在 Oracle 上 INSERT ... SELECT 后获取插入 ID

Question

如果我从 Oracle 客户端（SQL Developer）运行此 SQL 语句，则它可以工作：

insert into Person (Name) select 'Bob' from dual

如果我通过 Spring JDBC 发出它，而不使用 KeyHolder：

final PreparedStatementCreator psc = new PreparedStatementCreator() {

    @Override
    public PreparedStatement createPreparedStatement(Connection con)
        throws SQLException
    {
        return con.prepareStatement(
                "insert into Person (Name) select 'Bob' from dual");
    }
};
jdbcOperations.update(psc);

但是我需要使用 KeyHolder 才能获取新插入行的 ID 。如果我将上面的代码更改为使用 KeyHolder，如下所示：

final KeyHolder keyHolder = new GeneratedKeyHolder();
final PreparedStatementCreator psc = new PreparedStatementCreator() {

    @Override
    public PreparedStatement createPreparedStatement(Connection con)
        throws SQLException
    {
        return con.prepareStatement(
            "insert into Person (Name) select 'Bob' from dual",
            new String[] {"PersonID"});
    }
};
jdbcOperations.update(psc, keyHolder);

...然后我收到此错误：

Exception in thread "main" org.springframework.jdbc.BadSqlGrammarException: PreparedStatementCallback; bad SQL grammar []; nested exception is java.sql.SQLSyntaxErrorException: ORA-00933: SQL command not properly ended
    at org.springframework.jdbc.support.SQLExceptionSubclassTranslator.doTranslate(SQLExceptionSubclassTranslator.java:94)
    at org.springframework.jdbc.support.AbstractFallbackSQLExceptionTranslator.translate(AbstractFallbackSQLExceptionTranslator.java:72)
    at org.springframework.jdbc.support.AbstractFallbackSQLExceptionTranslator.translate(AbstractFallbackSQLExceptionTranslator.java:80)
    at org.springframework.jdbc.core.JdbcTemplate.execute(JdbcTemplate.java:602)
    at org.springframework.jdbc.core.JdbcTemplate.update(JdbcTemplate.java:842)
    at au.com.bisinfo.codecombo.logic.ImportServiceImpl.insertLoginRedirectRule(ImportServiceImpl.java:107)
    at au.com.bisinfo.codecombo.logic.ImportServiceImpl.runImport(ImportServiceImpl.java:68)
    at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
    at sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:39)
    at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:25)
    at java.lang.reflect.Method.invoke(Method.java:597)
    at org.springframework.aop.support.AopUtils.invokeJoinpointUsingReflection(AopUtils.java:309)
    at org.springframework.aop.framework.ReflectiveMethodInvocation.invokeJoinpoint(ReflectiveMethodInvocation.java:183)
    at org.springframework.aop.framework.ReflectiveMethodInvocation.proceed(ReflectiveMethodInvocation.java:150)
    at org.springframework.transaction.interceptor.TransactionInterceptor.invoke(TransactionInterceptor.java:110)
    at org.springframework.aop.framework.ReflectiveMethodInvocation.proceed(ReflectiveMethodInvocation.java:172)
    at org.springframework.aop.framework.JdkDynamicAopProxy.invoke(JdkDynamicAopProxy.java:202)
    at $Proxy8.runImport(Unknown Source)
    at au.com.bisinfo.codecombo.ui.Main.main(Main.java:39)
Caused by: java.sql.SQLSyntaxErrorException: ORA-00933: SQL command not properly ended
    at oracle.jdbc.driver.T4CTTIoer.processError(T4CTTIoer.java:439)
    at oracle.jdbc.driver.T4CTTIoer.processError(T4CTTIoer.java:395)
    at oracle.jdbc.driver.T4C8Oall.processError(T4C8Oall.java:802)
    at oracle.jdbc.driver.T4CTTIfun.receive(T4CTTIfun.java:436)
    at oracle.jdbc.driver.T4CTTIfun.doRPC(T4CTTIfun.java:186)
    at oracle.jdbc.driver.T4C8Oall.doOALL(T4C8Oall.java:521)
    at oracle.jdbc.driver.T4CPreparedStatement.doOall8(T4CPreparedStatement.java:205)
    at oracle.jdbc.driver.T4CPreparedStatement.executeForRows(T4CPreparedStatement.java:1008)
    at oracle.jdbc.driver.OracleStatement.doExecuteWithTimeout(OracleStatement.java:1307)
    at oracle.jdbc.driver.OraclePreparedStatement.executeInternal(OraclePreparedStatement.java:3449)
    at oracle.jdbc.driver.OraclePreparedStatement.executeUpdate(OraclePreparedStatement.java:3530)
    at oracle.jdbc.driver.OraclePreparedStatementWrapper.executeUpdate(OraclePreparedStatementWrapper.java:1350)
    at org.apache.commons.dbcp.DelegatingPreparedStatement.executeUpdate(DelegatingPreparedStatement.java:105)
    at org.apache.commons.dbcp.DelegatingPreparedStatement.executeUpdate(DelegatingPreparedStatement.java:105)
    at org.springframework.jdbc.core.JdbcTemplate$3.doInPreparedStatement(JdbcTemplate.java:844)
    at org.springframework.jdbc.core.JdbcTemplate$3.doInPreparedStatement(JdbcTemplate.java:1)
    at org.springframework.jdbc.core.JdbcTemplate.execute(JdbcTemplate.java:586)
    ... 15 more

FWIW，如果我执行 INSERT ... VALUES 而不是 INSERT ... SELECT，一切都很好（尽管这没有帮助）我，因为我需要选择东西）：

final KeyHolder keyHolder = new GeneratedKeyHolder();
final PreparedStatementCreator psc = new PreparedStatementCreator() {

    @Override
    public PreparedStatement createPreparedStatement(Connection con)
        throws SQLException
    {
        return con.prepareStatement(
            "insert into Person (Name) values ('Bob')",
            new String[] {"PersonID"});
    }
};
jdbcOperations.update(psc, keyHolder);

我正在使用：

• Spring JDBC 3.0.3.RELEASE
• JDBC 驱动程序：ojdbc6.jar 版本 11.2.0.1.0
• RDBMS：Oracle9i Release 9.2.0.5.0 - Production
• commons-dbcp

1.4我的应用程序需要使用标准 SQL 才能保持数据库中立，这排除了任何特定于 Oracle 的 SQL（在现实生活中我不会从“双重”中进行选择）。

感谢您的任何帮助。

Answer 1

java.sql.Connection.prepareStatement(java.lang.String, int) 接口清晰

创建一个默认的PreparedStatement对象，该对象具有检索自动生成的密钥的功能

因此您使用的是错误的方法。尝试

return con.prepareStatement(
        "insert into Person (Name) select 'Bob' from dual",
        Statement.RETURN_GENERATED_KEYS);

一下

Answer 2

我怀疑使用带有 INSERT SELECT 语句的 KeyHolder 不会也不会受支持，因为理论上该选择可以选择多行，并且如果这样做，则无法将这些多个键返回到单个 KeyHolder 中。对于您想要完成的任务，简单地使用 select 语句和 insert 语句可能会更容易。

Answer 3

怎么样

INSERT INTO blah b (blah1, blah2, blah3)
VALUES (?, ?, ?) RETURNING b.id INTO ?";

Answer 4

Oracle JDBC 驱动程序不支持此功能