与KeyHolder一起使用insert时的ArrayIndexOutOfBoundsException [英] ArrayIndexOutOfBoundsException when using insert with KeyHolder

问题描述

有一个示例，在此版本中适用:

There is an example, which works in this version:

    String sql = "insert into album (name) VALUES (:name)";
    Object[] params = new Object[] { al.getName() };
    jdbcTemplate.update(sql, params);

并在此版本中给出ArrayIndexOutOfBoundsException:

And gives ArrayIndexOutOfBoundsException when in this version:

    KeyHolder holder = new GeneratedKeyHolder();
    String sql = "insert into album (name) VALUES (:name)";
    Object[] params = new Object[] { al.getName() };
    jdbcTemplate.update(sql, params, holder);

其他信息:

@Autowired
public void setDataSource(DataSource dataSource) {
    this.jdbcTemplate = new JdbcTemplate(dataSource);
}

<bean id="dataSource" class="org.springframework.jdbc.datasource.DriverManagerDataSource">
    <property name="driverClassName" value="org.sqlite.JDBC"></property>
    <property name="url" value="jdbc:sqlite:db/springDB.db"></property>
    <property name="username" value=""></property>
    <property name="password" value=""></property>
</bean>

有一个堆栈跟踪:

Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: 1
at org.sqlite.core.CorePreparedStatement.batch(CorePreparedStatement.java:128)
at org.sqlite.jdbc3.JDBC3PreparedStatement.setObject(JDBC3PreparedStatement.java:388)
at org.springframework.jdbc.core.StatementCreatorUtils.setValue(StatementCreatorUtils.java:402)
at org.springframework.jdbc.core.StatementCreatorUtils.setParameterValueInternal(StatementCreatorUtils.java:235)
at org.springframework.jdbc.core.StatementCreatorUtils.setParameterValue(StatementCreatorUtils.java:166)
at org.springframework.jdbc.core.ArgumentPreparedStatementSetter.doSetValue(ArgumentPreparedStatementSetter.java:66)
at org.springframework.jdbc.core.ArgumentPreparedStatementSetter.setValues(ArgumentPreparedStatementSetter.java:47)
at org.springframework.jdbc.core.JdbcTemplate$2.doInPreparedStatement(JdbcTemplate.java:914)
at org.springframework.jdbc.core.JdbcTemplate$2.doInPreparedStatement(JdbcTemplate.java:909)
at org.springframework.jdbc.core.JdbcTemplate.execute(JdbcTemplate.java:644)
at org.springframework.jdbc.core.JdbcTemplate.update(JdbcTemplate.java:909)
at org.springframework.jdbc.core.JdbcTemplate.update(JdbcTemplate.java:970)
at org.springframework.jdbc.core.JdbcTemplate.update(JdbcTemplate.java:980)
at com.user.impls.SQLiteDAO.insertAlbum(SQLiteDAO.java:214)
at com.user.app.Start.main(Start.java:34)

这是怎么了?好像没有错误.第一个版本插入数据并创建一个ID.但是第二个给出了错误.谢谢

What is wrong here? Seems like there is no error. The first version inserts data and creates an id. But the second gives error. Thank you

推荐答案

因此，使用NamedParameterJdbcTemplate代替JdbcTemplate，作为 Mark Rotteveel建议，解决问题.

So, using NamedParameterJdbcTemplate instead of JdbcTemplate, as Mark Rotteveel suggested, solves the problem.

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