Apache spark Cassandra连接器--joinWithCassandraTable和leftJoinWithCassandraTable之间的差异--无法解析符号

我正试图通过使用datastax Cassandra连接器连接来访问来自Cassandra的数据。下面的代码适用于我。我试图在join之后总结RDD和Cassandra中的值列

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，当我尝试进行左连接时，它会给出一个“无法解析符号getLong” 我相信这是因为left join不能保证值，因为它可以为null，但我无法在scala中编写此代码

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"))

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感谢您的帮助。如果需要任何信息，请告诉我，我将尝试添加当您在Cassandra中没有获得数据时，您应该获得

选项[Row]

而不是

Row

对象

您可以编写以下内容，而不是

.map{case（ip，row）=>…}

：

.map{案例（ip，行）=>
排赛{
案例无=>ip
案例部分（数据）=>IP（..，IP.value1+data.getLong（“value1”））
}
}

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在这种情况下-如果没有数据（

None

），则只需返回

IP

对象本身，如果有数据，则构建新的

IP

对象。成功了。我必须将整个方法的返回类型更改为returnunit（我之前没有指定任何内容），因为它抛出了一个错误，表示块是递归的。不知道为什么它是递归的