Spark Structured Streaming from Files on S3/Disk - add batch filename to records/lines?
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I am implementing a spark structured streaming application that processes webserver log files from a folder on disk or perhaps S3.
Spark Structured Streaming fits the use case almost perfectly, with one wrinkle.
The filenames in the folder also contain the machine name eg. like:
/node1_20181101.json.gz
/node1_20181102.json.gz
/node2_20181101.json.gz
/node3_20181102.json.gz
/node4_20181102.json.gz
...and so on.
A (simplified) version of the source looks something like this ( I would turn the below to a continuous stream with windowing etc):
val inputDF = spark.read
.option("codec", classOf[GzipCodec].getName)
.option("maxFilesPerTrigger", 1.toString)
.json(config.directory)
.transform { ds =>
logger.info(ds.inputFiles)
ds
}.foreach(println(_))
I would like to transform the batch and add the node ID from the filename to each record line, - I can't seem to see any kind of an onBatch trigger that I could use to enrich the record schema with the node ID from the file name.
I have looked at the following and nothing seems to fit:
[FileStreamSource][https://jaceklaskowski.gitbooks.io/spark-structured-streaming/spark-sql-streaming-FileStreamSource.html#metadataLog]
Unfortunately getting a handle on the machine name from the file name is key to the analytics I do later, and I have no control over how the logs are populated
Any clues?
scala apache-spark spark-structured-streaming
edited Dec 3 '18 at 20:23
Jacek Laskowski
46.3k18138278
asked Nov 26 '18 at 15:38
ARocheARoche
61
add a comment |
I am implementing a spark structured streaming application that processes webserver log files from a folder on disk or perhaps S3.
Spark Structured Streaming fits the use case almost perfectly, with one wrinkle.
The filenames in the folder also contain the machine name eg. like:
/node1_20181101.json.gz
/node1_20181102.json.gz
/node2_20181101.json.gz
/node3_20181102.json.gz
/node4_20181102.json.gz
...and so on.
A (simplified) version of the source looks something like this ( I would turn the below to a continuous stream with windowing etc):
val inputDF = spark.read
.option("codec", classOf[GzipCodec].getName)
.option("maxFilesPerTrigger", 1.toString)
.json(config.directory)
.transform { ds =>
logger.info(ds.inputFiles)
ds
}.foreach(println(_))
I would like to transform the batch and add the node ID from the filename to each record line, - I can't seem to see any kind of an onBatch trigger that I could use to enrich the record schema with the node ID from the file name.
I have looked at the following and nothing seems to fit:
[FileStreamSource][https://jaceklaskowski.gitbooks.io/spark-structured-streaming/spark-sql-streaming-FileStreamSource.html#metadataLog]
Unfortunately getting a handle on the machine name from the file name is key to the analytics I do later, and I have no control over how the logs are populated
Any clues?
scala apache-spark spark-structured-streaming
edited Dec 3 '18 at 20:23
Jacek Laskowski
46.3k18138278
asked Nov 26 '18 at 15:38
ARocheARoche
61
Do you want to get access to the file name intransformmethod?
– Jacek Laskowski
Dec 3 '18 at 20:25
add a comment |
I am implementing a spark structured streaming application that processes webserver log files from a folder on disk or perhaps S3.
Spark Structured Streaming fits the use case almost perfectly, with one wrinkle.
The filenames in the folder also contain the machine name eg. like:
/node1_20181101.json.gz
/node1_20181102.json.gz
/node2_20181101.json.gz
/node3_20181102.json.gz
/node4_20181102.json.gz
...and so on.
A (simplified) version of the source looks something like this ( I would turn the below to a continuous stream with windowing etc):
val inputDF = spark.read
.option("codec", classOf[GzipCodec].getName)
.option("maxFilesPerTrigger", 1.toString)
.json(config.directory)
.transform { ds =>
logger.info(ds.inputFiles)
ds
}.foreach(println(_))
I would like to transform the batch and add the node ID from the filename to each record line, - I can't seem to see any kind of an onBatch trigger that I could use to enrich the record schema with the node ID from the file name.
I have looked at the following and nothing seems to fit:
[FileStreamSource][https://jaceklaskowski.gitbooks.io/spark-structured-streaming/spark-sql-streaming-FileStreamSource.html#metadataLog]
Unfortunately getting a handle on the machine name from the file name is key to the analytics I do later, and I have no control over how the logs are populated
Any clues?
scala apache-spark spark-structured-streaming
edited Dec 3 '18 at 20:23
Jacek Laskowski
46.3k18138278
asked Nov 26 '18 at 15:38
ARocheARoche
61
I am implementing a spark structured streaming application that processes webserver log files from a folder on disk or perhaps S3.
Spark Structured Streaming fits the use case almost perfectly, with one wrinkle.
The filenames in the folder also contain the machine name eg. like:
/node1_20181101.json.gz
/node1_20181102.json.gz
/node2_20181101.json.gz
/node3_20181102.json.gz
/node4_20181102.json.gz
...and so on.
A (simplified) version of the source looks something like this ( I would turn the below to a continuous stream with windowing etc):
val inputDF = spark.read
.option("codec", classOf[GzipCodec].getName)
.option("maxFilesPerTrigger", 1.toString)
.json(config.directory)
.transform { ds =>
logger.info(ds.inputFiles)
ds
}.foreach(println(_))
I would like to transform the batch and add the node ID from the filename to each record line, - I can't seem to see any kind of an onBatch trigger that I could use to enrich the record schema with the node ID from the file name.
I have looked at the following and nothing seems to fit:
[FileStreamSource][https://jaceklaskowski.gitbooks.io/spark-structured-streaming/spark-sql-streaming-FileStreamSource.html#metadataLog]
Unfortunately getting a handle on the machine name from the file name is key to the analytics I do later, and I have no control over how the logs are populated
Any clues?
scala apache-spark spark-structured-streaming
scala apache-spark spark-structured-streaming
edited Dec 3 '18 at 20:23
Jacek Laskowski
46.3k18138278
asked Nov 26 '18 at 15:38
ARocheARoche
61
edited Dec 3 '18 at 20:23
Jacek Laskowski
46.3k18138278
asked Nov 26 '18 at 15:38
ARocheARoche
61
edited Dec 3 '18 at 20:23
Jacek Laskowski
46.3k18138278
edited Dec 3 '18 at 20:23
Jacek Laskowski
46.3k18138278
edited Dec 3 '18 at 20:23
Jacek Laskowski
46.3k18138278
46.3k18138278
asked Nov 26 '18 at 15:38
ARocheARoche
61
asked Nov 26 '18 at 15:38
ARocheARoche
61
asked Nov 26 '18 at 15:38
ARocheARoche
61
61
Do you want to get access to the file name intransformmethod?
– Jacek Laskowski
Dec 3 '18 at 20:25
add a comment |
Do you want to get access to the file name intransformmethod?
– Jacek Laskowski
Dec 3 '18 at 20:25
Do you want to get access to the file name in
transform method?– Jacek Laskowski
Dec 3 '18 at 20:25
Do you want to get access to the file name in
transform method?– Jacek Laskowski
Dec 3 '18 at 20:25
add a comment |
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Do you want to get access to the file name in
transformmethod?– Jacek Laskowski
Dec 3 '18 at 20:25