Como criar tabelas particionadas por data no BigQuery avaliações

The issue still in the second "check my progress"

Alberto Josué C. · Revisado há almost 4 years

Task number 2 will not show completed

Raisa J. · Revisado há almost 4 years

Task2 score not updated

Vetrivel D. · Revisado há almost 4 years

not good - table created but can't check progress

Rudolf E. · Revisado há almost 4 years

At "Create a new partitioned table based on date" the "Check my progress" button does not work.

Anna U. · Revisado há almost 4 years

The partitioned table was created by using the provided query but the system didn't recognize it and passed the checkpoint.

Olivia C. · Revisado há almost 4 years

Divya S. · Revisado há almost 4 years

lab could not confirm I created partition_by_day table.

Samantha L. · Revisado há almost 4 years

Task 2 error

Alvin Z. · Revisado há almost 4 years

JUAN M. · Revisado há almost 4 years

Alberto Josué C. · Revisado há almost 4 years

It is not possible to finish the lab, I have reported it 5 times and I have not received a response

GONZALO ANDRES M. · Revisado há almost 4 years

Alberto Josué C. · Revisado há almost 4 years

JUAN M. · Revisado há almost 4 years

Rajdeep S. · Revisado há almost 4 years

JYOTI RANJAN P. · Revisado há almost 4 years

Task2 score not updated

Vetrivel D. · Revisado há almost 4 years

Naafi M. · Revisado há almost 4 years

after completing the assessment , it still says that complete the assessment

Priya N. · Revisado há almost 4 years

Jaime N. · Revisado há almost 4 years

had to change query

Marc R. · Revisado há almost 4 years

There is problem in the below query its not creating any record #standardSQL CREATE OR REPLACE TABLE ecommerce.partition_by_day PARTITION BY date_formatted OPTIONS( description="a table partitioned by date" ) AS SELECT DISTINCT PARSE_DATE("%Y%m%d", date) AS date_formatted, fullvisitorId FROM `data-to-insights.ecommerce.all_sessions_raw`

Prosenjit R. · Revisado há almost 4 years

Sudhanshu S. · Revisado há almost 4 years

Clément B. · Revisado há almost 4 years

Error in task 2, but is working with --- SELECT DISTINCT DATE('2022-09-06') AS date_formatted ---

Pedro E. · Revisado há almost 4 years