Cassandra 为什么是cqlsh；创建；语句创建一个；uuid“；字段为；timeuuid"；？

卡桑德拉3.11.0
在Ubuntu 16.04.4、JRE-8中运行
我正在尝试创建一个带有uuid字段的表，如下所示：

CREATE TABLE IF NOT EXISTS policy (
  tenant_id         text,
  policy_id         uuid,
  name              text,
  enabled           boolean,
  creation_time     bigint,
  PRIMARY KEY (tenant_id, name)
);

执行此操作后，模式将“policy_id”显示为“timeuuid”，而不是“uuid”
看到[类似问题]：
但这是否适用于cqlsh

放弃表/键空间并重试，但没有成功
这是间歇性问题
获取以下异常：

Exception while loading CQL script:
com.datastax.driver.core.exceptions.InvalidQueryException: Type error: cannot assign result of function system.uuid (type uuid) to id (type timeuuid)
        at com.datastax.driver.core.exceptions.InvalidQueryException.copy(InvalidQueryException.java:50)
        at com.datastax.driver.core.DriverThrowables.propagateCause(DriverThrowables.java:37)
        at com.datastax.driver.core.DefaultResultSetFuture.getUninterruptibly(DefaultResultSetFuture.java:245)
        at com.datastax.driver.core.AbstractSession.execute(AbstractSession.java:68)
        at com.datastax.driver.core.AbstractSession.execute(AbstractSession.java:43)

在卡桑德拉日志中也可以看到以下内容：

org.apache.cassandra.exceptions.ConfigurationException: Column family ID mismatch (found e686a660-8994-11e9-984c-2767f9f5fd28; expected e5d72c80-8994-11e9-b706-831d59206120)
    at org.apache.cassandra.config.CFMetaData.validateCompatibility(CFMetaData.java:808) ~[apache-cassandra-3.11.0.jar:3.11.0]
    at org.apache.cassandra.config.CFMetaData.apply(CFMetaData.java:770) ~[apache-cassandra-3.11.0.jar:3.11.0]
    at org.apache.cassandra.config.Schema.updateTable(Schema.java:621) ~[apache-cassandra-3.11.0.jar:3.11.0]
    at org.apache.cassandra.schema.SchemaKeyspace.updateKeyspace(SchemaKeyspace.java:1430) ~[apache-cassandra-3.11.0.jar:3.11.0]
    at org.apache.cassandra.schema.SchemaKeyspace.mergeSchema(SchemaKeyspace.java:1386) ~[apache-cassandra-3.11.0.jar:3.11.0]
    at org.apache.cassandra.schema.SchemaKeyspace.mergeSchemaAndAnnounceVersion(SchemaKeyspace.java:1336) ~[apache-cassandra-3.11.0.jar:3.11.0]
    at org.apache.cassandra.service.MigrationTask$1.response(MigrationTask.java:91) ~[apache-cassandra-3.11.0.jar:3.11.0]
    at org.apache.cassandra.net.ResponseVerbHandler.doVerb(ResponseVerbHandler.java:53) [apache-cassandra-3.11.0.jar:3.11.0]
    at org.apache.cassandra.net.MessageDeliveryTask.run(MessageDeliveryTask.java:66) [apache-cassandra-3.11.0.jar:3.11.0]
    at java.util.concurrent.Executors$RunnableAdapter.call(Executors.java:511) [na:1.8.0_131]
    at java.util.concurrent.FutureTask.run(FutureTask.java:266) [na:1.8.0_131]
    at java.util.concurrent.ThreadPoolExecutor.runWorker(ThreadPoolExecutor.java:1142) [na:1.8.0_131]
    at java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:617) [na:1.8.0_131]
    at org.apache.cassandra.concurrent.NamedThreadFactory.lambda$threadLocalDeallocator$0(NamedThreadFactory.java:81) [apache-cassandra-3.11.0.jar:3.11.0]
    at java.lang.Thread.run(Thread.java:748) ~[na:1.8.0_131]

策略id应具有数据类型“uuid”，但它具有数据类型“timeuuid”

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创建表需要cql语句中的任何其他关键字？

如何检查策略id是否为timeuuid？我试图复制您的问题，并在表上进行了描述，但策略id是uuid。感谢Simon，我从Devcenter连接并检查了架构，它显示了timeuuid。如何检查策略id是timeuuid？我试图复制您的问题，并在表上进行了描述，但policy_id是uuid。感谢Simon，我从Devcenter连接并检查了schema，它显示了timeuuid。