How do I pass parameters to selectExpr? SparkSQL-Scala - apache-spark-sql

Question

:)
When you have a data frame, you can add columns and fill their rows with the method selectExprt
Something like this:
scala> table.show
+------+--------+---------+--------+--------+
|idempr|tipperrd| codperrd|tipperrt|codperrt|
+------+--------+---------+--------+--------+
| OlcM| h|999999999| J| 0|
| zOcQ| r|777777777| J| 1|
| kyGp| t|333333333| J| 2|
| BEuX| A|999999999| F| 3|
scala> var table2 = table.selectExpr("idempr", "tipperrd", "codperrd", "tipperrt", "codperrt", "'hola' as Saludo")
tabla: org.apache.spark.sql.DataFrame = [idempr: string, tipperrd: string, codperrd: decimal(9,0), tipperrt: string, codperrt: decimal(9,0), Saludo: string]
scala> table2.show
+------+--------+---------+--------+--------+------+
|idempr|tipperrd| codperrd|tipperrt|codperrt|Saludo|
+------+--------+---------+--------+--------+------+
| OlcM| h|999999999| J| 0| hola|
| zOcQ| r|777777777| J| 1| hola|
| kyGp| t|333333333| J| 2| hola|
| BEuX| A|999999999| F| 3| hola|
My point is:
I define strings and call a method which use this String parameter to fill a column in the data frame. But I am not able to do the select expresion get the string (I tried $, +, etc..) . To achieve something like this:
scala> var english = "hello"
scala> def generar_informe(df: DataFrame, tabla: String) {
var selectExpr_df = df.selectExpr(
"TIPPERSCON_BAS as TIP.PERSONA CONTACTABILIDAD",
"CODPERSCON_BAS as COD.PERSONA CONTACTABILIDAD",
"'tabla' as PUNTO DEL FLUJO" )
}
scala> generar_informe(df,english)
.....
scala> table2.show
+------+--------+---------+--------+--------+------+
|idempr|tipperrd| codperrd|tipperrt|codperrt|Saludo|
+------+--------+---------+--------+--------+------+
| OlcM| h|999999999| J| 0| hello|
| zOcQ| r|777777777| J| 1| hello|
| kyGp| t|333333333| J| 2| hello|
| BEuX| A|999999999| F| 3| hello|
I tried:
scala> var result = tabl.selectExpr("A", "B", "$tabla as C")
scala> var abc = tabl.selectExpr("A", "B", ${tabla} as C)
<console>:31: error: not found: value $
var abc = tabl.selectExpr("A", "B", ${tabla} as C)
scala> var abc = tabl.selectExpr("A", "B", "${tabla} as C")
scala> sqlContext.sql("set tabla='hello'")
scala> var abc = tabl.selectExpr("A", "B", "${tabla} as C")
SAME ERROR:
java.lang.RuntimeException: [1.1] failure: identifier expected
${tabla} as C
^
at scala.sys.package$.error(package.scala:27)
Thanks in advance!

Answer 1

Can you try this.
val english = "hello"
generar_informe(data,english).show()
}
def generar_informe(df: DataFrame , english : String)={
df.selectExpr(
"transactionId" , "customerId" , "itemId","amountPaid" , s"""'${english}' as saludo """)
}
This is the output I got.
17/11/02 23:56:44 INFO CodeGenerator: Code generated in 13.857987 ms
+-------------+----------+------+----------+------+
|transactionId|customerId|itemId|amountPaid|saludo|
+-------------+----------+------+----------+------+
| 111| 1| 1| 100.0| hello|
| 112| 2| 2| 505.0| hello|
| 113| 3| 3| 510.0| hello|
| 114| 4| 4| 600.0| hello|
| 115| 1| 2| 500.0| hello|
| 116| 1| 2| 500.0| hello|
| 117| 1| 2| 500.0| hello|
| 118| 1| 2| 500.0| hello|
| 119| 2| 3| 500.0| hello|
| 120| 1| 2| 500.0| hello|
| 121| 1| 4| 500.0| hello|
| 122| 1| 2| 500.0| hello|
| 123| 1| 4| 500.0| hello|
| 124| 1| 2| 500.0| hello|
+-------------+----------+------+----------+------+
17/11/02 23:56:44 INFO SparkContext: Invoking stop() from shutdown hook