为什么从 Cassandra CLI 教程中进行剪切和粘贴不起作用？

Question

盲目关注http://wiki.apache.org/cassandra/Cas​​sandraCli，有人可以解释一下吗这？

aaron-mac:apache-cassandra-1.0.0 aaron$ bin/cassandra-cli -host localhost -port 9160
Connected to: "Test Cluster" on localhost/9160
Welcome to the Cassandra CLI.

Type 'help;' or '?' for help.
Type 'quit;' or 'exit;' to quit.

[default@unknown] use Keyspace1;
Authenticated to keyspace: Keyspace1

[default@Keyspace1] create column family User with comparator = UTF8Type; 
5ef4bad0-fb2a-11e0-0000-242d50cf1ffd
Waiting for schema agreement...
... schemas agree across the cluster
[default@Keyspace1] set User['jsmith']['first'] = 'John'; 
org.apache.cassandra.db.marshal.MarshalException: cannot parse 'jsmith' as hex bytes
[default@Keyspace1] 

Answer 1

在 Cassandra CLI 中运行任何 set 命令之前，始终建议执行以下操作：

assume <column_family> keys as utf8;
assume <column_family> comparator as utf8;
assume <column_family> validator as utf8;

这将确保您设置和列出的所有内容都将被理解

P.S：这仅适用于 Cassandra 新手

Answer 2

这是针对旧版本的 Cassandra。默认情况下，键现在被视为十六进制字节，因此您需要：

set User[utf8('jsmith')]['first'] = 'John';

或执行：

assume User keys as utf8;
set User['jsmith']['first'] = 'John';

或者，如文档中的注释所述：

注意：从 Cassandra 0.8 开始，我们需要为列族声明一个 key_validation_class：

update column family User with key_validation_class=UTF8Type;

Answer 3

set User['jsmith'][utf8('first')] = utf8('John');

如果您不确定 set 命令，您可以输入 help set;，它将显示完整的文档。