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Mysql very slow subquery optimizing


Mysql very slow subquery optimizing

By : Elias Salom
Date : November 17 2020, 11:55 AM
I hope this helps you . Assuming that (movie_id,category_id) is unique in movies_categories table, I'd get the specified result using join operations, rather than subqueries.
To exclude "skipped" movies, an anti-join pattern would suffice... that's a left outer join to find matching rows in skipped_movies, and then a predicate in the WHERE clause to exclude any matches found, leaving only rows that didn't have a match.
code :
SELECT SQL_NO_CACHE m.*
  FROM movies m
  JOIN movie_categories c 
    ON c.movie_id = m.id 
   AND c.category_id = 1
  LEFT
  JOIN skipped_movies s
    ON s.movie_id = m.id
   AND s.user_id = 1
 WHERE s.movie_id IS NULL
   AND m.score <= 9
 ORDER
    BY m.score DESC
 LIMIT 1
... ON movie_categories (category_id, movie_id)
... ON skipped_movies (user_id, movie_id)


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MySql select from subquery slow (even though subquery on it's own is fast)

MySql select from subquery slow (even though subquery on it's own is fast)


By : Piercing Light
Date : March 29 2020, 07:55 AM
I hope this helps . MySQL buffers the inline view before doing any operations on it.
0.0006 seconds you see for the first query is most probably the response time (the first row fetched), not the overall time (the last row fetched).
code :
order_item (order_id, checked)
order_item (order_id, last_update)
SELECT  *
FROM    order o
WHERE   o.canceled = 0
        AND
        (
        o.return_date > '2011-03-14'
        OR
        (
        SELECT  MIN(checked)
        FROM    order_item oi
        WHERE   order_id = o.id
        ) = 0
        OR
        (
        SELECT  MAX(last_update)
        FROM    order_item oi
        WHERE   oi.order_id = o.id
        ) > '2011-03-14'
        )
Optimizing MySQL subquery

Optimizing MySQL subquery


By : user3756418
Date : March 29 2020, 07:55 AM
I think the issue was by ths following , Working with MTA API.
code :
select trip_id 
from stop_times 
where stop_id in (111, 222)
group by trip_id
having count(distinct stop_id) = 2
MySQL, Optimizing Subquery

MySQL, Optimizing Subquery


By : Dj 2nyi
Date : March 29 2020, 07:55 AM
I think the issue was by ths following , Strugle with optimizing this subquery that count item:
code :
SELECT 
   `tblScan`.`crewId` AS `crewid`
   COUNT(`tblScan`.`id`) AS `number_documents`,
   SUM(IF(`tblScan`.`classification` IS NOT NULL,1,0)) AS `classified`,
   SUM(IF(`tblScan`.`classification` IS NULL,1,0)) AS `unclassified`,
   CONCAT(`tblcrew`.`name`, ' ', `tblcrew`.`surname`) AS `crewname`,
   MAX(`tblScan`.`uploadeddate`) AS `uploadeddate`,
   MAX(`tblScan`.`uploadedby`) AS `uploadedby`,
   MAX(IFNULL(`tblScan`.`updateddate`,'0000-01-01')) AS `updateddate`,
   MAX(IFNULL(`tblScan`.`updatedby`,'0')) AS `updatedby`
FROM tblScan
INNER JOIN tblcrew ON `tblcrew`.`crewId`=`tblScan`.`crewId`
GROUP BY `tblScan`.`crewId`
Optimizing mysql subquery for multiple INs

Optimizing mysql subquery for multiple INs


By : Tainá Zaqui
Date : March 29 2020, 07:55 AM
should help you out First, you should have indexes for article in table products and for item in table structure.
If you always have at least one row in structure for a given product, you could give this a try:
code :
SELECT s.final_item, p.article
    FROM  `structure` s INNER JOIN products p ON s.item = p.article
    WHERE s.article IN ('200101','200102')
    ORDER BY s.level DESC 
    LIMIT 1
MySQL Optimizing a subquery which uses a WHERE clause?

MySQL Optimizing a subquery which uses a WHERE clause?


By : user2772383
Date : March 29 2020, 07:55 AM
this will help Sub queries are often not great in MySQL due to not using indexes for the joins.
However it might be worth trying a sub query with a join, rather than a sub query with a sub query:-
code :
LEFT JOIN (SELECT s1.product_id, MAX(s1.offer_date) AS sec_last_date 
           FROM t_offers AS s1  
           INNER JOIN t_offers AS s2
           ON s2.product_id = s1.product_id
           AND s2.offer_date > s1.offer_date
           GROUP BY s1.product_id) AS t2 ON t2.product_id=p.id 
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