填充它，否则不会的SQLException的情况下，适当地填充异常链。

有没有办法解决这个问题，而无需使用方面或任何自定义代码。理想的解决方案只涉及配置文件中的变化。

Answer 1:

有没有需要编写任何自定义代码来实现这一点 - Hibernate会默认登录异常的原因。如果你不能看到这一点，Hibernate日志不能正确设置。下面是与SLF4J + log4j的，并使用Maven依赖管理的例子。

的src /主/爪哇/ pgextest / PGExceptionTest.java

public class PGExceptionTest {

    public static void main(String[] args) throws Exception {

        EntityManagerFactory entityManagerFactory = Persistence.createEntityManagerFactory(
                "pgextest");
        EntityManager entityManager = entityManagerFactory.createEntityManager();
        entityManager.getTransaction().begin();
        // here I attempt to persist an object with an ID that is already in use
        entityManager.persist(new PGExceptionTestBean(1));
        entityManager.getTransaction().commit();
        entityManager.close();
    }
}

SRC /主/资源/ log4j.properties

log4j.rootLogger=ERROR, stdout

log4j.appender.stdout=org.apache.log4j.ConsoleAppender
log4j.appender.stdout.layout=org.apache.log4j.PatternLayout
log4j.appender.stdout.layout.ConversionPattern=%5p [%t] - %m%n

SRC /主/资源/ META-INF / persistence.xml中

<persistence 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"
        version="2.0">
    <persistence-unit name="pgextest">
        <properties>
            <property name="javax.persistence.jdbc.driver" value="org.postgresql.Driver"/>
            <property name="javax.persistence.jdbc.url" value="jdbc:postgresql://localhost/pgextest"/>
            <property name="javax.persistence.jdbc.user" value="postgres"/>
            <property name="javax.persistence.jdbc.password" value="postgres"/>
            <property name="hibernate.dialect" value="org.hibernate.dialect.PostgreSQLDialect"/>
            <property name="hibernate.jdbc.batch_size" value="5"/>
        </properties>
    </persistence-unit>
</persistence>

的pom.xml

<project xmlns="http://maven.apache.org/POM/4.0.0" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
    xsi:schemaLocation="http://maven.apache.org/POM/4.0.0 http://maven.apache.org/xsd/maven-4.0.0.xsd">

    <modelVersion>4.0.0</modelVersion>

    <groupId>pgextest</groupId>
    <artifactId>pgextest</artifactId>
    <version>0.0.1-SNAPSHOT</version>

    <build>
        <plugins>
            <plugin>
                <artifactId>maven-compiler-plugin</artifactId>
                <configuration>
                    <source>1.6</source>
                    <target>1.6</target>
                </configuration>
            </plugin>
        </plugins>
    </build>

    <dependencies>
        <dependency>
            <groupId>org.hibernate</groupId>
            <artifactId>hibernate-entitymanager</artifactId>
            <version>3.6.9.Final</version>
        </dependency>   

        <dependency>
            <groupId>postgresql</groupId>
            <artifactId>postgresql</artifactId>
            <version>9.1-901.jdbc4</version>
            <scope>runtime</scope>
        </dependency>

        <dependency>
            <groupId>org.slf4j</groupId>
            <artifactId>slf4j-log4j12</artifactId>
            <version>1.6.1</version>
            <scope>runtime</scope>
        </dependency>

        <dependency>
            <groupId>log4j</groupId>
            <artifactId>log4j</artifactId>
            <version>1.2.15</version>
            <scope>runtime</scope>
        </dependency>
    </dependencies>
</project>

然后执行的主要方法将记录以下内容：

ERROR [main] - Batch entry 0 insert into PGExceptionTestBean (label, id) values (NULL, '1') was aborted.  Call getNextException to see the cause.
ERROR [main] - ERROR: duplicate key value violates unique constraint "pgexceptiontestbean_pkey"

这也许值得一提的是，您可以禁用JDBC批处理，通过属性设置包装原始异常hibernate.jdbc.batch_size到0 （不用说了你可能不希望这样做生产。）

Answer 2:

这个工作让我得到这导致了该问题（休眠3.2.5.ga）异常消息：

catch (JDBCException jdbce) {
    jdbce.getSQLException().getNextException().printStackTrace();
}

Answer 3:

对我来说，这个例外是一个PersistenceException下，所以我不得不这样做：

try {
//...
} catch (javax.persistence.PersistenceException e) {
    log.error(((java.sql.BatchUpdateException) e.getCause().getCause()).getNextException());
}

Answer 4:

我觉得看点编程是一个更好的解决方案来解决这类问题。

但是，如果你要编写自定义代码来做到这一点，你可以通过它赶上了SQLException并循环并记录每个异常。像这样的东西应该工作。

try {
 // whatever your code is
} catch (SQLException e) {
    while(e!= null) {
      logger.log(e);
      e = e.getNextException();
    }
}

Answer 5:

try {
 // code
} catch (SQLException e) {      
  for (Throwable throwable : e) {
        log.error("{}", throwable);
  }
}

Answer 6:

以防万一，如果恰巧你是从JUnit测试中得到这个例外您可以将JUnit的例外与TestRule （这是由源激发ExpectedException TestRule ）

public class HibernateBatchUnwindRule implements TestRule {

    private boolean handleAssumptionViolatedExceptions = false;

    private boolean handleAssertionErrors = false;

    private HibernateBatchUnwindRule() {
    }

    public static HibernateBatchUnwindRule create(){
        return new HibernateBatchUnwindRule();
    }

    public HibernateBatchUnwindRule handleAssertionErrors() {
        handleAssertionErrors = true;
        return this;
    }

    public HibernateBatchUnwindRule handleAssumptionViolatedExceptions() {
        handleAssumptionViolatedExceptions = true;
        return this;
    }

    public Statement apply(Statement base,
            org.junit.runner.Description description) {
        return new ExpectedExceptionStatement(base);
    }

    private class ExpectedExceptionStatement extends Statement {
        private final Statement fNext;

        public ExpectedExceptionStatement(Statement base) {
            fNext = base;
        }

        @Override
        public void evaluate() throws Throwable {
            try {
                fNext.evaluate();
            } catch (AssumptionViolatedException e) {
                optionallyHandleException(e, handleAssumptionViolatedExceptions);
            } catch (AssertionError e) {
                optionallyHandleException(e, handleAssertionErrors);
            } catch (Throwable e) {
                handleException(e);
            }
        }
    }

    private void optionallyHandleException(Throwable e, boolean handleException)
            throws Throwable {
        if (handleException) {
            handleException(e);
        } else {
            throw e;
        }
    }

    private void handleException(Throwable e) throws Throwable {
        Throwable cause = e.getCause();
        while (cause != null) {
            if (cause instanceof BatchUpdateException) {
                BatchUpdateException batchUpdateException = (BatchUpdateException) cause;
                throw batchUpdateException.getNextException();
            }
            cause = cause.getCause();
        };
        throw e; 
    }
}

然后添加规则测试用例

public class SomeTest {

    @Rule
    public HibernateBatchUnwindRule batchUnwindRule = HibernateBatchUnwindRule.create();

    @Test
    public void testSomething(){...}
  }