[SQL][MySQL]Problem z zapytaniem sql Opcje

[eminiasty]

[SQL] pobierz, plaintext 
SELECT mgr_recipes.*,mgr_recipescategories.*,mgr_image.image_codename, GROUP_CONCAT(DISTINCT mgr_recipescategories.categories_id,';' , mgr_categories.categories_name, ';' , mgr_categories.categories_colorText,';',mgr_categories.categories_colorBg) AS kat,mgr_users.users_login FROM mgr_recipes JOIN mgr_recipescategories ON mgr_recipes.recipes_id = mgr_recipescategories.recipes_id JOIN mgr_categories ON mgr_categories.categories_id = mgr_recipescategories.categories_id JOIN mgr_recipesimage ON mgr_recipes.recipes_id = mgr_recipesimage.recipes_id JOIN mgr_image ON mgr_image.image_id = mgr_recipesimage.image_id JOIN mgr_users ON mgr_users.users_id = mgr_recipes.users_id JOIN mgr_recipesingredients ON mgr_recipes.recipes_id = mgr_recipesingredients.recipes_id JOIN mgr_recipesmethod ON mgr_recipes.recipes_id = mgr_recipesmethod.recipes_id JOIN mgr_recipesrecommended ON mgr_recipes.recipes_id = mgr_recipesrecommended.recipes_id JOIN mgr_recipesseason ON mgr_recipes.recipes_id = mgr_recipesseason.recipes_id JOIN mgr_recipestimeofday ON mgr_recipes.recipes_id = mgr_recipestimeofday.recipes_id WHERE mgr_recipes.recipes_active = 1 AND mgr_recipes.recipes_id NOT IN (
 
    SELECT mgr_recipes.recipes_id FROM mgr_recipes JOIN mgr_recipescategories ON mgr_recipes.recipes_id = mgr_recipescategories.recipes_id JOIN mgr_categories ON mgr_categories.categories_id = mgr_recipescategories.categories_id JOIN mgr_recipesimage ON mgr_recipes.recipes_id = mgr_recipesimage.recipes_id JOIN mgr_image ON mgr_image.image_id = mgr_recipesimage.image_id JOIN mgr_users ON mgr_users.users_id = mgr_recipes.users_id JOIN mgr_recipesingredients ON mgr_recipes.recipes_id = mgr_recipesingredients.recipes_id JOIN mgr_recipesmethod ON mgr_recipes.recipes_id = mgr_recipesmethod.recipes_id JOIN mgr_recipesrecommended ON mgr_recipes.recipes_id = mgr_recipesrecommended.recipes_id JOIN mgr_recipesseason ON mgr_recipes.recipes_id = mgr_recipesseason.recipes_id JOIN mgr_recipestimeofday ON mgr_recipes.recipes_id = mgr_recipestimeofday.recipes_id WHERE mgr_recipes.recipes_active = 1 GROUP BY mgr_recipes.recipes_id LIMIT 3
)
GROUP BY mgr_recipes.recipes_id
[SQL] pobierz, plaintext 

Mam taki blad:
#1235 - This version of MariaDB doesn't yet support 'LIMIT & IN/ALL/ANY/SOME subquery'

Jak poprawic to zapytanie?

#Wersja klienta bazy danych: libmysql - mysqlnd 5.0.11-dev

Ten post edytował eminiasty 24.07.2016, 17:06:57

[eminiasty]

Zrobilem cos takiego ale nie do konca dziala, bez petli owszem jednak chcialbym zastosowac takie rozwiazanie:

pierwszy arg- limit
drugi arg - offset

[PHP] pobierz, plaintext 
for($i=3;$i<100;$i=i+3){	
echo '<div class="item">';
funkcja(3,$i,$prefix,$link);
echo '</div>'; 
}
[PHP] pobierz, plaintext 

Blad:

[PHP] pobierz, plaintext 
Notice: Use of undefined constant i - assumed 'i' in
 
[PHP] pobierz, plaintext 

i nie przeskakuje do kolejnego elementu karuzeli

Ten post edytował eminiasty 31.07.2016, 10:13:12