Код (Text): select mixbrands.* from mixbrands where id in (select idmix from mixbrandsrels where idmy in ( select code_brand from model where code_type in (select idmy from mixtypesrels where idmix=138608) group by code_brand)) Как этот запрос сделать без in-ов ?
Код (Text): SELECT mb.* FROM MIXBRANDS MB INNER JOIN MIXBRANDSRELS MBR ON MBR.IDMIX=MB.ID INNER JOIN MODEL M ON M.CODE_BRAND=MBR.IDMY INNER JOIN MIXTYPESRELS MTR ON MTR.IDMY=M.CODE_TYPE WHERE mtr.idmix=138608 попробуйте
а как использовать JOIN в таком запросе Код (Text): select * from ".$site."_posts where ID in (select object_id from ".$site."_term_relationships where term_taxonomy_id = ( select term_id from ".$site."_terms where name='"._iconv("НОВОСТИ")."')) and post_status='publish' заменить тот же in нужно
попробуйте Код (PHP): select p.* from ".$site."_posts p INNER join ".$site."_term_relationships rs ON rs.object_id=p.ID INNER JOIN ".$site."_terms t ON (t.select term_id=rs.term_taxonomy_id AND t.name='"._iconv("НОВОСТИ")."') Where p.post_status='publish'