WITH语句不是问题所在，而是INSERT INTO语句造成了问题。
下面是一个使用.insertInto（）样式而不是“INSERT INTO”SQL的工作示例：

val s = Seq((1,"foo"), (2, "bar"))
s: Seq[(Int, String)] = List((1,foo), (2,bar))
val df = s.toDF("id", "name")
df.registerTempTable("df")
sql("CREATE TABLE edf_final (id int, name string)")
val e = sql("WITH edf AS (SELECT id+1, name FROM df cook) SELECT * FROM edf")
e.insertInto("edf_final")

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另一个选项是使用df.write.mode("append").saveAsTable("edf_final")样式。
相关SO："INSERT INTO ..." with SparkSQL HiveContext

因为答案是针对Spark 2.x的，所以我以Spark 3的方式重写它。

%scala

import org.apache.spark.sql.functions.col

val s = Seq((1,"foo"), (2, "bar"))
val df = s.toDF("id", "name")
df.createOrReplaceTempView("df")
spark.sql("CREATE TABLE if not exists edf_final (id int, name string)")
val e = spark.sql("WITH edf AS (SELECT id+1 AS id, name FROM df) SELECT * FROM edf")
e.select(col("id"), col("name")).write.insertInto("edf_final")

spark.sql("select * from edf_final").show

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测试结果：

df:org.apache.spark.sql.DataFrame
id:integer
name:string
e:org.apache.spark.sql.DataFrame
id:integer
name:string
+---+----+
| id|name|
+---+----+
|  2| foo|
|  3| bar|
+---+----+

型