MYSQL按日期desc获取记录分组并按日期asc排序

Question

I've a mysql table with below data.

name        table_data       table_date 
name1         {count:2}       2020-01-24 02:23:41
name1         {count:5}       2020-01-24 05:45:07
name1         {count:8}       2020-01-24 12:45:18
name1         {count:20}      2020-01-30 08:40:07
name1         {count:28}      2020-01-30 15:08:12

I want a sql for getting the only one record for a date. The problem is that, I've to order the table based on table_date asc but need only the latest record for each date.

Expected result:

name        table_data      table_date  
name1         {count:8}       2020-01-24 12:45:18
name1         {count:28}      2020-01-30 15:08:12

My sql is like:

select name, table_data, table_date
 from my_table 
 where DATE(table_date) <= DATE('2020-04-01') group by DATE(table_date) order by table_date asc 

But this SQL will only give the record with minimum date.

Can anyone help me solve this?

Thanks in advance.

Answer 1

You don't mention the MySQL version you are using. In MySQL 8.x you can use ROW_NUMBER() to decide which rows to keep.

For example:

select name, table_date
from (
  select name, table_date
    row_number() over partition by name order by table_date desc) as rn
  from my_table
) x
where rn = 1

I think in MySQL 5.x you can use IN with tuples. For example:

select *
from my_table
where (name, table_date) in (
  select name, max(table_date) from my_table group by name, date(table_date)
)