Как выбрать записи из этих отношений?

Question

У меня эти четыре таблицы: фиды, feed_entries, entries_categorias и категориальные. Whith этих структур:

CREATE TABLE `categorias` (
    `id` int(11) NOT NULL auto_increment, 
    `nome` varchar(100) collate utf8_unicode_ci NOT NULL, 
    `slug` varchar(100) collate utf8_unicode_ci NOT NULL, 
    `principal` int(1) NOT NULL default '0', 
    `ordem` int(11) NOT NULL default '0', 
    PRIMARY KEY (`id`), 
    UNIQUE KEY `nome` (`nome`) 
) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci; 

CREATE TABLE `entries_categorias` (
    `id` int(11) NOT NULL auto_increment, 
    `entry_id` int(11) NOT NULL, 
    `categoria_id` int(11) NOT NULL, 
    PRIMARY KEY (`id`), 
    KEY `entry_id` (`entry_id`), 
    KEY `categoria_id` (`categoria_id`) 
) ENGINE=InnoDB DEFAULT CHARSET=latin1; 


CREATE TABLE `feeds` (
    `id` int(11) NOT NULL auto_increment, 
    `categoria_id` int(11) NOT NULL, 
    `titulo` varchar(255) collate utf8_unicode_ci NOT NULL, 
    `descricao` text collate utf8_unicode_ci NOT NULL, 
    `link` varchar(255) collate utf8_unicode_ci NOT NULL, 
    `link_comentarios` varchar(255) collate utf8_unicode_ci NOT NULL, 
    `url` varchar(255) collate utf8_unicode_ci NOT NULL, 
    PRIMARY KEY (`id`), 
    KEY `categoria_id` (`categoria_id`) 
) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci; 

CREATE TABLE `feed_entries` (
    `id` int(11) NOT NULL auto_increment, 
    `feed_id` int(11) NOT NULL, 
    `colunista_id` int(11) NOT NULL, 
    `titulo` varchar(255) collate utf8_unicode_ci NOT NULL, 
    `descricao` text collate utf8_unicode_ci NOT NULL, 
    `slug` varchar(255) collate utf8_unicode_ci NOT NULL, 
    `link` varchar(255) collate utf8_unicode_ci NOT NULL, 
    `permaLink` varchar(255) collate utf8_unicode_ci NOT NULL, 
    `html` text collate utf8_unicode_ci NOT NULL, 
    `tags` varchar(255) collate utf8_unicode_ci NOT NULL, 
    `imagemChamada` int(1) NOT NULL, 
    `imagem332x332` int(1) NOT NULL, 
    `imagem201x144` int(1) NOT NULL, 
    `imagem145x145` int(1) NOT NULL, 
    `imagem101x76` int(1) NOT NULL, 
    `date` datetime NOT NULL, 
    `created_at` datetime NOT NULL, 
    `comments` int(11) NOT NULL, 
    `comments_date` datetime NOT NULL, 
    `views` int(11) NOT NULL, 
    `deleted` int(1) NOT NULL, 
    PRIMARY KEY (`id`), 
    UNIQUE KEY `permaLink` (`permaLink`), 
    KEY `feed_id` (`feed_id`), 
    KEY `colunista_id` (`colunista_id`) 
) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci; 

ALTER TABLE `entries_categorias` 
    ADD CONSTRAINT `entries_categorias_ibfk_5` FOREIGN KEY (`entry_id`) REFERENCES `feed_entries` (`id`) ON DELETE CASCADE, 
    ADD CONSTRAINT `entries_categorias_ibfk_6` FOREIGN KEY (`categoria_id`) REFERENCES `categorias` (`id`) ON DELETE CASCADE; 

ALTER TABLE `feeds` 
    ADD CONSTRAINT `feeds_ibfk_2` FOREIGN KEY (`categoria_id`) REFERENCES `categorias` (`id`) ON DELETE CASCADE; 

ALTER TABLE `feed_entries` 
    ADD CONSTRAINT `feed_entries_ibfk_1` FOREIGN KEY (`feed_id`) REFERENCES `feeds` (`id`) ON DELETE CASCADE, 
    ADD CONSTRAINT `feed_entries_ibfk_2` FOREIGN KEY (`colunista_id`) REFERENCES `colunistas` (`id`) ON DELETE CASCADE; 

мне нужно сделать выбор, как это:

SELECT e.* 
FROM feed_entries AS e 
    INNER JOIN feeds AS f 
     ON e.feed_id =f.id INNER 
    JOIN entries_categorias AS ec 
     ON ec.entry_id =e.id 
    INNER JOIN categorias AS c 
     ON ec.categoria_id =c.id 
WHERE (c.nome ='Manchete') 
    AND (e.deleted =0) 
ORDER BY e.date DESC 

Но вместо especify только один cateogory от отношений, я хочу, чтобы указать два или более ...

Другими словами, как я могу выбрать все записи фида, которые были включены именно в эти две категории, например: Google и Apple. Эти записи могут иметь другие категории, но должны иметь эти два.

Answer 1

Попробуйте что-то вроде этого:

SELECT * 
FROM `feed_entries` 
WHERE id IN (
    SELECT e.id 
    FROM `feed_entries` AS `e` 
    INNER JOIN `feeds` AS `f` ON e.feed_id =f.id 
    INNER JOIN `entries_categorias` AS `ec` 
    ON ec.entry_id =e.id INNER JOIN `categorias` AS `c` 
    ON ec.categoria_id =c.id 
    WHERE c.nome IN ('Google','Apple') 
    AND (e.deleted =0) 
    GROUP BY e.id 
    HAVING COUNT(DISTINCT ec.id) = 2 
) 

Answer 2

Измени ИНЕКЕ использовать оператор IN:

SELECT `e`.* 
FROM `feed_entries` AS `e` 
INNER JOIN `feeds` AS `f` ON e.feed_id =f.id 
INNER JOIN `entries_categorias` AS `ec` ON ec.entry_id =e.id 
INNER JOIN `categorias` AS `c` ON ec.categoria_id =c.id 
WHERE (c.nome IN (<List of categories separated by commas>) AND (e.deleted =0) 
ORDER BY `e`.`date` DESC