SQLException上的sqlContext HiveDriver错误:不支持方法

Question

我一直在尝试使用sqlContext.read.format("jdbc").options(driver="org.apache.hive.jdbc.HiveDriver")Hive表进入Spark而没有任何成功.我做过研究并阅读如下:

如何从spark连接到远程配置单元服务器

Spark 1.5.1无法使用hive jdbc 1.2.0

http://belablotski.blogspot.in/2016/01/access-hive-tables-from-spark-using.html

我使用了最新的Hortonworks Sandbox 2.6并向社区询问了同样的问题:

https://community.hortonworks.com/questions/156828/pyspark-jdbc-py4jjavaerror-calling-o95load-javasql.html?childToView=156936#answer-156936

我想做的事情非常简单pyspark:

df = sqlContext.read.format("jdbc").options(driver="org.apache.hive.jdbc.HiveDriver", url="jdbc:hive2://localhost:10016/default", dbtable="sample_07",user="maria_dev", password="maria_dev").load()

这给了我这个错误:

17/12/30 19:55:14 INFO HiveConnection: Will try to open client transport with JDBC Uri: jdbc:hive2://localhost:10016/default
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/usr/hdp/current/spark-client/python/pyspark/sql/readwriter.py", line 139, in load
    return self._df(self._jreader.load())
  File "/usr/hdp/current/spark-client/python/lib/py4j-0.9-src.zip/py4j/java_gateway.py", line 813, in __call__
  File "/usr/hdp/current/spark-client/python/pyspark/sql/utils.py", line 45, in deco
    return f(*a, **kw)
  File "/usr/hdp/current/spark-client/python/lib/py4j-0.9-src.zip/py4j/protocol.py", line 308, in get_return_value
py4j.protocol.Py4JJavaError: An error occurred while calling o119.load.
: java.sql.SQLException: Method not supported
at org.apache.hive.jdbc.HiveResultSetMetaData.isSigned(HiveResultSetMetaData.java:143)
at org.apache.spark.sql.execution.datasources.jdbc.JDBCRDD$.resolveTable(JDBCRDD.scala:136)
at org.apache.spark.sql.execution.datasources.jdbc.JDBCRelation.<init>(JDBCRelation.scala:91)
at org.apache.spark.sql.execution.datasources.jdbc.DefaultSource.createRelation(DefaultSource.scala:57)
at org.apache.spark.sql.execution.datasources.ResolvedDataSource$.apply(ResolvedDataSource.scala:158)
at org.apache.spark.sql.DataFrameReader.load(DataFrameReader.scala:119)
at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
at sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:62)
at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:43)
at java.lang.reflect.Method.invoke(Method.java:498)
at py4j.reflection.MethodInvoker.invoke(MethodInvoker.java:231)
at py4j.reflection.ReflectionEngine.invoke(ReflectionEngine.java:381)
at py4j.Gateway.invoke(Gateway.java:259)
at py4j.commands.AbstractCommand.invokeMethod(AbstractCommand.java:133)
at py4j.commands.CallCommand.execute(CallCommand.java:79)
at py4j.GatewayConnection.run(GatewayConnection.java:209)
at java.lang.Thread.run(Thread.java:748)

使用直线,它工作正常

beeline> !connect jdbc:hive2://localhost:10016/default maria_dev maria_dev
Connecting to jdbc:hive2://localhost:10016/default
Connected to: Spark SQL (version 2.1.1.2.6.1.0-129)
Driver: Hive JDBC (version 1.2.1000.2.6.1.0-129)
Transaction isolation: TRANSACTION_REPEATABLE_READ
0: jdbc:hive2://localhost:10016/default> select * from sample_07 limit 2;
+----------+-------------------------+------------+---------+--+
|   code   |       description       | total_emp  | salary  |
+----------+-------------------------+------------+---------+--+
| 00-0000  | All Occupations         | 134354250  | 40690   |
| 11-0000  | Management occupations  | 6003930    | 96150   |
+----------+-------------------------+------------+---------+--+

我也可以这样做:

spark = SparkSession.Builder().appName("testapp").enableHiveSupport().??getOrCreate()
spark.sql("select * from default.sample_07").collect()

但是这会直接读入Hive元数据.我想将JDBC用于Spark Thrift Server以实现细粒度的安全性.

我可以像这样做PostgreSQL:

sqlContext.read.format("jdbc").options(driver="org.postgresql.Driver")

我也可以使用Scala java.sql.{DriverManager, Connection, Statement, ResultSet}创建JDBC Connection作为客户端来获取Spark.但这基本上将所有数据放入内存,然后手动重新创建Dataframe.

所以问题是:有没有办法用Hive表数据创建Spark数据帧而不将数据加载到像Scala这样的JDBC客户端内存而不使用SparkSession.Builder()上面的例子？我的用例是我需要处理细粒度的安全性.

Answer 1

我不确定我是否正确理解你的问题，但据我了解，你需要将配置单元表放入数据框架中，因为你不需要 JDBC 连接，在你的示例链接中它们是尝试连接到不同的数据库（RDBMS），而不是 Hive。

请参阅下面的方法，使用 hive 上下文，您可以将表放入数据框中。

import org.apache.spark.SparkConf
import org.apache.spark.SparkContext
import org.apache.spark.sql.{DataFrame, SQLContext}

def main(args: Array[String]): Unit = {

val sparkConf = new SparkConf().setAppName("APPName")
    val sc = new SparkContext(sparkConf)
    val hiveContext = new org.apache.spark.sql.hive.HiveContext(sc)
    val sqlContext = new SQLContext(sc)

val hive_df = hiveContext.sql("select * from schema.table").first()

//other way
// val hive_df= hiveContext.table ("SchemaName.TableName")

//Below will print the first line
df.first()
//count on dataframe
df.count()

}

如果您确实想使用 JDBC 连接，我有以下用于 Oracle 数据库的示例，它可能会对您有所帮助。

val oracle_data = sqlContext.load("jdbc", Map("url" -> "jdbc:oracle:thin:username/password//hostname:2134/databaseName", "dbtable" -> "Your query tmp", "driver" -> "oracle.jdbc.driver.OracleDriver"));