现在已经很晚了，但是今天我尝试使用PySpark SQL实现cte递归查询。
这里，我有一个简单的 Dataframe ，我想做的是找到每个ID的最新ID。
原始 Dataframe ：

+-----+-----+
|OldID|NewID|
+-----+-----+
|    1|    2|
|    2|    3|
|    3|    4|
|    4|    5|
|    6|    7|
|    7|    8|
|    9|   10|
+-----+-----+

我想要的结果：

+-----+-----+
|OldID|NewID|
+-----+-----+
|    1|    5|
|    2|    5|
|    3|    5|
|    4|    5|
|    6|    8|
|    7|    8|
|    9|   10|
+-----+-----+

下面是我的代码：

df = sqlContext.createDataFrame([(1, 2), (2, 3), (3, 4), (4, 5), (6, 7), (7, 8),(9, 10)], "OldID integer,NewID integer").checkpoint().cache()

dfcheck = df.drop('NewID')
dfdistinctID = df.select('NewID').distinct()
dfidfinal = dfdistinctID.join(dfcheck, [dfcheck.OldID == dfdistinctID.NewID], how="left_anti") #We find the IDs that have not been replaced

dfcurrent = df.join(dfidfinal, [dfidfinal.NewID == df.NewID], how="left_semi").checkpoint().cache() #We find the the rows that are related to the IDs that have not been replaced, then assign them to the dfcurrent dataframe.
dfresult = dfcurrent
dfdifferentalias = df.select(df.OldID.alias('id1'), df.NewID.alias('id2')).checkpoint().cache()

while dfcurrent.count() > 0:
  dfcurrent = dfcurrent.join(broadcast(dfdifferentalias), [dfcurrent.OldID == dfdifferentalias.id2], how="inner").select(dfdifferentalias.id1.alias('OldID'), dfcurrent.NewID.alias('NewID')).cache()
  dfresult = dfresult.unionAll(dfcurrent)

display(dfresult.orderBy('OldID'))

Databricks notebook screenshot
我知道表演很糟糕，但至少，它给予了我需要的答案。
这是我第一次发布StackOverflow的答案，所以如果我犯了任何错误，请原谅我。

使用SPARK SQL是不可能的，WITH子句是存在的，但不适用于CONNECT BY，比如ORACLE，或者DB2中的递归。

The Spark documentation提供了“CTE定义中的CTE”。复制如下：

-- CTE in CTE definition
WITH t AS (
    WITH t2 AS (SELECT 1)
    SELECT * FROM t2
)
SELECT * FROM t;
+---+
|  1|
+---+
|  1|
+---+

您可以将其扩展到多个嵌套查询，但语法很快就会变得很笨拙。我的建议是使用注解来明确下一个select语句的来源。本质上，从第一个查询开始，并根据需要在其上方和下方放置其他CTE语句：

WITH t3 AS (
WITH t2 AS (
WITH t1 AS (SELECT distinct b.col1
            FROM data_a as a, data_b as b
            WHERE a.col2 = b.col2
            AND a.col3 = b.col3
-- select from t1
            )
            SELECT distinct b.col1, b.col2, b.col3
            FROM t1 as a, data_b as b
            WHERE a.col1 = b.col1
-- select from t2
            )
            SELECT distinct b.col1
            FROM t2 as a, data_b as b
            WHERE a.col2 = b.col2
            AND a.col3 = b.col3
-- select from t3
            )
            SELECT distinct b.col1, b.col2, b.col3
            FROM t3 as a, data_b as b
            WHERE a.col1 = b.col1;

你可以递归地使用createOrReplaceTempView来构建一个递归查询，它不会很快，也不会很漂亮，但是它很有效，下面是@Prade的例子，PySpark：

def recursively_resolve(df):
    rec = df.withColumn('level', F.lit(0))
    
    sql = """
        select this.oldid
             , coalesce(next.newid, this.newid) as newid
             , this.level + case when next.newid is not null then 1 else 0 end as level
             , next.newid is not null as is_resolved
          from rec this
          left outer
          join rec next
            on next.oldid = this.newid
    """
    find_next = True
    while find_next:
        rec.createOrReplaceTempView("rec")
        rec = spark.sql(sql)
        # check if any rows resolved in this iteration
        # go deeper if they did
        find_next = rec.selectExpr("ANY(is_resolved = True)").collect()[0][0]
        
    return rec.drop('is_resolved')

然后：

src = spark.createDataFrame([(1, 2), (2, 3), (3, 4), (4, 5), (6, 7), (7, 8),(9, 10)], "OldID integer,NewID integer")
result = recursively_resolve(src)
result.show()

图纸：

+-----+-----+-----+
|oldid|newid|level|
+-----+-----+-----+
|    2|    5|    2|
|    4|    5|    0|
|    3|    5|    1|
|    7|    8|    0|
|    6|    8|    1|
|    9|   10|    0|
|    1|    5|    2|
+-----+-----+-----+