Dear Sirs/Madames
Could anyone here help me to figureout the way to use scala to query an
select SQL against kylin cube via API then turn that table result into a
dataframe in scala for other purpose?
Thank you so much for your time!
Best regards
On Fri, 29 Mar 2024 at 17:52 Nam Đỗ Duy <na...@vnpay.vn> wrote:
> Hi Xiaoxiang,
> Sir & Madames,
>
> I use the following code to query the cube via API but I cannot use the
> result as a dataframe, could you suggest a way to do that because it is
> very important for our project.
>
> Thanks and best regards
>
> ===================================
>
> import org.apache.spark.sql.{DataFrame, SparkSession}
> import org.apache.spark.sql.functions._
>
> object APICaller {
> def main(args: Array[String]): Unit = {
> val spark = SparkSession.builder()
> .appName("APICaller")
> .master("local[*]")
> .getOrCreate()
>
> import spark.implicits._
>
> val username = "namdd"
> val password = "eer123"
> val urlString = "http://localhost:7070/kylin/api/query";
> val project = "learn_kylin"
> val query = "select count(*) from HIVE_DWH_STANDARD.factuserEvent"
>
> val response: String = callAPI(urlString, username, password, project,
> query)
>
> // Convert response to DataFrame
> val df = spark.read.json(Seq(response).toDS())
>
> // Show DataFrame
> df.show()
>
> // Stop Spark session
> spark.stop()
> }
>
> def callAPI(url: String, username: String, password: String, project:
> String, query: String): String = {
> val encodedAuth =
> java.util.Base64.getEncoder.encodeToString(s"$username:$password".getBytes)
>
> val connection = scalaj.http.Http(url)
> .postData(s"""{"project": "$project", "sql": "$query"}""")
> .header("Content-Type", "application/json")
> .header("Accept", "application/json")
> .auth(username, password)
> .asString
>
> if (connection.isError)
> throw new RuntimeException(s"Error calling API: ${connection.body}")
>
> connection.body
> }
> }
>
>
