com.datastax.driver.core.exceptions.InvalidQueryException:未配置的表用户”
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【中文标题】com.datastax.driver.core.exceptions.InvalidQueryException:未配置的表用户”【英文标题】:com.datastax.driver.core.exceptions.InvalidQueryException: unconfigured table user" 【发布时间】:2017-12-28 05:18:42 【问题描述】:我是 Cassandra 的新手,这一直给我带来问题。我下载了 apache-Cassandra 3.11,我正在使用 spring boot 1.5.4.RELEASE。我做了一些研究,发现了一个source,它说这可能是因为 Spring 数据使用了不同的 Cassandra 驱动程序核心版本?但是最新的使用 cql 3 正确吗?我还制作了一个java类配置文件。问题可能就在这里。
import org.springframework.cassandra.config.CassandraCqlClusterFactoryBean;
import org.springframework.cassandra.config.DataCenterReplication;
import org.springframework.cassandra.core.keyspace.CreateKeyspaceSpecification;
import org.springframework.context.annotation.Bean;
import org.springframework.context.annotation.Configuration;
import org.springframework.data.cassandra.config.SchemaAction;
import org.springframework.data.cassandra.config.java.AbstractCassandraConfiguration;
import org.springframework.data.cassandra.repository.config.EnableCassandraRepositories;
import java.util.ArrayList;
import java.util.List;
@Configuration
@EnableCassandraRepositories
public class CassandraConfig extends AbstractCassandraConfiguration
private static final String KEYSPACE = "my_keyspace";
@Bean
@Override
public CassandraCqlClusterFactoryBean cluster()
CassandraCqlClusterFactoryBean bean = new CassandraCqlClusterFactoryBean();
bean.setKeyspaceCreations(getKeyspaceCreations());
return bean;
/**
* if it dont exist , create it
* @return
*/
@Override
public SchemaAction getSchemaAction()
return SchemaAction.CREATE_IF_NOT_EXISTS;
@Override
protected String getKeyspaceName()
return KEYSPACE;
@Override
public String[] getEntityBasePackages()
return new String[]"com.cassandra";
protected List<CreateKeyspaceSpecification> getKeyspaceCreations()
List<CreateKeyspaceSpecification> createKeyspaceSpecifications = new ArrayList<>();
createKeyspaceSpecifications.add(getKeySpaceSpecification());
return createKeyspaceSpecifications;
// Below method creates "my_keyspace" if it doesnt exist.
private CreateKeyspaceSpecification getKeySpaceSpecification()
CreateKeyspaceSpecification pandaCoopKeyspace = new CreateKeyspaceSpecification();
DataCenterReplication dcr = new DataCenterReplication("dc1", 3L);
pandaCoopKeyspace.name(KEYSPACE);
pandaCoopKeyspace.ifNotExists(true).createKeyspace().withNetworkReplication(dcr);
return pandaCoopKeyspace;
@Override
public String getContactPoints()
return "localhost";
这是我正在尝试使用的 bean
import org.springframework.data.cassandra.mapping.PrimaryKey;
import org.springframework.data.cassandra.mapping.Table;
import java.util.UUID;
@Table("user")
public class User
@PrimaryKey
private UUID id;
private String firstName;
private String lastName;
public User()
super();
public User(UUID id, String firstName, String lastName)
this.id = id;
this.firstName = firstName;
this.lastName = lastName;
public User(String firstName, String lastName)
this.firstName = firstName;
this.lastName = lastName;
@Override
public String toString()
return "User" +
"id=" + id +
", firstName='" + firstName + '\'' +
", lastName='" + lastName + '\'' +
'';
如果您需要更多信息,请告诉我,我终生无法弄清楚出了什么问题。谢谢。
【问题讨论】:
如果您可以共享完整的堆栈跟踪或异常,将会很有帮助。 【参考方案1】:在键空间my_keyspace
中创建user
表
CREATE TABLE user(
id uuid PRIMARY KEY,
firstName text,
lastName text
);
【讨论】:
为什么@Override protected String getKeyspaceName() return KEYSPACE;
不起作用?
@Alexandr 问题出在桌子上。以上是关于com.datastax.driver.core.exceptions.InvalidQueryException:未配置的表用户”的主要内容,如果未能解决你的问题,请参考以下文章
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