Question

我想知道如何从全名解析出名字，姓氏，同时考虑在中间删除中间名，先生，女士和JR，最后是JR。

            private def loaduserdata(spark: SparkSession): Unit = {
          import spark.implicits._
           // Create an RDD of User objects from a text file, convert it to a Dataframe
          val userDF = spark.sparkContext
            .textFile("examples/src/main/resources/users.csv")
            .map(_.split("::"))
            .map(attributes => users(attributes(0).trim.toInt,  attributes(1), attributes(2).trim.toInt, attributes(3), attributes(4)))
            .toDF()
          // Register the DataFrame as a temporary view
          userDF.createOrReplaceTempView("Users")
          // SQL statements can be run by using the sql methods provided by Spark
         val zipcodesDF = spark.sql("SELECT distinct zipcode, substr(zipcode,1,5) as state FROM Users ORDER BY zipcode desc") //  zipcodesDF.map(zipcodes => "zipcode: " + zipcodes.getAs[String]("zipcode") + getstate(zipcodes.getAs[String]("zipcode"))).show()
        val colNames = zipcodesDF.columns
      val cols = colNames.map(cName => zipcodesDF.col(cName))
      val theColumn = zipcodesDF("state")
      val mappedCols = cols.map(c => 
        if (c.toString() == theColumn.toString()) getstate(c).as("state") else c)
        val geoDF = zipcodesDF.select(mappedCols:_*)//.show()
      geoDF.createOrReplaceTempView("Geo")
        }
       val getstate = udf {(zipcode: String) => 
           val url = "http://maps.googleapis.com/maps/api/geocode/json?address="+zipcode
           val result = scala.io.Source.fromURL(url).mkString
           val address = parse(result) 
          val statenm = for {
                      JObject(statename) <- address
                      JField("types", JArray(types)) <- statename
                      JField("short_name", JString(short_name)) <- statename
                     if types.toString().equals("List(JString(administrative_area_level_1), JString(political))")
                     //  if types.head.equals("JString(administrative_area_level_1)")
                    } yield short_name
           val str = if (statenm.isEmpty.toString().equals("true")) "N/A" else statenm.head        
           }

谢谢

Answer 1

首先选择左（名称，CHARINDEX（＆＃39;＆＃39;，名称））， substring（name，CHARINDEX（＆＃39;＆＃39;，name）+1，len（name） - （CHARINDEX（＆＃39;＆＃39;，name）-1））来自table-name;

Answer 2

我找到了一个名为PARSE_NAME_UDF here

的漂亮函数

我在数据迁移和集成工作中不时使用此功能：

以下是一些返回两列First和Last名称的示例：

  SELECT 'First:' = dbo.PARSE_NAME_UDF('Jon Doe Jr', 'F'), 'Last:' = dbo.PARSE_NAME_UDF('Jon Doe Jr', 'L')
      UNION ALL SELECT 'First:' = dbo.PARSE_NAME_UDF('Jon J Doe', 'F'), 'Last:' = dbo.PARSE_NAME_UDF('Jon J Doe', 'L')
      UNION ALL SELECT 'First:' = dbo.PARSE_NAME_UDF('Jon Doe', 'F'), 'Last:' = dbo.PARSE_NAME_UDF('Jon Doe', 'L')

以下是更多示例：

  SELECT dbo.PARSE_NAME_UDF('Fred J Muggs Jr', 'F M L S') -- Returns Fred J Muggs Junior
  SELECT dbo.PARSE_NAME_UDF('Fred J Muggs Jr', 'F M L s') -- Returns Fred J Muggs Jr
  SELECT dbo.PARSE_NAME_UDF('Fred J Muggs Jr', 'f. M. L s') -- Returns F. J. Muggs Jr

从SQL Server中解析出名字，姓氏

2 个答案: