CODEWITHSUNDEEP
sqlapache-sparkpysparkapache-spark-sqldate-format
Fisher Coder
Fisher Coder

Reputation: 3576

How to format date in Spark SQL?

I need to transform this given date format: 2019-10-22 00:00:00 to this one: 2019-10-22T00:00:00.000Z

I know this could be done in some DB via:

In AWS Redshift, you can achieve this using the following:

TO_DATE('{RUN_DATE_YYYY/MM/DD}', 'YYYY/MM/DD') || 'T00:00:00.000Z' AS VERSION_TIME

But my platform is Spark SQL, so neither above two work for me, the best I could get is using this:

concat(d2.VERSION_TIME, 'T00:00:00.000Z') as VERSION_TIME

which is a bit hacky, but still not completely correct, with this, I got this date format: 2019-10-25 00:00:00T00:00:00.000Z, but this part 00:00:00 in the middle of the string is redundant and I cannot leave it there.

Anyone has any insight here would be greatly appreciated!

Upvotes: 9

Views: 38802

Answers (2)

Lamanus
Lamanus

Reputation: 13541

This is the natural way I think.

spark.sql("""SELECT date_format(to_timestamp("2019-10-22 00:00:00", "yyyy-MM-dd HH:mm:ss"), "yyyy-MM-dd'T'HH:mm:ss.SSS'Z'") as date""").show(false)

The result is:

+------------------------+
|date                    |
+------------------------+
|2019-10-22T00:00:00.000Z|
+------------------------+

Upvotes: 10

chlebek
chlebek

Reputation: 2451

Maybe something like this? It's a bit different approach.

scala> val df = spark.range(1).select(current_date.as("date"))
scala> df.show()
+----------+
|      date|
+----------+
|2019-11-09|
+----------+

scala> 

df.withColumn("formatted",
    concat(
    regexp_replace(date_format('date,"yyyy-MM-dd\tHH:mm:ss.SSS"),"\t","T"),
    lit("Z")
    )
).show(false)

+----------+------------------------+
|date      |formatted               |
+----------+------------------------+
|2019-11-09|2019-11-09T00:00:00.000Z|
+----------+------------------------+

Upvotes: 3

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