我正在尝试在SPARK SQL中编写一个查询,执行三个表的连接。但查询输出实际上是null
。它适用于单个表。我的 Join 查询是正确的,因为我已经在 oracle 数据库中执行了它。我需要在这里应用什么更正?Spark版本是2.0.0。
from pyspark.sql import SQLContext, Row
sqlContext = SQLContext(sc)
lines = sc.textFile("/Users/Hadoop_IPFile/purchase")
lines2 = sc.textFile("/Users/Hadoop_IPFile/customer")
lines3 = sc.textFile("/Users/Hadoop_IPFile/book")
parts = lines.map(lambda l: l.split("t"))
purchase = parts.map(lambda p: Row(year=p[0],cid=p[1],isbn=p[2],seller=p[3],price=int(p[4])))
schemapurchase = sqlContext.createDataFrame(purchase)
schemapurchase.registerTempTable("purchase")
parts2 = lines.map(lambda l: l.split("t"))
customer = parts2.map(lambda p: Row(cid=p[0],name=p[1],age=p[2],city=p[3],sex=p[4]))
schemacustomer = sqlContext.createDataFrame(customer)
schemacustomer.registerTempTable("customer")
parts3 = lines.map(lambda l: l.split("t"))
book = parts3.map(lambda p: Row(isbn=p[0],name=p[1]))
schemabook = sqlContext.createDataFrame(book)
schemabook.registerTempTable("book")
result_purchase = sqlContext.sql("""SELECT DISTINCT customer.name AS name FROM purchase JOIN book ON purchase.isbn = book.isbn JOIN customer ON customer.cid = purchase.cid WHERE customer.name != 'Harry Smith' AND purchase.isbn IN (SELECT purchase.isbn FROM customer JOIN purchase ON customer.cid = purchase.cid WHERE customer.name = 'Harry Smith')""")
result = result_purchase.rdd.map(lambda p: "name: " + p.name).collect()
for name in result:
print(name)
DataSet
---------
Purchase
1999 C1 B1 Amazon 90
2001 C1 B2 Amazon 20
2008 C2 B2 Barnes Noble 30
2008 C3 B3 Amazon 28
2009 C2 B1 Borders 90
2010 C4 B3 Barnes Noble 26
Customer
C1 Jackie Chan 50 Dayton M
C2 Harry Smith 30 Beavercreek M
C3 Ellen Smith 28 Beavercreek F
C4 John Chan 20 Dayton M
Book
B1 Novel
B2 Drama
B3 Poem
我在某个网页中找到了以下说明,但它仍然不起作用:
schemapurchase.join(schemabook, schemapurchase.isbn == schemabook.isbn)
schemapurchase.join(schemacustomer, schemapurchase.cid == schemacustomer.cid)
给定这个输入数据帧,就像在你的例子中一样(抱歉,如果某些列名是错误的,我猜到了(:
购买:
+----+---+----+------------+-----+
|year|cid|isbn| shop|price|
+----+---+----+------------+-----+
|1999| C1| B1| Amazon| 90|
|2001| C1| B2| Amazon| 20|
|2008| C2| B2|Barnes Noble| 30|
|2008| C3| B3| Amazon| 28|
|2009| C2| B1| Borders| 90|
|2010| C4| B3|Barnes Noble| 26|
+----+---+----+------------+-----+
客户:
+---+-----------+---+-----------+-----+
|cid| name|age| city|genre|
+---+-----------+---+-----------+-----+
| C1|Jackie Chan| 50| Dayton| M|
| C2|Harry Smith| 30|Beavercreek| M|
| C3|Ellen Smith| 28|Beavercreek| F|
| C4| John Chan| 20| Dayton| M|
+---+-----------+---+-----------+-----+
书:
+----+-----+
|isbn|genre|
+----+-----+
| B1|Novel|
| B2|Drama|
| B3| Poem|
+----+-----+
可以使用数据帧函数转换该 sql 查询,如下所示:
val result = purchase.join(book, purchase("isbn")===book("isbn"))
.join(customer, customer("cid")===purchase("cid"))
.where(customer("name") !== "Harry Smith")
.join(temp, purchase("isbn")===temp("purchase_isbn"))
.select(customer("name").as("NAME")).distinct()
其中">temp">是"SELECT IN">的结果,可以认为是另一个连接的结果:
val temp = customer.join(purchase, customer("cid")===purchase("cid") )
.where(customer("name")==="Harry Smith")
.select(purchase("isbn").as("purchase_isbn"))
+-------------+
|purchase_isbn|
+-------------+
| B2|
| B1|
+-------------+
所以最终的结果是:
+-----------+
| NAME|
+-----------+
|Jackie Chan|
+-----------+
将此答案视为您可以开始思考的要点(例如,过多的连接会对性能产生不良影响(。