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Cassandra phpcassa对LongType和dateOf(TimeUUIDType)抛出警告_Cassandra_Phpcassa - Fatal编程技术网

Cassandra phpcassa对LongType和dateOf(TimeUUIDType)抛出警告

Cassandra phpcassa对LongType和dateOf(TimeUUIDType)抛出警告,cassandra,phpcassa,Cassandra,Phpcassa,我在卡桑德拉有下表 CREATE TABLE reports ( c_date text, c_n int, c_id timeuuid, report_id bigint, report text, PRIMARY KEY ((c_date, c_n), c_id) ) c_date用于按日期查询报表 c_n是防止热点的节点数(平均分布数据的节点数) c_id是插入的timeuuid 我的选择查询(cql3)如下所示 select report, dateOf(c_i

我在卡桑德拉有下表

CREATE TABLE reports (
  c_date text,
  c_n int,
  c_id timeuuid,
  report_id bigint,
  report text,
  PRIMARY KEY ((c_date, c_n), c_id)
)
c_date
用于按日期查询报表

c_n
是防止热点的节点数(平均分布数据的节点数)

c_id
是插入的timeuuid

我的选择查询(cql3)如下所示

select report, dateOf(c_id), report_id 
from keyspace.reports 
where c_date = '2013-08-02' and 
      c_n = 1 and 
      c_id > minTimeuuid('2013-08-02 02:52:10-0400');
  [0] => Array
        (
            [reportid] => 281474976712782
            [report]   => some_report
            [dateOf(c_id)] => d:1375426331.32100009918212890625;
        )
我已成功获得结果集; 但是,当我使用在另一个示例()上实现的
cql\u get\u rows()函数时

无法正确解析时间戳(dateOf(id)),并且bigint字段会产生以下警告

PHP Warning:  unpack(): Type N: not enough input, need 4, have 0 
in /home/arascan/my-project/tools/vendor/phpcassa/lib/phpcassa/Schema/DataType/LongType.php on line 47
cql\u get\u rows()
返回的数据如下:

select report, dateOf(c_id), report_id 
from keyspace.reports 
where c_date = '2013-08-02' and 
      c_n = 1 and 
      c_id > minTimeuuid('2013-08-02 02:52:10-0400');
  [0] => Array
        (
            [reportid] => 281474976712782
            [report]   => some_report
            [dateOf(c_id)] => d:1375426331.32100009918212890625;
        )
如何防止此函数抛出警告并以日期格式获取时间戳? (请不要建议@usage)