用于创建的Spark SQL查询类似于this-
CREATE [TEMPORARY] TABLE [IF NOT EXISTS] [db_name.]table_name
[(col_name1 col_type1 [COMMENT col_comment1], ...)]
USING datasource
[OPTIONS (key1=val1, key2=val2, ...)]
[PARTITIONED BY (col_name1, col_name2, ...)]
[CLUSTERED BY (col_name3, col_name4, ...) INTO num_buckets BUCKETS]
[LOCATION path]
[COMMENT table_comment]
[TBLPROPERTIES (key1=val1, key2=val2, ...)]
[AS select_statement]
其中[x]
表示x
是可选的。如果通过CREATE
sql查询,我希望输出为以下顺序的元组-
(db_name, table_name, [(col1 name, col1 type), (col2 name, col2 type), ...])
那么有什么方法可以使用pyspark sql函数或者需要正则表达式的帮助吗?
如果正则表达式可以帮助正则表达式?
答案 0 :(得分:2)
可以通过通过java_gateway
访问非官方API来完成:
plan = spark_session._jsparkSession.sessionState().sqlParser().parsePlan("CREATE TABLE foobar.test (foo INT, bar STRING) USING json")
print(f"database: {plan.tableDesc().identifier().database().get()}")
print(f"table: {plan.tableDesc().identifier().table()}")
# perhaps there is a better way to convert the schemas, using JSON string hack here
print(f"schema: {StructType.fromJson(json.loads(plan.tableDesc().schema().json()))}")
输出:
database: foobar
table: test
schema: StructType(List(StructField(foo,IntegerType,true),StructField(bar,StringType,true)))
请注意,如果未定义数据库并且应该正确处理Scala选项,database().get()
将会失败。另外,如果使用CREATE TEMPORARY VIEW
,则访问器的名称也不同。命令可以在这里找到
https://github.com/apache/spark/blob/master/sql/core/src/main/scala/org/apache/spark/sql/execution/datasources/ddl.scala#L38
https://github.com/apache/spark/blob/master/sql/core/src/main/scala/org/apache/spark/sql/execution/datasources/ddl.scala#L58