Dynamic stage path in snowflake - snowflake-cloud-data-platform

I have a stage path as below
copy into table1 as (
select $1:InvestorID::varchar as Investor_ID from #company_stage/pbook/2022-03-10/Invor/part-00000-33cbc68b-69c1-40c0-943c-f586dfab3f49-c000.snappy.parquet
)
This is my S3 location company_stage/pbook/2022-03-10/Invor,
I need to make this dynamic:
I) I need to change this "2022-03-10" folder to current date
II)it must take all parquet files in the folder automatically, without me mentioning of filename. How to achieve this?

Here is one approach. Your stage shouldn't include the date as part of the stage name because if it did, you would need a new stage every day. Better to define the stage as company_stage/pbook/.
To make it dynamic, I suggest using the pattern option together with the COPY INTO command. You could create a variable with the regex pattern expression using current_date(), something like this:
set mypattern = '\.*'||to_char(current_date(), 'YYYY-MM-DD')||'\.*';
Then use this variable in your COPY INTO command like this:
copy into table1 as (
select $1:InvestorID::varchar as Investor_ID from #company_stage/pbook/ pattern = $mypattern
)
Of course you can adjust your pattern matching as you see fit.

Related

using pattern while loading csv file from s3 to snowflake

below copy command is not working , please correct me if something wrong.
copy into mytable from #mystage pattern='20.*csv.gz'
here i am trying to load the files which are starts with 20, there are mix of files which are having the name as 2021myfile.csv.gz, myfile202109.csv.gz, above command is not loading any files though there are files which starts with 20.
if i use pattern as pattern='.*20.*csv.gz'`` it is taking all the files which is wrong, i need to load only the files that are starts with 20`.
Thanks!
This is because the pattern clause is a Regex expression.
Try this:
copy into mytable from #mystage pattern = '[20]*.*csv.gz'
Reference: Loading Using Pattern Matching

COPY INTO with partitioned ADLS

I have a container with partitioned parquet files that I want to use with the copy into command. My directories look like the below.
ABC_PARTITIONED_ID=1 (directory)
1-snappy.parquet
2-snappy.parquet
3-snappy.parquet
4-snappy.parquet
ABC_PARTITIONED_ID=2 (directory)
1-snappy.parquet
2-snappy.parquet
3-snappy.parquet
ABC_PARTITIONED_ID=3 (directory)
1-snappy.parquet
2-snappy.parquet
....
Each partitioned directory can contain multiple parquet files. I do not have a hive partition column that matches the pattern of the directories (ID1, ID2 etc).
How do I properly use the pattern parameter in the copy into command to write to a SF table from my ADLS? I am using this https://www.snowflake.com/blog/how-to-load-terabytes-into-snowflake-speeds-feeds-and-techniques/ as an example.
I do not think that you have anything to do with the pattern parameter.
You said you do not have a hive partition column that matches the pattern of the directories. If you do not have a column to use these partitions, then they are probably not beneficial for querying the data. Maybe they were generated to help maintenance. If this is the case, ignore the partition, and read all files with the COPY command.
If you think having such a column would help, then the blog post (you mentioned) already shows how you can parse the filenames to generate the column value. Add the partition column to your table (and even you may define it as the clustering key), and run the COPY command to read all files in all partitions/directories, parse the value of the column from the file name.
For parsing the partition value, I would use this one which seems easier:
copy into TARGET_TABLE from (
select
REGEXP_SUBSTR (
METADATA$FILENAME,
'.*\ABC_PARTITIONED_ID=(.*)\/.*',
1,1,'e',1
) partitioned_column_value,
$1:column_name,
...
from #your_stage/data_folder/);
If the directory/partition name doesn't matter to you, then you can use some of the newer functions in Public Preview that support Parquet format to create the table and ingest the data. Your question on how to construct the pattern would be PATTERN='*.parquet' as all subfolders would be read.
//create file format , only required to create one time
create file format my_parquet_format
type = parquet;
//EXAMPLE CREATE AND COPY INTO FOR TABLE1
//create an empty table using this file format and location. name the table table1
create or replace table ABC
using template (
select array_agg(object_construct(*))
from table(
infer_schema(
location=>'#mystage/ABC_PARTITIONED_ROOT',
file_format=>'my_parquet_format'
)
));
//copy parquet files in folder /table1 into table TABLE1
copy into ABC from #mystage/ABC_PARTITIONED_ROOT pattern = '*.parquet' file_format=my_parquet_format match_by_column_name=case_insensitive;
This should be possible by creating a storage integration, granting access in Azure for Snowflake to access the storage location, and then creating an external stage.
Alternatively you can generate a shared access signature (SAS) token to grant Snowflake (limited) access to objects in your storage account. You can then access an external (Azure) stage that references the container using the SAS token.
Snowflake metadata provides
METADATA$FILENAME - Name of the staged data file the current row belongs to. Includes the path to the data file in the stage.
METADATA$FILE_ROW_NUMBER - Row number for each record
We could do something like this:
select $1:normal_column_1, ..., METADATA$FILENAME
FROM
'#stage_name/path/to/data/' (pattern => '.*.parquet')
limit 5;
For example: it would give something like:
METADATA$FILENAME
----------
path/to/data/year=2021/part-00020-6379b638-3f7e-461e-a77b-cfbcad6fc858.c000.snappy.parquet
we need to handle deducing the column from it. We could do a regexp_replace and get the partition value as column like this:
select
regexp_replace(METADATA$FILENAME, '.*\/year=(.*)\/.*', '\\1'
) as year
$1:normal_column_1,
FROM
'#stage_name/path/to/data/' (pattern => '.*.parquet')
limit 5;
In the above regexp, we give the partition key.
Third parameter \\1 is the regex group match number. In our case, first group match - this holds the partition value.
More detailed answer and other approaches to solve this issue is available on this stackoverflow answer

query snowflake s3 external file

I've created an S3 [external] stage and uploaded csv files into \stage*.csv folder.
I can see stage content by doing list #my_stage.
if I query the stage
select $1,$2,$3,$4,$5,$6 from #my_s3_stage it looks like I'm randomly picking up files.
So I'm trying to select from specific file by adding a pattern
PATTERN => job.csv
This returns no results.
Note: I've used snowflake for all of 5 hours so pretty new to syntax
For a pattern you can use
select t.$1, t.$2 from #mystage1 (file_format => 'myformat', pattern=>'.*data.*[.]csv.gz') t;
The pattern is a regex expression.
For a certain file you have to add the file name to the query like this:
select t.$1, t.$2 from #mystage/data1.csv.gz;
If your file format is set in your stage definition, you don't need the file format-parameter.
More info can be found here: https://docs.snowflake.com/en/user-guide/querying-stage.html

Auto-generating destinations of split files in SSIS

I am working on my first SSIS package. I have a view with data that looks something like:
Loc Data
1 asd
1 qwe
2 zxc
3 jkl
And I need all of the rows to go to different files based on the Loc value. So all of the data rows where Loc = 1 should end up in the file named Loc1.txt, and the same for each other Loc.
It seems like this can be accomplished with a conditional split to flat file, but that would require a destination for each Location. I have a lot of Locations, and they all will be handled the same way other than being split in to different files.
Is there a built in way to do this without creating a bunch of destination components? Or can I at least use the script component to act as a way?
You should be able to set an expression using a variable. Define your path up to the directory and then set the variable equal to that column.
You'll need an Execute SQL task to return a Single Row result set, and loop that in a container for every row in your original result set.
I don't have access at the moment to post screenshots, but this link should help outline the steps.
So when your package runs the expression will look like:
'C:\Documents\MyPath\location' + #User::LocationColumn + '.txt'
It should end up feeding your directory with files according to location.
Set the User::LocationColumn equal to the Location Column in your result set. Write your result set to group by Location, so all your records write to a single file per Location.
I spent some time try to complete this task using the method #Phoenix suggest, but stumbled upon this video along the way.
I ended up going with the method shown in the video. I was hoping I wouldn't have to separate it in to multiple select statements for each location and an extra one to grab the distinct locations, but I thought the SSIS implementation in the video was much cleaner than the alternative.
Change the connection manager's connection string, in which you have to use variable which should be changed.
By varying the variable, destination file also changes
and connection string is :
'C:\Documents\ABC\Files\' + #User::data + '.txt'
vote this if it helps you

How to Dynamically render Table name and File name in pentaho DI

I have a requirement in which one source is a table and one source is a file. I need to join these both on a column. The problem is that I can do this for one table with one transformation but I need to do it for multiple set of files and tables to load into another set of specific files as target using the same transformation.
Breaking down my requirement more specifically :
Source Table Source File Target File
VOICE_INCR_REVENUE_PROFILE_0 VoiceRevenue0 ProfileVoice0
VOICE_INCR_REVENUE_PROFILE_1 VoiceRevenue1 ProfileVoice1
VOICE_INCR_REVENUE_PROFILE_2 VoiceRevenue2 ProfileVoice2
VOICE_INCR_REVENUE_PROFILE_3 VoiceRevenue3 ProfileVoice3
VOICE_INCR_REVENUE_PROFILE_4 VoiceRevenue4 ProfileVoice4
VOICE_INCR_REVENUE_PROFILE_5 VoiceRevenue5 ProfileVoice5
VOICE_INCR_REVENUE_PROFILE_6 VoiceRevenue6 ProfileVoice6
VOICE_INCR_REVENUE_PROFILE_7 VoiceRevenue7 ProfileVoice7
VOICE_INCR_REVENUE_PROFILE_8 VoiceRevenue8 ProfileVoice8
VOICE_INCR_REVENUE_PROFILE_9 VoiceRevenue9 ProfileVoice9
The table and file names are always corresponding i.e. VOICE_INCR_REVENUE_PROFILE_0 should always join with VoiceRevenue0 and the result should be stored in ProfileVoice0. There should be no mismatches in this case. I tried setting the variables with table names and file names, but it only takes on value at a time.
All table names and file names are constant. Is there any other way to get around this. Any help would be appreciated.
Try using "Copy rows to result" step. It will store all the incoming rows (in your case the table and file names) into a memory. And for every row, it will try to execute your transformation. In this way, you can read multiple filenames at one go.
Try reading this link. Its not the exact answer, but similar.
I have created a sample here. Please check if this is what is required.
In the first transformation, i read the tablenames and filenames and loaded it in the memory. After that i have used the get variable step to read all the files and table names to generate the output. [Note: I have not used table input as source anywhere, instead used TablesNames. You can replace the same with the table input data.]
Hope it helps :)

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