如何从 Spark SQL Query [PySpark] 中获取表名？

Question

要从 SQL 查询中获取表名，

select *
from table1 as t1
full outer join table2 as t2
  on t1.id = t2.id

我在 Scala How to get table names from SQL 查询中找到了解决方案？

def getTables(query: String): Seq[String] = {
  val logicalPlan = spark.sessionState.sqlParser.parsePlan(query)
  import org.apache.spark.sql.catalyst.analysis.UnresolvedRelation
  logicalPlan.collect { case r: UnresolvedRelation => r.tableName }
}

当我遍历返回序列getTables(query).foreach(println)时，这给了我正确的表名

table1
table2

PySpark 的等效语法是什么？ 我遇到的最接近的是How to extract column name and column type from SQL in pyspark

plan = spark_session._jsparkSession.sessionState().sqlParser().parsePlan(query)
print(f"table: {plan.tableDesc().identifier().table()}")

回溯失败

Py4JError: An error occurred while calling o78.tableDesc. Trace:
py4j.Py4JException: Method tableDesc([]) does not exist
    at py4j.reflection.ReflectionEngine.getMethod(ReflectionEngine.java:318)
    at py4j.reflection.ReflectionEngine.getMethod(ReflectionEngine.java:326)
    at py4j.Gateway.invoke(Gateway.java:274)
    at py4j.commands.AbstractCommand.invokeMethod(AbstractCommand.java:132)
    at py4j.commands.CallCommand.execute(CallCommand.java:79)
    at py4j.GatewayConnection.run(GatewayConnection.java:238)
    at java.base/java.lang.Thread.run(Thread.java:835)

我了解，问题源于我需要过滤所有UnresolvedRelation类型的计划项目，但我在 python/pyspark 中找不到等效符号

Answer 1

我有一个方法，但相当复杂。 It dumps the Java Object and JSON (a poor man's serialization process), deserializes it to python object, filter and parse the table names

import json
def get_tables(query: str):
    plan = spark._jsparkSession.sessionState().sqlParser().parsePlan(query)
    plan_items = json.loads(plan.toJSON())
    for plan_item in plan_items:
        if plan_item['class'] == 'org.apache.spark.sql.catalyst.analysis.UnresolvedRelation':
            yield plan_item['tableIdentifier']['table']

当我遍历 function list(get_tables(query))时，会产生['fast_track_gv_nexus', 'buybox_gv_nexus']

注意不幸的是，这对CTE 不利

例子

with delta as (
   select *
    group by id
    cluster by id
 )
select   *
  from ( select  *
         FROM
          (select   *
            from dmm
            inner join delta on dmm.id = delta.id
           )
  )

为了解决它，我必须通过正则表达式来破解

import json
import re
def get_tables(query: str):
    plan = spark._jsparkSession.sessionState().sqlParser().parsePlan(query)
    plan_items = json.loads(plan.toJSON())
    plan_string = plan.toString()
    cte = re.findall(r"CTE \[(.*?)\]", plan_string)
    for plan_item in plan_items:
        if plan_item['class'] == 'org.apache.spark.sql.catalyst.analysis.UnresolvedRelation':
            tableIdentifier = plan_item['tableIdentifier']
            table =  plan_item['tableIdentifier']['table']   
            database =  tableIdentifier.get('database', '')
            table_name = "{}.{}".format(database, table) if database else table
            if table_name not in cte:
                yield table_name