I'm trying to upload file to s3a using distcp.
distcp writes to temporary file first and than renames it to proper filename.
But user does not allow for update/delete. So I have file with proper size, wrong name.
-rw-rw-rw- 1 3738 2021-05-24 12:04 s3a://testbucket/.distcp.tmp.attempt_1621587961870_0001_m_000000_0
on s3 and receive an error:
Error: java.io.IOException: File copy failed: file:///testfile.json --> s3a://testbucket/testfile.json
Is it possible to omit renaming and write directly to final filename?
I've found it here:
https://hadoop.apache.org/docs/current/hadoop-distcp/DistCp.html
there is a parameter :
-direct
Write directly to destination paths Useful for avoiding potentially very expensive temporary file rename operations when the destination is an object store
example
hadoop distcp -direct hdfs://nn1:8020/datasets/set1 s3a://bucket/datasets/set1
Unfortunately my distcp version is too old and hasn't got this feature.
Related
I need to restore a lot of object from aws s3 glacier deep archive. So i try to use a s3 batch jobs. For that i use a python code to create a manifest as a csv with to columns Bucket,Key.
But my first issue : some Key contain a comma so the job failed.
To solve (partialy) this issue i just cut the csv file to keep only the first two columns hoping that there are not many files involved.
But now i have another issue:
ErrorMessage: Task target couldn't be URL decoded
Any Idea ?
As mentioned on https://docs.aws.amazon.com/AmazonS3/latest/userguide/batch-ops-create-job.html#specify-batchjob-manifest, the manifest CSV file must be URL encoded. The , character in a key name gets converted to %2C with URL encoding so the resulting file will be valid CSV even with commas in the key name
I am facing an odd issue where my lookup is returning a filenotfound error when I use a wildcard path. If I specify and exact file path, the lookup runs without error. However, if I replace the filename with a *, I get a filenotfound error.
The file is Data_643.json, located in my Azure Data Lake Storage Gen2, under the labournavigatorfile system. The exact file path is:
labournavigatorfile/raw_data/Scraped/HeadHunter/Saudi_Arabia/Data_643.json.
If I put this exact path into the Integration dataset configuration, the pipeline runs without issue. However, as soon as I replace the 'Data_643.json' with a '*', the pipeline crashes with a filenotfound error.
What am I doing wrong? Many Thanks for any support. This must be something very simple that I am missing.
Exact path works:
Wildcrad path throws error:
I have 3 files in my container as file1.json, file2.json, file3.json as shown below:
The following is how I configured my dataset to read using wildcard with configuration same as in the image provided in the question.
When I used this in lookup I got the same error:
To overcome this, go to your lookup activity. When you want to use wildcards to read a file/files, check the wildcard file path option. Then specify the folder structure and use wildcard where required. The following is an image for reference.
The following is the debug output when I run the pipeline (Each of my files had 10 rows):
I try to read a dbc file in databricks (mounted from an s3 bucket)
the file path is:
file_location="dbfs:/mnt/airbnb-dataset-ml/dataset/airbnb.dbc"
how to read this file using spark?
I tried the code below:
df=spark.read.parquet(file_location)
But it generates and error:
AnalysisException: Unable to infer schema for Parquet. It must be specified manually.
thanks for help !
I tried the code below: df=spark.read.parquet(file_location) But it
generates and error:
You are using spark.read.parquet but want to read dbc file. It won't work this way.
Don't use parquet but use load. Give file path with file name (without .dbc extension) in path parameter and dbc in format paramter.
Try below code:
df=spark.read.load(path='<file_path_with_filename>', format='dbc')
Eg: df=spark.read.load(path='/mnt/airbnb-dataset-ml/dataset/airbnb', format='dbc')
I am trying to generate a file from a Dataframe that I have created in AWS-Glue, I am trying to give it a specific name, I see most answers on stack overflow actually uses Filesystem modules, but here this particular csv file is generated in S3, also I want to give the file a name while generating it, and not rename it after it is generated, is there any way to do that?
I have tried using df.save(s3:://PATH/filename.csv) which actually generates a new directory in S3 named filename.csv and then generates part-*.csv inside that directory
df.repartition(1).write.mode('append').format('csv').save('s3://PATH').option("header", "true")
I'm trying to load simple file:
log = load 'file_1.gz' using TextLoader AS (line:chararray);
dump log
And I get an error:
2014-04-08 11:46:19,471 [main] ERROR org.apache.pig.tools.pigstats.SimplePigStats - ERROR 2997: Unable to recreate exception from backend error: org.apache.pig.backend.executionengine.ExecException: ERROR 2118: Input Pattern hdfs://hadoop1:8020/pko/file*gz matches 0 files
at org.apache.pig.backend.hadoop.executionengine.mapReduceLayer.PigInputFormat.getSplits(PigInputFormat.java:288)
at org.apache.hadoop.mapred.JobClient.writeNewSplits(JobClient.java:1054)
Is is possible to manage such situation before error appears?
Input Pattern hdfs://hadoop1:8020/pko/file*gz matches 0 files
The error is the input file doesn't exist in the given hdfs path.
log = load 'file_1.gz' using TextLoader AS (line:chararray);
as you haven’t mentioned the absolute path of file_1.gz , it will taken the home hdfs dir of the user with which you are running your pig-script
Unfortunately in the current version of Pig (0.15.0) it is impossible to manage these errors without using UDF's.
I suggest creating a Java or Python script using try and catch to take care of this.
Here's a good website that might be of some use to you: https://wiki.apache.org/pig/PigErrorHandlingInScripts
Good luck learning Pig!
I'm facing this issue as well. My load command is:
DATA = LOAD '${qurwf_folder_input}/data/*/' AS (...);
I want to load all files from the data subfolders, but the data folder is empty and I got the same error as you. What I did, in my particular case, was to create an empty folder in the data directory. So the LOAD returns an empty dataset and the script did not fail.
By the way, I'm using Oozie workflow to run the scripts, and in the prepare, I create the empty folders.