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我正在使用 pyspark==2.4.3，我只想运行一个 hql 文件

use myDatabaseName;
show tables;

这就是我尝试过的

from os.path import expanduser, join, abspath

from pyspark.sql import SparkSession
from pyspark.sql import Row

# warehouse_location points to the default location for managed databases and tables
warehouse_location = abspath('spark-warehouse')

spark = SparkSession \
    .builder \
    .appName("Python Spark SQL Hive integration example") \
    .config("spark.sql.warehouse.dir", warehouse_location) \
    .enableHiveSupport() \
    .getOrCreate()

with open('full/path/to/my/hqlfile') as t:
    q=t.read()

print q
'use myDatabaseName;show tables;\n'
spark.sql(q)

但我明白了

Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/some/path/python2.7/site-packages/pyspark/sql/session.py", line 767, in sql
    return DataFrame(self._jsparkSession.sql(sqlQuery), self._wrapped)
  File "/some/path/python2.7/site-packages/py4j/java_gateway.py", line 1257, in __call__
    answer, self.gateway_client, self.target_id, self.name)
  File "/some/path/python2.7/site-packages/pyspark/sql/utils.py", line 73, in deco
    raise ParseException(s.split(': ', 1)[1], stackTrace)
pyspark.sql.utils.ParseException: u"\nmismatched input ';' expecting <EOF>(line 1, pos 11)\n\n== SQL ==\nuse myDatabaseName;show tables;\n-----------^^^\n"

我究竟做错了什么 ？

4

1 回答 1

2

就像建议的错误一样，;spark.sql 中的语法无效，

其次，您不能在一个 spark.sql 调用中调用两个命令。

我将把它修改q为一个查询字符串列表，;然后再for循环。

query_lt = q.split(";")[:-1]
for qs in query_lt:
    spark.sql(qs)
于 2020-01-26T00:36:36.100 回答