How can I unload snowflake data to s3 without using any file format?
For unloading the data into a specific extension we use file format in snowflake.
E.g. code
copy into 's3://mybucket/unload/'
from mytable
storage_integration = myint
file_format = (format_name = my_csv_format);
But what I want is to store data without any extension.
SINGLE is what I was looking for. It is one of parameters we can use with COPY command which creates the file without extension.
Code:
copy into 's3://mybucket/unload/'
from mytable
storage_integration = myint
file_format = (format_name = my_csv_format)
SINGLE = TRUE;
Go through note of below link for better understanding:
https://docs.snowflake.com/en/sql-reference/sql/create-file-format.html#:~:text=comma%20(%2C)-,FILE_EXTENSION,-%3D%20%27string%27%20%7C%20NONE
You can add the parameter FILE_EXTENSION = NONE to your file format. With this parameter Snowflake is not adding a file extension based on your file format (in this case .csv), but is using the passed extension (NONE or any other).
copy into 's3://mybucket/unload/'
from mytable
storage_integration = myint
file_format = (format_name = my_csv_format file_extension = NONE);
https://docs.snowflake.com/en/sql-reference/sql/copy-into-location.html
we are facing below issue while loading csv file from S3 to Snowflake.
SQL Compilation error: Insert column value list does not match column list expecting 7 but got 6
we have tried removing the column from table and tried again but this time it is showing expecting 6 but got 5
below are the the commands that we have used for stage creation and copy command.
create or replace stage mystage
url='s3://test/test'
STORAGE_INTEGRATION = test_int
file_format = (type = csv FIELD_OPTIONALLY_ENCLOSED_BY='"' COMPRESSION=GZIP);
copy into mytable
from #mystage
MATCH_BY_COLUMN_NAME = CASE_INSENSITIVE;
FILE_FORMAT = (TYPE = CSV FIELD_OPTIONALLY_ENCLOSED_BY='"' COMPRESSION=GZIP error_on_column_count_mismatch=false TRIM_SPACE=TRUE NULL_IF=(''))
FORCE = TRUE
ON_ERROR = Continue
PURGE=TRUE;
You can not use MATCH_BY_COLUMN_NAME for the CSV files, this is why you get this error.
This copy option is supported for the following data formats:
JSON
Avro
ORC
Parquet
https://docs.snowflake.com/en/sql-reference/sql/copy-into-table.html
In a file, few of the rows have \ in a column value for example, i have rows in below format.
101,Path1,Z:\VMC\PSPS,abc
102,Path5,C:\wintm\PSPS,abc
I was wondering how to load \ character
COPY INTO TEST_TABLE from #database.schema.stage_name FILE_FORMAT = ( TYPE = CSV FIELD_OPTIONALLY_ENCLOSED_BY = '\"' SKIP_HEADER = 1 );
is there any thing that i can mention the file_format line?
Are you still getting this error? I just tried to recreate it by creating a CSV based off your sample data and a test table. I loaded the CSV into an internal stage and then ran your COPY command. It worked for me. Please see the screenshot below.
Could you provide more details on the error you are facing? Perhaps there was something off with your table definition.
I am trying to load data from azure blob storage.
The data has already been staged.
But, the issue is when I try to run
copy into random_table_name
from #stage_name_i_created
file_format = (type='csv')
pattern ='*.csv'
Below is the error I encounter:
raise error_class(
snowflake.connector.errors.ProgrammingError: 001757 (42601): SQL compilation error:
Table 'random_table_name' does not exist
Basically, it says table does not exist, which it does not, but the syntax on website is the same as mine.
COPY INTO query on Snowflake returns TABLE does not exist error
In my case the table name is case-sensitive. Snowflake seems to convert everything to upper case. I changed the database/schema/table names to all upper-case and it started working.
First run the below query to fetch the column headers
select $1 FROM #stage_name_i_created/filename.csv limit 1
Assuming below are the header lines from your csv file
id;first_name;last_name;email;age;location
Create a file_format csv
create or replace file format semicolon
type = 'CSV'
field_delimiter = ';'
skip_header=1;
Then you should define the datatype and field name as below
create or replace table <yourtable> as
select $1::varchar as id
,$2::varchar as first_name
,$3::varchar as last_name
,$4::varchar as email
,$5::int as age
,$6::varchar as location
FROM #stage_name_i_created/yourfile.csv
(file_format => semicolon );
The table must exist prior to running a COPY INTO command. In your post, you say that the table does not exist...so that is your issue.
If your table exist, try by forcing the table path like this:
copy into <database>.<schema>.<random_table_name>
from #stage_name_i_created
file_format = (type='csv')
pattern ='*.csv'
or by steps like this:
use database <database_name>;
use schema <schema_name>;
copy into database.schema.random_table_name
from #stage_name_i_created
file_format = (type='csv')
pattern ='*.csv';
rbachkaniwala, what do you mean by 'How do I create a table?( according to snowflake syntax it is not possible to create empty tables)'.
You can just do below to create a table
CREATE TABLE random_table_name (FIELD1 VARCHAR, FIELD2 VARCHAR)
The table does need to exist. You should check the documentation for COPY INTO.
Other areas to consider are
do you have the right context set for the database & schema
does the user / role have access to the table or object.
It basically seems like you don't have the table defined yet. You should
ensure the table is created
ensure all columns in the CSV exist as columns in the table
ensure the order of the columns are the same as in the CSV
I'd check data types too.
"COPY INTO" is not a query command, it is the actual data transfer execution from source to destination, which both must exist as others commented here but If you want just to query without loading the files then run the following SQL:
//Display list of files in the stage to verify stage
LIST #stage_name_i_created;
//Create a file format
CREATE OR REPLACE FILE FORMAT RANDOM_FILE_CSV
type = csv
COMPRESSION = 'GZIP' FIELD_DELIMITER = ',' RECORD_DELIMITER = '\n' SKIP_HEADER = 0 FIELD_OPTIONALLY_ENCLOSED_BY = '\042'
TRIM_SPACE = FALSE ERROR_ON_COLUMN_COUNT_MISMATCH = FALSE ESCAPE = 'NONE' ESCAPE_UNENCLOSED_FIELD = 'NONE' DATE_FORMAT = 'AUTO' TIMESTAMP_FORMAT = 'AUTO'
NULL_IF = ('\\N');
//Now select the data in the files
Select $1 as first_col,$2 as second_col //can add as necessary number of columns ...etc
from #stage_name_i_created
(FILE_FORMAT => RANDOM_FILE_CSV)
More information can be found in the documentation link here
https://docs.snowflake.com/en/user-guide/querying-stage.html
(Submitted on behalf of a Snowflake User)
I have a test s3 folder called s3://bucket/path/test=integration_test_sanity/file.parquet
I want to be able to load this into snowflake using the COPY INTO command but I want to be able to load all the test folders which have a structure like test=*/file.parquet.
I've tried:
COPY INTO raw.test_sanity_test_parquet
FROM 's3://bucket/path/'
CREDENTIALS=(AWS_KEY_ID='XXX' AWS_SECRET_KEY='XXX')
PATTERN='test=(.*)/.*'
FILE_FORMAT = (TYPE = parquet)
and also
COPY INTO raw.test_sanity_test_parquet
FROM 's3://bucket/path/'
CREDENTIALS=(AWS_KEY_ID='XXX' AWS_SECRET_KEY='XXX')
PATTERN='test=.*/.*'
FILE_FORMAT = (TYPE = parquet)
Neither of these works. I was wondering what regex parser is used by Snowflake and which regex I should use to get this to work.
This works but I can't filter on just test folders which can cause issues
COPY INTO raw.test_sanity_test_parquet
FROM 's3://bucket/path/'
CREDENTIALS=(AWS_KEY_ID='XXX' AWS_SECRET_KEY='XXX')
PATTERN='.*/.*'
FILE_FORMAT = (TYPE = parquet)
Any recommendations? Thanks!
Try this:
COPY INTO raw.test_sanity_test_parquet
FROM 's3://bucket/path/'
CREDENTIALS=(AWS_KEY_ID='XXX' AWS_SECRET_KEY='XXX')
PATTERN='.*/test.*[.]parquet'
FILE_FORMAT = (TYPE = parquet)