Question

我正在尝试使用三个表的连接在SPARK SQL中编写查询。但查询输出为null。它适用于单桌。我的Join查询是正确的，因为我已经在oracle数据库中执行了它。我需要在这里进行哪些更正？ Spark Version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

我在某个网页上找到了以下说明，但它仍然无效：schemapurchase.join（schemabook，schemapurchase.isbn == schemabook.isbn）schemapurchase.join（schemacustomer，schemapurchase.cid == schemacustomer.cid）

Answer 1

在你的例子中给出了这样的输入DataFrame（对不起，如果有些列名错了，我猜对了）：

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|
+----+-----+

您可以使用DataFrame函数翻译该SQL查询，例如：

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＆＃34; temp＆＃34; 是＆＃34; SELECT IN＆＃34; 的结果，可以认为是另一个的结果一个加入：

val temp = customer.join(purchase, customer("cid")===purchase("cid") )
                   .where(customer("name")==="Harry Smith")
                   .select(purchase("isbn").as("purchase_isbn"))    


+-------------+
|purchase_isbn|
+-------------+
|           B2|
|           B1|
+-------------+

所以最后的结果是：

+-----------+
|       NAME|
+-----------+
|Jackie Chan|
+-----------+

将此答案视为您可以开始思考的一个问题（例如，过多的联接会对性能造成不良影响）。

在Spark SQL中加入2个以上的表

1 个答案: