MySql 按 item_id 分组的最后一条记录，按日期排序

Question

My database table name is ledgers and fields are id, item_id, date, ...other fields

I Want the last record from ( groupBy item_id order by date ASC ). from each group. I tried below query

select 
    `id`, 
    `item_id`, 
    `date`, 
    `opening_quantity`, 
    `closing_quantity`, 
    `item_rate`, 
    `item_value`, 
    `previous_rate` 
from `ledgers` 
where date(`date`) >= ? and date(`date`) <= ? 
group by `item_id` 
order by `date` desc

Can you guys please help.

Answer 1

You can filter with a correlated subquery:

select t.*
from `ledgers` t
where 
    date(t.`date`) >= ? 
    and date(t.`date`) <= ?
    and t.`date` = (
        select max(t1.`date`)
        from `ledgers` t1
        where t1.`item_id` = t.`item_id`
    )

For performance, consider an index on (item_id, date) .

Another option is to use rank() (available in MySQ 8.0 only):

select * 
from (
    select 
        t.*,
        rank() over(partition by `item_id` order by `date` desc) rn
    from `ledgers` t
    where  date(t.`date`) >= ? and date(t.`date`) <= ?
) t
where rn = 1