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我有这四个表:feeds、feed_entries、entries_categorias 和 categorias。这些结构:
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;
我需要进行这样的选择:
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
但是我不想只指定关系中的一个类别,而是想指定两个或多个...
换句话说,我如何选择恰好包含在这两个类别中的所有提要条目,例如:Google 和 Apple。这些条目可以有其他类别,但需要有这两个。
最佳答案
尝试这样的事情:
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
)
关于sql - 如何从这种关系中选择条目?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/3318726/
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