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My SQL Life Cycle vs Version and release dates

LewisLL12257Oct 25 2019 — edited Oct 28 2019

Has anyone see a version chart on information on what is the next version of MySQL Enterprise Edition after version 8?

I am looking to do a road map for the next 3 years and I want to plan training and upgrade activities in the future.

This post has been answered by Dave Stokes-MySQL Community Team-Oracle on Oct 28 2019
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Comments

Solomon Yakobson
with t as (
           select  date_start + offset dt
             from  get_dates,
                   lateral(
                           select  level - 1 offset
                             from  dual
                             connect by level <= date_end - date_start + 1
                          )
          )
select  case grouping(to_char(dt,'mm')) when 1 then to_char(dt,'yyyy') || ' Total' end year,
        to_char(dt,'mm') month,
        sum(case dt - trunc(dt,'iw') when 5 then 1 end) saturday,
        sum(case dt - trunc(dt,'iw') when 6 then 1 end) sunday,
        sum(case dt - trunc(dt,'iw') when 0 then 1 end) monday,
        sum(case dt - trunc(dt,'iw') when 1 then 1 end) tuesday,
        sum(case dt - trunc(dt,'iw') when 2 then 1 end) wednesday,
        sum(case dt - trunc(dt,'iw') when 3 then 1 end) thursday,
        sum(case dt - trunc(dt,'iw') when 4 then 1 end) friday,
        count(*) grand_total
  from  t
  group by grouping sets((to_char(dt,'yyyy')),(to_char(dt,'yyyy'),to_char(dt,'mm')))
  order by to_char(dt,'yyyy'),
           to_char(dt,'mm')
/

YEAR       MO   SATURDAY     SUNDAY     MONDAY    TUESDAY  WEDNESDAY   THURSDAY     FRIDAY GRAND_TOTAL
---------- -- ---------- ---------- ---------- ---------- ---------- ---------- ---------- -----------
           05          3          3          3          3          3          3          3          21
           06                     1          1          1          1                                 4
2014 Total             3          4          4          4          4          3          3          25
           01          1          2          2          2          1          1          1          10
           02          4          4          4          4          4          4          4          28
           03          3          3          3          3          4          3          3          22
2017 Total             8          9          9          9          9          8          8          60


7 rows selected.


SQL>

SY.

Shaik_Muhammad
Answer

Yes Sir, you suggested me to post in this forum, ..

  1. This forum is for discussion of the use of the SQL Developer GUI tool. Your question is purely a SQL question. Better asked at https://community.oracle.com/tech/developers/categories/3063-general_questions1
    Am testing code suggested by Mr. Soloman..
Marked as Answer by Shaik_Muhammad · Jun 8 2021
EdStevens

Yes Sir, you suggested me to post in this forum, ..
Ah, so I did. Mea culpa.

Shaik_Muhammad

Thanks for the reply.
We are using 11g database.
Attached is the query execution output.
query_execution.jpg (67.13 KB)

Solomon Yakobson

Next time provide version upfront, expecially when you are on very old version:

with t as (
           select  date_start + level - 1 dt
             from  get_dates
             connect by rowid = prior rowid
                    and level <= date_end - date_start + 1
                    and prior sys_guid() is not null
          )
select  case grouping(to_char(dt,'mm')) when 1 then to_char(dt,'yyyy') || ' Total' end year,
        to_char(dt,'mm') month,
        sum(case dt - trunc(dt,'iw') when 5 then 1 end) saturday,
        sum(case dt - trunc(dt,'iw') when 6 then 1 end) sunday,
        sum(case dt - trunc(dt,'iw') when 0 then 1 end) monday,
        sum(case dt - trunc(dt,'iw') when 1 then 1 end) tuesday,
        sum(case dt - trunc(dt,'iw') when 2 then 1 end) wednesday,
        sum(case dt - trunc(dt,'iw') when 3 then 1 end) thursday,
        sum(case dt - trunc(dt,'iw') when 4 then 1 end) friday,
        count(*) grand_total
  from  t
  group by grouping sets((to_char(dt,'yyyy')),(to_char(dt,'yyyy'),to_char(dt,'mm')))
  order by to_char(dt,'yyyy'),
           to_char(dt,'mm')
/


YEAR       MONTH   SATURDAY     SUNDAY     MONDAY    TUESDAY  WEDNESDAY   THURSDAY     FRIDAY GRAND_TOTAL
---------- ----- ---------- ---------- ---------- ---------- ---------- ---------- ---------- -----------
           05             3          3          3          3          3          3          3          21
           06                        1          1          1          1                                 4
2014 Total                3          4          4          4          4          3          3          25
           01             1          2          2          2          1          1          1          10
           02             4          4          4          4          4          4          4          28
           03             3          3          3          3          4          3          3          22
2017 Total                8          9          9          9          9          8          8          60


7 rows selected.


SQL>

SY.

Shaik_Muhammad

Many Thanks to you Mr. Solomon Yakobson Sir, Mr. EdSteven Sir, report working perfectly.

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Added on Oct 25 2019
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