无法使用PySpark插入SQL，但可以在SQL中使用 [英] Cannot Insert into SQL using PySpark, but works in SQL

问题描述

我使用以下代码在SQL中创建了一个表:

I have created a table below in SQL using the following:

CREATE TABLE [dbo].[Validation](
    [RuleId] [int] IDENTITY(1,1) NOT NULL,
    [AppId] [varchar](255) NOT NULL,
    [Date] [date] NOT NULL,
    [RuleName] [varchar](255) NOT NULL,
    [Value] [nvarchar](4000) NOT NULL
)

注意身份密钥(RuleId)

NOTE the identity key (RuleId)

在SQL中按如下所示将值插入表中时，它会起作用:

When inserting values into the table as below in SQL it works:

注意:如果表为空且递增，则不按原样插入主键会自动填充

Note: Not inserting the Primary Key as is will autofill if table is empty and increment

INSERT INTO dbo.Validation VALUES ('TestApp','2020-05-15','MemoryUsageAnomaly','2300MB')

但是，在数据块上创建临时表并在下面的PySpark上运行以下查询时，执行相同的查询时，如下所示:

However when creating a temp table on databricks and executing the same query below running this query on PySpark as below:

       %python

        driver = <Driver>
        url = "jdbc:sqlserver:<URL>"
        database = "<db>"
        table = "dbo.Validation"
        user = "<user>"
        password = "<pass>"

        #import the data
        remote_table = spark.read.format("jdbc")\
        .option("driver", driver)\
        .option("url", url)\
        .option("database", database)\
        .option("dbtable", table)\
        .option("user", user)\
        .option("password", password)\
        .load()

        remote_table.createOrReplaceTempView("YOUR_TEMP_VIEW_NAMES")

        sqlcontext.sql("INSERT INTO YOUR_TEMP_VIEW_NAMES VALUES ('TestApp','2020-05-15','MemoryUsageAnomaly','2300MB')")

我收到以下错误:

AnalysisException:'未知要求插入的数据与目标表具有相同的列数:目标表具有5列，但是插入的数据具有4列，包括0个分区列)具有恒定值.'

AnalysisException: 'unknown requires that the data to be inserted have the same number of columns as the target table: target table has 5 column(s) but the inserted data has 4 column(s), including 0 partition column(s) having constant value(s).;'

为什么它在SQL上有效，但在通过数据块传递查询时却不起作用?如何通过pyspark插入而不会出现此错误?

Why does it work on SQL but not when passing the query through databricks? How can I insert through pyspark without getting this error?

推荐答案

这里最直接的解决方案是使用Scala单元中的JDBC.EG

The most straightforward solution here is use JDBC from a Scala cell. EG

%scala

import java.util.Properties
import java.sql.DriverManager

val jdbcUsername = dbutils.secrets.get(scope = "kv", key = "sqluser")
val jdbcPassword = dbutils.secrets.get(scope = "kv", key = "sqlpassword")
val driverClass = "com.microsoft.sqlserver.jdbc.SQLServerDriver"

// Create the JDBC URL without passing in the user and password parameters.
val jdbcUrl = s"jdbc:sqlserver://xxxx.database.windows.net:1433;database=AdventureWorks;encrypt=true;trustServerCertificate=false;hostNameInCertificate=*.database.windows.net;loginTimeout=30;"

// Create a Properties() object to hold the parameters.

val connectionProperties = new Properties()

connectionProperties.put("user", s"${jdbcUsername}")
connectionProperties.put("password", s"${jdbcPassword}")
connectionProperties.setProperty("Driver", driverClass)

val connection = DriverManager.getConnection(jdbcUrl, jdbcUsername, jdbcPassword)
val stmt = connection.createStatement()
val sql = "INSERT INTO dbo.Validation VALUES ('TestApp','2020-05-15','MemoryUsageAnomaly','2300MB')"

stmt.execute(sql)
connection.close()

您也可以使用pyodbc，但是默认情况下未安装SQL Server ODBC驱动程序，而JDBC驱动程序已安装.

You could use pyodbc too, but the SQL Server ODBC drivers aren't installed by default, and the JDBC drivers are.

Spark解决方案是在SQL Server中创建一个视图，然后针对该视图进行插入.例如

A Spark solution would be to create a view in SQL Server and insert against that. eg

create view Validation2 as
select AppId,Date,RuleName,Value
from Validation

然后

tableName = "Validation2"
df = spark.read.jdbc(url=jdbcUrl, table=tableName, properties=connectionProperties)
df.createOrReplaceTempView(tableName)
sqlContext.sql("INSERT INTO Validation2 VALUES ('TestApp','2020-05-15','MemoryUsageAnomaly','2300MB')")

如果要封装Scala并从另一种语言(例如Python)调用它，则可以使用Scala 包装单元格.

If you want to encapsulate the Scala and call it from another language (like Python), you can use a scala package cell.

例如

%scala

package example

import java.util.Properties
import java.sql.DriverManager

object JDBCFacade 
{
  def runStatement(url : String, sql : String, userName : String, password: String): Unit = 
  {
    val connection = DriverManager.getConnection(url, userName, password)
    val stmt = connection.createStatement()
    try
    {
      stmt.execute(sql)  
    }
    finally
    {
      connection.close()  
    }
  }
}

然后您可以这样称呼它:

and then you can call it like this:

jdbcUsername = dbutils.secrets.get(scope = "kv", key = "sqluser")
jdbcPassword = dbutils.secrets.get(scope = "kv", key = "sqlpassword")

jdbcUrl = "jdbc:sqlserver://xxxx.database.windows.net:1433;database=AdventureWorks;encrypt=true;trustServerCertificate=false;hostNameInCertificate=*.database.windows.net;loginTimeout=30;"

sql = "select 1 a into #foo from sys.objects"

sc._jvm.example.JDBCFacade.runStatement(jdbcUrl,sql, jdbcUsername, jdbcPassword)

这篇关于无法使用PySpark插入SQL，但可以在SQL中使用的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！