Question

我正在使用CDH5.5

我在HIVE默认数据库中创建了一个表，并且能够从HIVE命令中查询它。

输出

hive> use default;

OK

Time taken: 0.582 seconds


hive> show tables;

OK

bank
Time taken: 0.341 seconds, Fetched: 1 row(s)

hive> select count(*) from bank;

OK

542

Time taken: 64.961 seconds, Fetched: 1 row(s)

但是，我无法从pyspark查询该表，因为它无法识别该表。

from pyspark.context import SparkContext

from pyspark.sql import HiveContext

sqlContext = HiveContext(sc)


sqlContext.sql("use default")

DataFrame[result: string]

sqlContext.sql("show tables").show()

+---------+-----------+

|tableName|isTemporary|

+---------+-----------+

+---------+-----------+


sqlContext.sql("FROM bank SELECT count(*)")

16/03/16 20:12:13 INFO parse.ParseDriver: Parsing command: FROM bank SELECT count(*)
16/03/16 20:12:13 INFO parse.ParseDriver: Parse Completed
Traceback (most recent call last):
    File "<stdin>", line 1, in <module>
    File "/usr/lib/spark/python/pyspark/sql/context.py", line 552, in sql
      return DataFrame(self._ssql_ctx.sql(sqlQuery), self)
    File "/usr/lib/spark/python/lib/py4j-0.8.2.1-src.zip/py4j/java_gateway.py",   line 538, in __call__
    File "/usr/lib/spark/python/pyspark/sql/utils.py", line 40, in deco
      raise AnalysisException(s.split(': ', 1)[1])
  **pyspark.sql.utils.AnalysisException: no such table bank; line 1 pos 5**

新错误

>>> from pyspark.sql import HiveContext
>>> hive_context = HiveContext(sc)
>>> bank = hive_context.table("default.bank")
16/03/22 18:33:30 INFO DataNucleus.Persistence: Property datanucleus.cache.level2 unknown - will be ignored
16/03/22 18:33:30 INFO DataNucleus.Persistence: Property hive.metastore.integral.jdo.pushdown unknown - will be ignored
16/03/22 18:33:44 INFO DataNucleus.Datastore: The class "org.apache.hadoop.hive.metastore.model.MFieldSchema" is tagged as "embedded-only" so does not have its own datastore table.
16/03/22 18:33:44 INFO DataNucleus.Datastore: The class "org.apache.hadoop.hive.metastore.model.MOrder" is tagged as "embedded-only" so does not have its own datastore table.
16/03/22 18:33:48 INFO DataNucleus.Datastore: The class "org.apache.hadoop.hive.metastore.model.MFieldSchema" is tagged as "embedded-only" so does not have its own datastore table.
16/03/22 18:33:48 INFO DataNucleus.Datastore: The class "org.apache.hadoop.hive.metastore.model.MOrder" is tagged as "embedded-only" so does not have its own datastore table.
16/03/22 18:33:50 INFO DataNucleus.Datastore: The class "org.apache.hadoop.hive.metastore.model.MResourceUri" is tagged as "embedded-only" so does not have its own datastore table.
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/usr/lib/spark/python/pyspark/sql/context.py", line 565, in table
    return DataFrame(self._ssql_ctx.table(tableName), self)
  File "/usr/lib/spark/python/lib/py4j-0.8.2.1-src.zip/py4j/java_gateway.py", line 538, in __call__
  File "/usr/lib/spark/python/pyspark/sql/utils.py", line 36, in deco
    return f(*a, **kw)
  File "/usr/lib/spark/python/lib/py4j-0.8.2.1-src.zip/py4j/protocol.py", line 300, in get_return_value
py4j.protocol.Py4JJavaError: An error occurred while calling o22.table.
: org.apache.spark.sql.catalyst.analysis.NoSuchTableException
    at org.apache.spark.sql.hive.client.ClientInterface$$anonfun$getTable$1.apply(ClientInterface.scala:123)
    at org.apache.spark.sql.hive.client.ClientInterface$$anonfun$getTable$1.apply(ClientInterface.scala:123)
    at scala.Option.getOrElse(Option.scala:120)
    at org.apache.spark.sql.hive.client.ClientInterface$class.getTable(ClientInterface.scala:123)
    at org.apache.spark.sql.hive.client.ClientWrapper.getTable(ClientWrapper.scala:60)
    at org.apache.spark.sql.hive.HiveMetastoreCatalog.lookupRelation(HiveMetastoreCatalog.scala:406)
    at org.apache.spark.sql.hive.HiveContext$$anon$1.org$apache$spark$sql$catalyst$analysis$OverrideCatalog$$super$lookupRelation(HiveContext.scala:422)
    at org.apache.spark.sql.catalyst.analysis.OverrideCatalog$$anonfun$lookupRelation$3.apply(Catalog.scala:203)
    at org.apache.spark.sql.catalyst.analysis.OverrideCatalog$$anonfun$lookupRelation$3.apply(Catalog.scala:203)
    at scala.Option.getOrElse(Option.scala:120)
    at org.apache.spark.sql.catalyst.analysis.OverrideCatalog$class.lookupRelation(Catalog.scala:203)
    at org.apache.spark.sql.hive.HiveContext$$anon$1.lookupRelation(HiveContext.scala:422)
    at org.apache.spark.sql.SQLContext.table(SQLContext.scala:739)
    at org.apache.spark.sql.SQLContext.table(SQLContext.scala:735)
    at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
    at sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:57)
    at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:43)
    at java.lang.reflect.Method.invoke(Method.java:606)
    at py4j.reflection.MethodInvoker.invoke(MethodInvoker.java:231)
    at py4j.reflection.ReflectionEngine.invoke(ReflectionEngine.java:379)
    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:207)
    at java.lang.Thread.run(Thread.java:745)

感谢

Answer 1

我们不能将Hive表名直接传递给Hive上下文sql方法，因为它不了解Hive表名。在pyspark shell中读取Hive表的一种方法是：

from pyspark.sql import HiveContext
hive_context = HiveContext(sc)
bank = hive_context.table("default.bank")
bank.show()

在hive表上运行SQL：首先，我们需要注册从读取hive表中获得的数据框。然后我们可以运行SQL查询。

bank.registerTempTable("bank_temp")
hive_context.sql("select * from bank_temp").show()

Answer 2

SparkSQL附带了自己的Metastore（derby），因此即使系统上没有安装hive，它也能正常工作。这是默认模式。

在上面的问题中，您在配置单元中创建了一个表。您收到table not found错误，因为SparkSQL正在使用其默认的Metastore，它没有您的hive表的元数据。

如果您希望SparkSQL使用hive Metastore并访问hive表，则必须在spark conf文件夹中添加hive-site.xml。

Answer 3

在我的问题上，将hp-site.xml转到你的$ SPARK_HOME / conf，并将cp的mysql-connect-java - * .jar转到你的$ SPARK_HOME / jars，这个解决方案解决了我的问题。

Answer 4

不确定，如果还没有解决，我正在检查带有Livy集成的pyspark内核，这就是我测试hive配置的方法

from pyspark.sql import Row
from pyspark.sql import HiveContext
sqlContext = HiveContext(sc)
test_list = [('A', 25),('B', 20),('C', 25),('D', 18)]
rdd = sc.parallelize(test_list)
people = rdd.map(lambda x: Row(name=x[0], age=int(x[1])))
schemaPeople = sqlContext.createDataFrame(people)
# Register it as a temp table
sqlContext.registerDataFrameAsTable(schemaPeople, "test_table")
sqlContext.sql("show tables").show()


Output:
--------
+--------+----------+-----------+
|database| tableName|isTemporary|
+--------+----------+-----------+
|        |test_table|       true|
+--------+----------+-----------+

 Now one can query it in many different ways,
 1. jupyter kernel(sparkmagic syntax): 
    %%sql 
    SELECT * FROM test_table limit 4
 2. Using default HiveContext:
    sqlContext.sql("Select * from test_table").show()

Answer 5

你可以使用sqlCtx.sql。应将hive-site.xml复制到spark conf路径。

my_dataframe = sqlCtx.sql（“从类别中选择*”） my_dataframe.show（）

查询pyspark中的HIVE表

5 个答案: