Reputation: 73
I have created the following DataFrame:
val data = spark.sparkContext.parallelize(Seq(("SnVsZXMgTmV3b25l"), ("Jason Kidd"), ("TXIgUm9uYWxkIE0=")))
val df_data = data.toDF()
val decoded_got = df_data.withColumn("xxx", unbase64(col("value")).cast("String"))
And I get the following:
+----------------+------------+
|name |xxx |
+----------------+------------+
|SnVsZXMgTmV3b25l|Jules Newone|
|Jason Kidd |%�(��� |
|TXIgUm9uYWxkIE0=|Mr Ronald M |
+----------------+------------+
What I want to do is avoid the values of the column name that are not in base 64. For example, get the following Df
:
+----------------+------------+
|name |xxx |
+----------------+------------+
|SnVsZXMgTmV3b25l|Jules Newone|
|Jason Kidd |Jason Kidd |
|TXIgUm9uYWxkIE0=|Mr Ronald M |
+----------------+------------+
I am trying something like this but is not working for me:
val regex1 = """^([A-Za-z0-9+/]{4})*([A-Za-z0-9+/]{3}=|[A-Za-z0-9+/]{2}==)?$"""
val check = df_data.withColumn("xxx", when(regex1 matches col("value"), unbase64(col("value"))).otherwise(col("value")))
Is there an option in Spark Scala to check if the value is in base64 or how could I do this?
Upvotes: 2
Views: 1961
Reputation: 32690
To check whether the value is a valid base64 encoded string or not, you can try to decode it and encode it again, you should get the initial value. If not, then it's not a base64 string:
val decoded_got = df_data.withColumn(
"xxx",
when(
base64(unbase64(col("value"))) === col("value"),
unbase64(col("value")).cast("string")
).otherwise(col("value"))
)
decoded_got.show
//+----------------+------------+
//| value| xxx|
//+----------------+------------+
//|SnVsZXMgTmV3b25l|Jules Newone|
//| Jason Kidd| Jason Kidd|
//|TXIgUm9uYWxkIE0=| Mr Ronald M|
//+----------------+------------+
Upvotes: 1