Thanks for the contribution.

Although the function does not show all the partitions, it can still be useful. Please address the comments and

IMPORTANT!

add tests!

Thanks for the contribution.

Although the function does not show all the partitions, it can still be useful. Please address the comments and

IMPORTANT!

add tests!

@ideoma in the description it was said all_table_partitions() function should be considered to be something like :

select 'table_1' as table_name,'false' as walEnabled,'DAY' as partitionBy,count(*) partition_count,sum(numrows) row_count,(sum(disksize)) as size_b from table_partitions('table_1') group by table_name union all
select 'table_2' as table_name,'false' as walEnabled,'DAY' as partitionBy,count(*) partition_count,sum(numrows) row_count,(sum(disksize)) as size_b from table_partitions('table_2') group by table_name union all
select 'table_3' as table_name,'false' as walEnabled,'DAY' as partitionBy,count(*) partition_count,sum(numrows) row_count,(sum(disksize)) as size_b from table_partitions('table_3') group by table_name

Which is giving cumulative metrics of table partitions
Is there anything else expected in this issue? any understanding gap?

@ideoma in the description it was said all_table_partitions() function should be considered to be something like :

select 'table_1' as table_name,'false' as walEnabled,'DAY' as partitionBy,count(*) partition_count,sum(numrows) row_count,(sum(disksize)) as size_b from table_partitions('table_1') group by table_name union all
select 'table_2' as table_name,'false' as walEnabled,'DAY' as partitionBy,count(*) partition_count,sum(numrows) row_count,(sum(disksize)) as size_b from table_partitions('table_2') group by table_name union all
select 'table_3' as table_name,'false' as walEnabled,'DAY' as partitionBy,count(*) partition_count,sum(numrows) row_count,(sum(disksize)) as size_b from table_partitions('table_3') group by table_name

Which is giving cumulative metrics of table partitions Is there anything else expected in this issue? any understanding gap?

@nwoolmer can you please check this

@ideoma in the description it was said all_table_partitions() function should be considered to be something like :

select 'table_1' as table_name,'false' as walEnabled,'DAY' as partitionBy,count(*) partition_count,sum(numrows) row_count,(sum(disksize)) as size_b from table_partitions('table_1') group by table_name union all
select 'table_2' as table_name,'false' as walEnabled,'DAY' as partitionBy,count(*) partition_count,sum(numrows) row_count,(sum(disksize)) as size_b from table_partitions('table_2') group by table_name union all
select 'table_3' as table_name,'false' as walEnabled,'DAY' as partitionBy,count(*) partition_count,sum(numrows) row_count,(sum(disksize)) as size_b from table_partitions('table_3') group by table_name

Which is giving cumulative metrics of table partitions Is there anything else expected in this issue? any understanding gap?

@nwoolmer can you please check this

Hey @siddharth0815 ,

I can see why there is confusion about this, you've not done anything wrong here.

In the original issue, the requester names the function all_tables_partitions. At face value, this sounds like a function which should return you the same results as table_partitions('tbl') but for all your tables, so you don't have to run individual queries and then concatenate them. Certainly, that was my first impression.

But then they provide a query in which they take a summary of these outputs. That could be achieved with a 'true' all_tables_partitions function, by returning all partition rows and then using a GROUP BY.

Your function above matches the request but their naming choice is probably not ideal. Your function is great and returns the summary they asked for - which is the stats for each table aggregated over all its partitions. And therefore, table_storage() seems like a good name for it.

We should ping the requestor to let them know their query is now the function table_storage, and then circle back to an all_tables_partitions function to run table_partitions for every table.