Get column after group by lastest date - mysql

Question

Please help me to get table 2 from table 1

Answer 1

A simple summarisation can be achieved using GROUP BY
select
code1
, max(code2) as code2
, count(*) as times
, max(`date`) as max_dt
from table1
group by
code1
However the result is different to the image:
+-------+-------+-------+------------+
| code1 | code2 | times | max_dt |
+-------+-------+-------+------------+
| 1 | D | 4 | 2020-02-21 |
| 2 | NNNN | 2 | 2021-01-21 |
+-------+-------+-------+------------+
Note:
The maximum of code2 may not be what you expect it to be. e.g. "D" is after "BBBBB" as the data is alphabetical. not based on length of string.
maximum of date for code1 isn't 16/2
I would not recommend naming any column "date" as it is usually a reserved word and can cause difficulties when developing queries.
for MySQL prior to version 8 a technique to get "the most recent" row may be used as follows:
SELECT code1, code2, `date`
, (select count(*) from mytable t2 where d.code1 = t2.code1) as times
FROM (
SELECT
#row_num :=IF(#prev_value = t.code1, #row_num + 1, 1) AS rn
, t.code1
, t.code2
, t.`date`
, #prev_value := t.code1
FROM mytable t
CROSS JOIN (SELECT #row_num :=1, #prev_value :='') vars
ORDER BY
t.code1
, t.`date` DESC
) as d
WHERE rn = 1
;
Or, for MySQL version 8 or later it is possible to use a much simpler query as there are windowing functions available through the over() clause:
SELECT code1, code2, `date`, times
FROM (
SELECT
row_number() over(partition by t.code1
order by t.`date` DESC) AS rn
, t.code1
, t.code2
, t.`date`
, count(*) over(partition by t.code1) as times
FROM mytable t
) as d
WHERE rn = 1
;
The results of the 2nd and 3rd queries are the same:
+-------+-------+------------+-------+
| code1 | code2 | date | times |
+-------+-------+------------+-------+
| 1 | D | 2020-02-21 | 4 |
| 2 | NNNN | 2021-01-21 | 2 |
+-------+-------+------------+-------+
solution demonstrated at db<>fiddle here