解析&；使用Spark&；在文本文件中展平JSON对象；将Scala转换为数据帧

我有一个结构如下的文本文件

(employeeID: Int, Name: String, ProjectDetails: JsonObject{[{ProjectName, Description, Duriation, Role}]})

例如：

有人能帮我用scala解析和展平数据帧下面的记录吗

employeeID, Name, ProjectName, Description, Duration, Role
123456, Employee1, Web Develoement, Online Sales website, 6 Months , Developer
123456, Employee1, Spark Develoement, Online Sales Analysis, 6 Months, Data Engineer
123456, Employee1, Scala Training, Training, 1 Month, null

---

你可以试试这个。。但是稍微修改了输入结构，因为前两列不是Json格式

scala>import org.apache.spark.SparkConf
scala>导入org.apache.spark.SparkContext
scala>import org.apache.spark.sql.SQLContext
scala>import org.apache.spark.sql_
scala>val sqlSC=neworg.apache.spark.sql.SQLContext（sc）
scala>导入sqlSC.implicits_
scala>val emp_DF=sqlSC.jsonFile（“file:///C:/Users/ABCD/Desktop/Examples/Spark/Mailing List/Employee_嵌套_Projects.json”）
scala>case类ProjectInfo（ProjectName:String，Description:String，Duration:String，Role:String）
scala>case类项目（employeeID:Int，Name:String，ProjectDetails:Seq[ProjectInfo]）
scala>val emp_projects_DF=emp_DF.explode（emp_DF（“ProjectDetails”））{
案例行（x:Seq[Row]）=>x.map（x=>ProjectInfo（x（0）.asInstanceOf[String]，x（1）.asInstanceOf[String]，x（2）.asInstanceOf[String]，x（3）.asInstanceOf[String]））
scala>emp\u projects\u DF.选择（$“employeeID”、$“Name”、$“ProjectName”、$“Description”、$“Duration”、$“Role”）.show（）

employeeID, Name, ProjectName, Description, Duration, Role
123456, Employee1, Web Develoement, Online Sales website, 6 Months , Developer
123456, Employee1, Spark Develoement, Online Sales Analysis, 6 Months, Data Engineer
123456, Employee1, Scala Training, Training, 1 Month, null