I am using local windows and trying to load the XML
file with the following code on python, and i am having this error, do anyone knows how to resolve it,
I've commit a similar but slightly different error: forgot the "s3://" prefix to file path. After adding this prefix to form "s3://path/to/object" the following code works:
my_data = spark.read.format("com.databricks.spark.csv")\
.option("header", "true")\
.option("inferSchema", "true")\
.option("delimiter", ",")\
.load("s3://path/to/object")
The error message says it all: you cannot use dataframe reader & load
to access files on the web (http
or htpps
). I suggest you first download the file locally.
See the pyspark.sql.DataFrameReader docs for more on the available sources (in general, local file system, HDFS, and databases via JDBC).
Irrelevantly to the error, notice that you seem to use the format
part of the command incorrectly: assuming that you use the XML Data Source for Apache Spark package, the correct usage should be format('com.databricks.spark.xml')
(see the example).
Somehow pyspark is unable to load the http or https, one of my colleague found the answer for this so here is the solution,
before creating the spark context and sql context we need to load this two line of code
import os
os.environ['PYSPARK_SUBMIT_ARGS'] = '--packages com.databricks:spark-xml_2.11:0.4.1 pyspark-shell'
after creating the sparkcontext and sqlcontext from sc = pyspark.SparkContext.getOrCreate
and sqlContext = SQLContext(sc)
add the http or https url into the sc by using sc.addFile(url)
Data_XMLFile = sqlContext.read.format("xml").options(rowTag="anytaghere").load(pyspark.SparkFiles.get("*_public.xml")).coalesce(10).cache()
this solution worked for me