cassandra连接器--joinwithcassandratable和leftjoinwithcassandratable之间的差异--无法解析符号

qyyhg6bp 于 2021-06-10 发布在 Cassandra

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我正在尝试通过使用datastax cassandra连接器连接从cassandra访问数据。下面的代码正在为我工作。我试图在join之后总结rdd和cassandra的值列

tm(a.joinWithCassandraTable("ks","tbl").on(SomeColumns("key","key2","key3","key4","key5","key6","key7","key8","key9","key10","key11","key12","key13","key14","key15","column1","column2","column3","column4","column5")).select("value1").map { case (ip, row) => IP(ip.key, ip.key2, ip.key3,ip.key4,ip.key5,ip.key6,ip.key7,ip.key8,ip.key9,ip.key10,ip.key11,ip.key12,ip.key13,ip.key14,ip.key15,ip.column1,ip.column2,ip.column3,ip.column4,ip.column5,ip.value1 + row.getLong("value1")) }.saveToCassandra("ks", "tbl"))

但是，当我尝试执行左连接时，它会给出一个“cannot resolve symbol getlong”，我认为这是因为左连接不能保证值，因为它可能为null，但我无法用scala编写此代码。

tm(a.leftJoinWithCassandraTable("ks","tbl").on(SomeColumns("key","key2","key3","key4","key5","key6","key7","key8","key9","key10","key11","key12","key13","key14","key15","column1","column2","column3","column4","column5")).select("value1").map { case (ip, row) => IP(ip.key, ip.key2, ip.key3,ip.key4,ip.key5,ip.key6,ip.key7,ip.key8,ip.key9,ip.key10,ip.key11,ip.key12,ip.key13,ip.key14,ip.key15,ip.column1,ip.column2,ip.column3,ip.column4,ip.column5,ip.value1 + row.getLong("value1")) }.saveToCassandra("ks", "tbl"))

感谢您的帮助。如果有任何需要的信息，让我知道，我会尽量补充

cassandra apache-spark spark-cassandra-connector datastax

来源：https://stackoverflow.com/questions/60174724/cassandra-connector-difference-between-joinwithcassandratable-and-leftjoinwit

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按热度按时间

qq24tv8q1#

当你在Cassandra没有数据时，你应该得到一个 Option[Row] 而不是 Row 对象。
而不是 .map { case (ip, row) => ...} 你可以写：

.map { case (ip, row) => 
  row match {
    case None => ip
    case Some(data) => IP(...., ip.value1 + data.getLong("value1"))
  }
}

在这种情况下-当你没有数据的时候( None )，然后你就回来了 IP 对象本身，如果有数据，则构造新的 IP 对象

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cassandra连接器--joinwithcassandratable和leftjoinwithcassandratable之间的差异--无法解析符号

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