1. SQL / Говнокод #26859

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    SELECT 
        `1000, 1000000 field`
        (SELECT group_concat(band_category_languages.name SEPARATOR ', ')
        FROM calendar_entry_categories
        LEFT JOIN band_category_languages ON band_category_languages.band_category_id = calendar_entry_categories.band_category_id
        WHERE calendar_entry_categories.calendar_entry_id = calendar_entries.id
        AND band_category_languages.language_id = 1
        GROUP BY calendar_entry_categories.calendar_entry_id) AS band_category_string,
        (SELECT entry.id
        FROM calendar_entries AS entry
        WHERE entry.entry_kind_parent_id = calendar_entries.id
        AND entry.profile_type_id = 10869
        AND entry.profile_type = 1) AS move_to_personal_calendar,
        (SELECT count(moved_entry.id)
        FROM calendar_entries AS moved_entry
        WHERE moved_entry.entry_kind_parent_id = calendar_entries.id
        AND moved_entry.status = 3) AS count_moved,
        (SELECT user_images.image
        FROM calendar_entries AS calendar_entries1
        LEFT JOIN user_images ON user_images.id = calendar_entries1.image
        WHERE calendar_entries1.id = calendar_entries.entry_kind_parent_id
        AND calendar_entries.entry_kind = 4) AS festival_image,
        (SELECT countries1.code
        FROM calendar_entries AS calendar_entries1
        LEFT JOIN locations AS locations1 ON locations1.id = calendar_entries1.location_id
        LEFT JOIN countries AS countries1 ON countries1.id = locations1.country_id
        WHERE calendar_entries1.id = calendar_entries.entry_kind_parent_id
        AND calendar_entries.entry_kind = 4) AS festival_country_code,
        (SELECT count(*)
        FROM calendar_entry_likes
        WHERE calendar_entry_likes.calendar_entry_id = calendar_entries.id) AS count_of_likes,
        (SELECT count(*)
        FROM calendar_entry_likes
        WHERE calendar_entry_likes.calendar_entry_id = calendar_entries.id
        AND calendar_entry_likes.user_id = 10869 ) AS is_like,
        (SELECT group_concat(entry_type_languages.name SEPARATOR ', ')
        FROM calendar_entry_types
        LEFT JOIN entry_type_languages ON entry_type_languages.entry_type_id = calendar_entry_types.entry_type_id
        WHERE (calendar_entry_types.calendar_entry_id = calendar_entries.id
        OR calendar_entry_types.calendar_entry_id = calendar_entries.entry_kind_parent_id)
        AND entry_type_languages.language_id = 1
        GROUP BY calendar_entry_types.calendar_entry_id) AS entry_types_string,
        (SELECT band_addresses.city_name
        FROM band_addresses
        WHERE band_addresses.band_id = calendar_entries.profile_type_id) AS band_city_name,
        (SELECT countries.code
        FROM band_addresses
        LEFT JOIN countries ON countries.id = band_addresses.country_id
        WHERE band_addresses.band_id = calendar_entries.profile_type_id) AS band_country_code,
        (SELECT group_concat(genre_languages.name SEPARATOR ', ')
        FROM calendar_entry_genres
        LEFT JOIN genre_languages ON genre_languages.genre_id = calendar_entry_genres.genre_id
        WHERE calendar_entry_genres.calendar_entry_id = calendar_entries.id
        AND genre_languages.language_id = 1
        GROUP BY calendar_entry_genres.calendar_entry_id) AS genre_string
    FROM `calendar_entries`
    LEFT JOIN `locations` ON `locations`.`id` = `calendar_entries`.`location_id`
    LEFT JOIN `calendar_entries` AS `festival` ON `festival`.`id` = `calendar_entries`.`entry_kind_parent_id`
    LEFT JOIN `locations` AS `festival_location` ON `festival_location`.`id` = ( 
        SELECT calendar_entries1.id
        FROM calendar_entries AS calendar_entries1
        WHERE calendar_entries1.id = calendar_entries.entry_kind_parent_id
        AND calendar_entries.entry_kind = 4
    )
    LEFT JOIN `countries` ON `countries`.`id` = `locations`.`country_id`
    LEFT JOIN `country_languages` ON `country_languages`.`country_id` = `locations`.`country_id`
    LEFT JOIN `states` ON `states`.`id` = `locations`.`state_id`
    LEFT JOIN `bands` ON `bands`.`id` = `calendar_entries`.`profile_type_id`
    AND `bands`.`soft_delete` = 0
    AND `bands`.`status` = 1
    LEFT JOIN `band_images` ON `band_images`.`id` = `calendar_entries`.`image`
    LEFT JOIN `band_privacy_settings` ON `band_privacy_settings`.`band_id` = `calendar_entries`.`profile_type_id`
    WHERE (`bands`.`soft_delete` = 0 OR `bands`.`soft_delete` IS NULL)
    AND (`bands`.`status` = 1 OR `bands`.`status` IS NULL)
    AND ((
            (SELECT count(*)
            FROM band_administrators
            WHERE band_administrators.band_id = bands.id
            AND band_administrators.user_id = 10869 ) > ? OR `band_privacy_settings`.`band_privacy_level` != ?
        ) AND (
            (SELECT user_type
            FROM users
            WHERE users.id = 10869 ) = ? OR `band_privacy_settings`.`band_privacy_level` != ?
        ) AND (
            (SELECT count(*)
            FROM user_bands
            WHERE user_bands.band_id = bands.id
            AND user_bands.status = 1
            AND user_bands.user_id = 10869 ) > ? OR `band_privacy_settings`.`band_privacy_level` != ?
        ) OR `band_privacy_settings`.`band_privacy_level` IS NULL
    ) AND (`festival`.`entry_kind` = ? OR `festival`.`id` IS NULL)
    AND (date(`calendar_entries`.`date_from`) >= ?)
    AND `calendar_entries`.`soft_delete` = ?
    AND (`country_languages`.`language_id` = ? OR `country_languages`.`language_id` IS NULL)
    AND `calendar_entries`.`entry_kind` IN (?, ?)
    AND `calendar_entries`.`event_class` IN (?, ?)
    AND `calendar_entries`.`status` = ?
    GROUP BY `1000, 1000000 field`
    ORDER BY calendar_entries.date_from
    LIMIT 1000, 20

    Сидить замовник такий не тужить тай думку гадає, а що це я буду платити півтори куска спеціалусту за роботу, як є "профі" за 500 доларів, а тут ше і Вася програміст каже, а шо це ти будеш платити таку суму, як я зроблю дешевше за 300! Настрій хороший, угода укладена, робота кипить. Проходить не так багато часу, як замовник каже, а що це сторінка 2хв грузиться, а Вася каже, то нічо ми то все оптимізуємо! Проходить ще декілька місяців, а результату нема. Тоді замовнек розуміє, що Вася не профі і шукає профі в Х рази дорощого! Не так багато часу пройшло, як профі за 1000 каже, я в два рази оптимізував, більше не реально! Замовник пригнічений, шукає профі, і в один прекрасний момент знаходиться дійсно профі, який по домовленості в зарплату X оптимізував, повикидував, переписав, дописав і в результаті 50 мс при 500000 записів. Підсумок: 300 + 1000 + X != 300 :)

    silverreve23, 12 Августа 2020

    Комментарии (10)
  2. SQL / Говнокод #26417

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    SET @from = 11836;
    SET @to = 11840;
    
    INSERT INTO `sprinter_catalog_tree` (
        `sprinter_catalog_tree`.`level`,
        `sprinter_catalog_tree`.`position`,
        `sprinter_catalog_tree`.`parent_id`,
        `sprinter_catalog_tree`.`catalog`,
        `sprinter_catalog_tree`.`info_id`,
    --  ...
    )
    SELECT
        `sprinter_catalog_tree`.`level`,
        `sprinter_catalog_tree`.`position`,
        @to,
        `sprinter_catalog_tree`.`catalog`,
        `sprinter_catalog_tree`.`info_id`,
        `sprinter_catalog_tree`.`format_id`,
        `sprinter_catalog_tree`.`linked_id`,
        `sprinter_catalog_tree`.`linked_shablon_id`,
    --  ...
    FROM `sprinter_catalog_tree` where parent_id = @from order by id;
    
    INSERT INTO `sprinter_catalog_tree` (
        `sprinter_catalog_tree`.`level`,
        `sprinter_catalog_tree`.`position`,
        `sprinter_catalog_tree`.`parent_id`,
        `sprinter_catalog_tree`.`catalog`,
        `sprinter_catalog_tree`.`info_id`,
    --  ...
    )
    SELECT
        a.`level`,
        a.`position`,
        (SELECT id from sprinter_catalog_tree as b where b.parent_id = @to and b.name like (SELECT name from sprinter_catalog_tree where id = a.parent_id)),
        a.`catalog`,
        a.`info_id`,
        a.`format_id`,
        a.`linked_id`,
        a.`linked_shablon_id`,
    --  ...
    FROM `sprinter_catalog_tree` as a where parent_id in (SELECT id FROM sprinter_catalog_tree where parent_id = @from) order by id;
    
    INSERT INTO `sprinter_catalog_info`
    (`name`,
    `eng_name`,
    `text`,
    --  ..
    )
    SELECT
        `sprinter_catalog_info`.`name`,
        `sprinter_catalog_info`.`eng_name`,
        `sprinter_catalog_info`.`text`,
        `sprinter_catalog_info`.`short_text`,
    --  ...
    FROM `sprinter_catalog_info` where id in (SELECT info_id from `sprinter_catalog_tree` where parent_id = @from or parent_id in (SELECT id from `sprinter_catalog_tree` where parent_id = @from));
    
    CREATE temporary table if not exists ids
    SELECT id from `sprinter_catalog_tree` where parent_id = @to or parent_id in (SELECT id from `sprinter_catalog_tree` where parent_id = @to);
    
    UPDATE sprinter_catalog_tree as a SET info_id = (SELECT id from sprinter_catalog_info as b where a.name like b.name order by id desc limit 1) where a.id in (SELECT id from ids);
    
    DROP table ids;

    Высрал вот такое говно в качестве write-n-throw скрипта.

    Дано: есть элементы дерева, хранящиеся в таблице sprinter_catalog_tree, связаны друг с другом через parent_id. Каждому из них соответствует указанный в info_id элемент таблицы sprinter_catalog_info.

    Задача: скопировать все вложенные в раздел @from каталога элементы и его подразделы (вложенность не более 1 уровня) в раздел @to, так чтобы у них были новые id, и также скопировать соответствующие им sprinter_catalog_info. Скопированные sprinter_catalog_tree должны указывать на корректные sprinter_catalog_info, id которых заранее не известны.

    Вот такое говно получилось, расскажите как надо было?

    vistefan, 05 Февраля 2020

    Комментарии (194)
  3. SQL / Говнокод #26334

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    SELECT SQL_NO_CACHE IT.itemCode,
                                    SUBSTRING_INDEX(GROUP_CONCAT(IT.ean),",", 1) as ean,
                                    SUBSTRING_INDEX(GROUP_CONCAT(IT.artCode),",", 1) as artCode,
                                    SUBSTRING_INDEX(GROUP_CONCAT(IT.description),",", 1) as description,
                                    SUBSTRING_INDEX(GROUP_CONCAT(IT.siteName),",", 1) as siteName,
                                    SUBSTRING_INDEX(GROUP_CONCAT(IT.familyName),",", 1) as familyName,
                                    SUBSTRING_INDEX(GROUP_CONCAT(IT.familyCode),",", 1) as familyCode,
                                    SUBSTRING_INDEX(GROUP_CONCAT(IT.SSfamilyCode),",", 1) as SSfamilyCode,
                                    SUBSTRING_INDEX(GROUP_CONCAT(IT.SSfamilyName),",", 1) as SSfamilyName,
                                    SUBSTRING_INDEX(GROUP_CONCAT(IT.sectorCode),",", 1) as sectorCode,
                                    SUBSTRING_INDEX(GROUP_CONCAT(IT.sectorName),",", 1) as sectorName,
                                    SUBSTRING_INDEX(GROUP_CONCAT(IT.radiusCode),",", 1) as radiusCode,
                                    SUBSTRING_INDEX(GROUP_CONCAT(IT.radiusName),",", 1) as radiusName,
                                    FI.areaName,
                                    FI.areaCode,
                                    ROUND(SUM(IT.price), 2) as price,
                                    SUM(IT.quantity) as theoreticalQty,
                                    IFNULL(FI.countedQty, 0) as countedQty,
                                    ROUND(SUM(IT.quantity) * ROUND(SUM(IT.price), 2), 2) as theoreticalQtyValue,
                                    ROUND(IFNULL(FI.countedQty, 0) * ROUND(SUM(IT.price), 2), 2) as countedQtyValue,
                                    IFNULL(FI.countedQty, 0) - SUM(IT.quantity) as stockGapQty,
                                    ROUND((IFNULL(FI.countedQty, 0) - SUM(IT.quantity)) * SUM(IT.price), 2) as stockGapValue,
                                    DATE_FORMAT(ST.lastImport, "%d-%m-%Y %H:%i:%s") as dateOfLastImport,
                                    DATE_FORMAT(NOW(), "%d-%m-%Y %H:%i:%s") as dateOfExport
                            FROM Items as IT
                            LEFT JOIN ( 
                                SELECT MIN(FI.ean) as ean, SUM(FI.quantity) as countedQty, IT.itemCode,
                                GROUP_CONCAT(DISTINCT AR.name) as areaName,
                                GROUP_CONCAT(DISTINCT AR.code) as areaCode
                                FROM FoundItems as FI
                                INNER JOIN Items as IT ON FI.ean = IT.ean
                                LEFT JOIN Areas as AR ON AR.id = FI.areaId
                                
                                INNER JOIN Stores as ST ON ST.id = IT.storeId
                                
                                WHERE eventId = ${args.eventId}
                                AND IT.storeId = (SELECT storeId FROM Events WHERE id = ${args.eventId})
                                GROUP BY IT.itemCode
                            ) as FI ON IT.itemCode = FI.itemCode
                            INNER JOIN Stores as ST ON ST.id = IT.storeId
                            WHERE IT.storeId = (SELECT storeId FROM Events WHERE id = ${args.eventId})
                            GROUP BY IT.itemCode, FI.areaName, FI.areaCode, FI.countedQty, dateOfLastImport

    а що ця конструкція вміє, а ета що, про та ще й ето потрібно впихнути сюди

    silverreve23, 09 Января 2020

    Комментарии (6)
  4. SQL / Говнокод #25628

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    SELECT country, QTY, MIN(launched)
    FROM (SELECT country, launched, COUNT(name) QTY
     FROM Classes c JOIN 
     Ships s ON c.class = s.class
     GROUP BY country,launched
     HAVING COUNT(name) = (SELECT MAX(qty) 
     FROM (SELECT country,launched,COUNT(name) qty
     FROM Classes c1 JOIN 
     Ships s1 ON c1.class = s1.class
     WHERE country = c.country
     GROUP BY country,launched
     )e
     )
     )T
    GROUP BY t.qty, t.country;

    -2 sql ex rating
    я так понял необходимо ещё как-то добавить проверку на NULL?

    5252mmr, 20 Мая 2019

    Комментарии (6)
  5. SQL / Говнокод #25627

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    select Outcomes.ship,count(*)
    from Outcomes
    where outcomes.result!='sunk'
    group by ship
    HAVING count(*)>=2
    union
    select ships.name,count(*)
    from ships
    where ships.launched!=NULL
    group by name
    HAVING count(*)>=2

    -9 sql ex...

    неверно на тренировочной

    коммент :
    парсить имя на '%NAME%' ? или как ?

    5252mmr, 20 Мая 2019

    Комментарии (1)
  6. SQL / Говнокод #25626

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    select distinct pass_in_trip.trip_no, (pass_in_trip.trip_no/5+pass_in_trip.trip_no/25+pass_in_trip.trip_no/125+pass_in_trip.trip_no/625+pass_in_trip.trip_no/3125+pass_in_trip.trip_no/15625+pass_in_trip.trip_no/78125)
    from pass_in_trip
    union
    select distinct trip.trip_no, (trip.trip_no/5+trip.trip_no/25+trip.trip_no/125+trip.trip_no/625+trip.trip_no/3125+trip.trip_no/15625+trip.trip_no/78125)
    from Trip

    -12 рейтинговое sql ex... на тренировочной что-то не так...
    наверно что-то связанное с NULL ?
    добавить проверку ?

    5252mmr, 20 Мая 2019

    Комментарии (0)
  7. SQL / Говнокод #25625

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    select model,(CAST(substring(model, 1, 1) as int)+CAST(substring(model, 2, 1) as int)+CAST(substring(model, 3, 1) as int)+CAST(substring(model, 4, 1) as int)) as qty
    from product
    where  
    (substring(model, 1, 1) like '0' or substring(model, 1, 1) like '1' or substring(model, 1, 1) like '2' 
    or substring(model, 1, 1) like '3' or substring(model, 1, 1) like '4' or substring(model, 1, 1) like '5' 
    or substring(model, 1, 1) like '6' or substring(model, 1, 1) like '7' or substring(model, 1, 1) like '8' 
    or substring(model, 1, 1) like '9')
     AND (substring(model, 2, 1) like '0' or substring(model, 2, 1) like '1' or substring(model, 2, 1) like '2' 
     or substring(model, 2, 1) like '3' or substring(model, 2, 1) like '4' or substring(model, 2, 1) like '5' 
     or substring(model, 2, 1) like '6' or substring(model, 2, 1) like '7' or substring(model, 2, 1) like '8' 
     or substring(model, 2, 1) like '9') 
     AND (substring(model, 3, 1) like '0' or substring(model, 3, 1) like '1' or substring(model, 3, 1) like '2' 
    or substring(model, 3, 1) like '3' or substring(model, 3, 1) like '4' or substring(model, 3, 1) like '5'
     or substring(model, 3, 1) like '6' or substring(model, 3, 1) like '7' or substring(model, 3, 1) like '8' 
     or substring(model, 3, 1) like '9') 
     AND (substring(model, 4, 1) like '0' or substring(model, 4, 1) like '1' or substring(model, 4, 1) like '2'
     or substring(model, 4, 1) like '3' or substring(model, 4, 1) like '4' or substring(model, 4, 1) like '5' 
     or substring(model, 4, 1) like '6' or substring(model, 4, 1) like '7' or substring(model, 4, 1) like '8' 
     or substring(model, 4, 1) like '9')
    
    
    
    select model, 
    ASCII(SUBSTRING(model,1,1))+ 
    ASCII(SUBSTRING(model,2,1))+ 
    ASCII(SUBSTRING(model,3,1))+ 
    ASCII(SUBSTRING(model,4,1)) 
    -48*4 as qty from Product
    тоже не помогло.

    sql ex рейтинговое 4. показывает что на тренировочной что-то не правильно.
    лайк же вроде проверяет '1' как цифру а не как строку.
    любые хинты как тут

    5252mmr, 20 Мая 2019

    Комментарии (0)
  8. SQL / Говнокод #25624

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    select trip_no,concat(CAST(trip_no &1024 aS bit),CAST(trip_no &512 aS bit),CAST(trip_no &256 aS bit), CAST(trip_no &128 aS bit),CAST(trip_no &64 aS bit),
    cast(trip_no & 32 as bit)
     , cast(trip_no & 16 as bit) 
        , cast(trip_no & 8 as bit) 
        , cast(trip_no & 4 as bit) 
        , cast(trip_no & 2 as bit)  
        ,cast(trip_no & 1 as bit)) as trip_no_bit
    from pass_in_trip
    where trip_no<=2048 and trip_no>=1024
    
    union 
    
    select trip_no,concat(CAST(trip_no &1024 aS bit),CAST(trip_no &512 aS bit),CAST(trip_no &256 aS bit), CAST(trip_no &128 aS bit),CAST(trip_no &64 aS bit),
    cast(trip_no & 32 as bit)
     , cast(trip_no & 16 as bit) 
        , cast(trip_no & 8 as bit)
        , cast(trip_no & 4 as bit) 
        , cast(trip_no & 2 as bit)  
        ,cast(trip_no & 1 as bit))  as trip_no_bit
    from trip
    where trip_no<=2048 and trip_no>=1024
    union

    В конкате первый аргумент concat(CAST(trip_no &1024 aS bit),
    &1024 имеется ввиду

    а внизу проверка where trip_no<=2048 and trip_no>=1024 - правильно ли ? или where trip_no<=1024and trip_no>=512 должно быть ?

    и есть ли более профитный вариант решения рейтинговой задачи sqlex`a ? а то какой-то говнокод получается около 11к строк а там лимит 8к...
    или лучше заменить на %2 и /2)%2 , /2)/2)%2 ????

    5252mmr, 20 Мая 2019

    Комментарии (0)
  9. SQL / Говнокод #25504

    −101

    1. 1
    SELECT AVG(len) from huis

    bormandyan, 03 Апреля 2019

    Комментарии (21)
  10. SQL / Говнокод #25503

    −102

    1. 1
    SELECT MIN(len) from huis

    bormandyan, 03 Апреля 2019

    Комментарии (5)