How to extract urls from HYPERLINKS in excel file when reading into scala spark dataframe - excel

Question

I have an Excel file with Column A containing HYPERLINKS like this:
=HYPERLINK("https://google.com","View Link")
I can load the Excel file in scala spark dataframe using com.crealytics.spark.excel library but only with the 'View Link' text which DOES NOT contain the url
import org.apache.spark.sql._
import org.apache.spark.sql.types._
object Tut {
def main(args: Array[String]): Unit = {
println("started")
val spark = SparkSession
.builder()
.appName("MySpark")
.config("spark.master", "local")
.getOrCreate()
val customSchema = StructType(Array(
StructField("A", StringType, nullable = false),
StructField("B", IntegerType, nullable = false)))
val df = spark.read.format("com.crealytics.spark.excel")
.option("useHeader", "true").schema(customSchema)
.option("dataAddress", "A1")
.load("/MY_PATH/src/main/resources/SampFile.xlsx")
df.printSchema()
df.show()
}
}
My goal is to load the entire content of the HYPERLINK as a string:
=HYPERLINK("https://google.com","View Link")
and then extract the url
https://google.com.
Do you know if there is a way to do this using com.crealytics.spark.excel library or any other spark library? Thanks in advance!

Answer 1

About the other question link you provided in the comments, they're trying to read the column as BinaryType, and cast it out of the box into StringType, well, such thing is not possible (even in scala itself), since you need to know how to read the bytes and represent it as a human readable string, right? for instance the encoding, etc.
Now we know that we need to define a custom approach. I used a sample in-code dataframe, and this approach worked:
scala> import spark.implicits._
import spark.implicits._
scala> val df = Seq(
| ("ddd".getBytes, 1)
| ).toDF("A", "B")
df: org.apache.spark.sql.DataFrame = [A: binary, B: int]
scala> val btos: Array[Byte] => String = bytes => new String(bytes) // short fot bytes to string
btos: Array[Byte] => String = $Lambda$2322/665683021#738f6e44
scala> spark.udf.register("btos", btos)
res0: org.apache.spark.sql.expressions.UserDefinedFunction = SparkUserDefinedFunction($Lambda$2322/665683021#738f6e44,StringType,List(Some(class[value[0]: binary])),Some(btos),true,true)
scala> df.withColumn("C", expr("btos(A)")).show
+----------+---+---+
| A| B| C|
+----------+---+---+
|[64 64 64]| 1|ddd|
+----------+---+---+
Hope this works for you.